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1, , LECTURE NOTES, Course No: STCA-101, STATISTICS, , TIRUPATI, , Karl Pearson, , R. A. Fisher, , Sir Ronald Aylmer Fisher (1890-1962), Karl Pearson (né Carl Pearson), Born: 27 March 1857, Islington, London, England, , Born: 17 February 1890, East Finchley, London , England, , Prepared By, Dr. G. MOHAN NAIDU M.Sc., Ph.D.,, Assistant Professor & Head, Dept. of Statistics & Mathematics, S.V. Agricultural College, TIRUPATI, ACHARYA N.G. RANGA AGRICULTURAL UNIVERSITY, , BS Publication

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2, LECTURE OUTLINE, Course No. STCA-101, Course Title: STATISTICS, , Credits: 2 (1+1), THEORY, , S. No., , Topic/Lesson, , 1, , Introduction to Statistics, Definition, Advantages and Limitations., , 2, , Frequency distribution: Construction of Frequency Distribution table., , 3, , Measures of Central Tendency: Definition, Characteristics of Satisfactory, average., , 4, , Arithmetic Mean, Median, Mode for grouped and ungrouped data – Merits and, Demerits of Arithmetic Mean., , 5, , Measures of Dispersion: Definition, standard deviation, variance and, coefficient of variation., , 6, , Normal Distribution and its properties. Introduction to Sampling: Random, sampling, concept of standard error of Mean., , 7, , Tests of Significance: Introduction, Types of errors, Null hypothesis, level of, significance and degrees of freedom, steps in testing of hypothesis., , 8, , Large sample tests: Test for Means – Z-test, One sample and Two samples, with population S.D. known and Unknown., , 9, , Small sample tests: Test for Means – One sample t – test, Two samples t-test, and Paired t-test., , 10, , Chi-Square test in 2x2 contingency table with Yate‟s correction, F-test., , 11, , Correlation: Definition, types, properties, Scatter diagram, calculation, , and, , testing., 12, , Regression: Definition, Fitting of two lines Y on X and X on Y, Properties,, inter relation between correlation and regression., , 13, , Introduction to Experimental Designs, Basic Principles, ANOVA its, assumptions., , 14, , Completely Randomized Design: Layout, Analysis with equal and unequal, replications., , 15, , Randomized Block Design: Layout and Analysis., , 16, , Latin Square Design: Layout and Analysis., , BS Publication

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3, PRACTICALS, S .No., , Topic, , 1, , Construction of Frequency Distribution tables, , 2, , Computation of Arithmetic Mean for Grouped and Un-grouped data, , 3, , Computation of Median for Grouped and Un-grouped data, , 4, , Computation of Mode for Grouped and Un-grouped data, , 5, , Computation of Standard Deviation and variance for grouped and, ungrouped data, , 6, , Computation of coefficient of variation for grouped and ungrouped data, , 7, , SND (Z) test for single sample, Population SD known and Unknown, , 8, , SND (Z) test for two samples, Population SD known and Unknown, , 9, , Student‟s t-test for single and two samples, , 10, , Paired t-test and F-test, , 11, , Chi-square test – 2x2 contingency table with Yate‟s correction, , 12, , Computation of correlation coefficient and its testing, , 13, , Fitting of simple regression equations Y on X and X on Y, , 14, , Completely Randomized Design: Analysis with equal and unequal, replications, , 15, , Randomized Block Design: Analysis, , 16, , Latin Square Design: Analysis, , BS Publication

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1, STATISTICS, Statistics has been defined differently by different authors from time to time. One, can find more than hundred definitions in the literature of statistics., Statistics can be used either as plural or singular. When it is used as plural, it is a, systematic presentation of facts and figures. It is in this context that majority of people, use the word statistics. They only meant mere facts and figures. These figures may be, with regard to production of food grains in different years, area under cereal crops in, different years, per capita income in a particular state at different times etc., and these are, generally published in trade journals, economics and statistics bulletins, news papers, etc.,, When statistics is used as singular, it is a science which deals with collection,, classification, tabulation, analysis and interpretation of data., The following are some important definition of statistics., Statistics is the branch of science which deals with the collection, classification and, tabulation of numerical facts as the basis for explanations, description and comparison of, phenomenon - Lovitt, The science which deals with the collection, analysis and interpretation of numerical data, - Corxton & Cowden, The science of statistics is the method of judging collective, natural or social phenomenon, from the results obtained from the analysis or enumeration or collection of estimates, , -, , King, Statistics may be called the science of counting or science of averages or statistics is the, science of the measurement of social organism, regarded as whole in all its manifestations, - Bowley, Statistics is a science of estimates and probabilities -Boddington, Statistics is a branch of science, which provides tools (techniques) for decision making in, the face of uncertainty (probability) - Wallis and Roberts, This is the modern definition of statistics which covers the entire body of statistics, All definitions clearly point out the four aspects of statistics collection of data,, analysis of data, presentation of data and interpretation of data., Importance: Statistics plays an important role in our daily life, it is useful in almost all, sciences – social as well as physical – such as biology, psychology, education, economics,, business management, agricultural sciences etc., . The statistical methods can be and are, , BS Publication

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2, being followed by both educated and uneducated people. In many instances we use sample, data to make inferences about the entire population., 1. Planning is indispensable for better use of nation‟s resources. Statistics are, indispensable in planning and in taking decisions regarding export, import, and, production etc., Statistics serves as foundation of the super structure of, planning., 2. Statistics helps the business man in the formulation of polices with regard to, business. Statistical methods are applied in market and production research,, quality control of manufactured products, 3. Statistics is indispensable in economics. Any branch of economics that require, comparison, correlation requires statistical data for salvation of problems, 4. State. Statistics is helpful in administration in fact statistics are regarded as, eyes of administration. In collecting the information about population, military, strength etc., Administration is largely depends on facts and figures thud it, needs statistics, 5. Bankers, stock exchange brokers, insurance companies all make extensive use, of statistical data. Insurance companies make use of statistics of mortality and, life premium rates etc., for bankers, statistics help in deciding the amount, required to meet day to day demands., 6. Problems relating to poverty, unemployment, food storage, deaths due to, diseases, due to shortage of food etc., cannot be fully weighted without the, statistical balance. Thus statistics is helpful in promoting human welfare, 7. Statistics are a very important part of political campaigns as they lead up to, elections. Every time a scientific poll is taken, statistics are used to calculate, and illustrate the results in percentages and to calculate the margin for error., In agricultural research, Statistical tools have played a significant role in the, analysis and interpretation of data., 1. In making data about dry and wet lands, lands under tanks, lands under, irrigation projects, rainfed areas etc.,, 2. In determining and estimating the irrigation required by a crop per day, per, base period., 3. In determining the required doses of fertilizer for a particular crop and crop, land., , BS Publication

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3, 4. In soil chemistry also statistics helps classifying the soils basing on their, analysis results, which are analyzed with statistical methods., 5. In estimating the losses incurred by particular pest and the yield losses due to, insect, bird, or rodent pests statistics is used in entomology., 6. Agricultural economists use forecasting procedures to determine the future, demand and supply of food and also use regression analysis in the empirical, estimation of function relationship between quantitative variables., 7. Animal scientists use statistical procedures to aid in analyzing data for decision, purposes., 8., , Agricultural engineers use statistical procedures in several areas, such as for, irrigation research, modes of cultivation and design of harvesting and, cultivating machinery and equipment., , Limitations of Statistics:, 1. Statistics does not study qualitative phenomenon, 2. Statistics does not study individuals, 3. Statistics laws are not exact laws, 4. Statistics does not reveal the entire information, 5. Statistics is liable to be misused, 6. Statistical conclusions are valid only on average base, Types of data: The data are of two types i) Primary Data and ii) Secondary Data, i), , Primary Data: It is the data collected by the primary source of information i.e, by the investigator himself., , ii), , Secondary Data: It is the data collected from secondary sources of information,, like news papers, trade journals and statistical bulletins, etc.,, , Variables and Attributes:, Variability is a common characteristic in biological sciences. A quantitative or qualitative, characteristic that varies from observation to observation in the same group is called a, variable. In case of quantitative variables, observations are made using interval scales, whereas in case of qualitative variables nominal scales are used. Conventionally, the, quantitative variables are termed as variables and qualitative variables are termed as, attributes. Thus, yield of a crop, available nitrogen in soil, daily temperature, number of, leaves per plant and number of eggs laid by insects are all variables. The crop varieties,, soil types, shape of seeds, seasons and sex of insects are attributes., , BS Publication

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4, The variable itself can be classified as continuous variable and discrete variable., The variables for which fractional measurements are possible, at least conceptually, are, called continuous variables. For example, in the range of 7 kg to 10 yield of a crop, yield, might be 7.15 or 7.024kg. Hence, yield is a continuous variable. The variables for which, such factional measurements are not possible are called discrete or discontinuous, variables. For example, the number of grains per panicle of paddy can be counted in full, numbers like 79, 80, 81 etc. Thus, number of grains per panicle is a discrete variable. The, variables, discrete or continuous are dented by capital letters like X and Y., Construction of Frequency Distribution Table:, In statistics, a frequency distribution is a tabulation of the values that one or more, variables take in a sample. Each entry in the table contains the frequency or count of the, occurrences of values within a particular group or interval, and in this way the table, summarizes the distribution of values in the sample., The following steps are used for construction of frequency table, Step-1: The number of classes are to be decided, The appropriate number of classes may be decided by Yule‟s formula, which is as follows:, Number of classes = 2.5 x n1/4. where „n‟ is the total number of observations, Step-2: The class interval is to be determined. It is obtain by using the relationship, Maximum value in the given data – Minimum value in the given data, C.I = ------------------------------------------------------------------------------------Number of classes, Step-3: The frequencies are counted by using Tally marks, Step-4: The frequency table can be made by two methods, a) Exclusive method, b) Inclusive method, a) Exclusive method: In this method, the upper limit of any class interval is kept the, same as the lower limit of the just higher class or there is no gap between upper limit of, one class and lower limit of another class. It is continuous distribution, , BS Publication

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5, , Ex:, C.I., , Tally marks, , Frequency (f), , 0-10, 10-20, 20-30, b) Inclusive method: There will be a gap between the upper limit of any class and the, lower limit of the just higher class. It is discontinuous distribution, Ex:, C.I., , Tally marks, , Frequency (f), , 0-9, 10-19, 20-29, To convert discontinuous distribution to continuous distribution by subtracting 0.5 from, lower limit and by adding 0.5 to upper limit, Note: The arrangement of data into groups such that each group will have some numbers., These groups are called class and number of observations against these groups are called, frequencies., Each class interval has two limits 1. Lower limit and 2. Upper limit, The difference between upper limit and lower limit is called length of class interval., Length of class interval should be same for all the classes. The average of these two limits, is called mid value of the class., Example: Construct a frequency distribution table for the following data, 25, 32, 45, 8, 24, 42, 22, 12, 9, 15, 26, 35, 23, 41, 47, 18, 44, 37, 27, 46, 38, 24, 43,, 46, 10, 21, 36, 45, 22, 18., Solution:, , Number of observations (n) = 30, Number of classes, , = 2.5 x n1/4, = 2.5 x 301/4, = 2.5 x 2.3, = 5.8  6.0, , BS Publication

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6, , Class interval =, , Max.value  Min.value, No.of .classes, , =, , 46  8, 6, , =, , 38, = 6.3  6.0, 6, , Inclusive method:, C.I., , Tally marks, , Frequency (f), , 8-14, , ||||, , 4, , 15-21, , ||||, , 4, , 22-28, , 8, , |||| |||, , 29-35, , ||, , 2, , 36-42, , ||||, , 5, , 42-49, , |||| ||, , 7, , Total, , 30, , Exclusive method:, C.I., , Tally marks, , Frequency (f), , 7.5-14.5, , ||||, , 4, , 14.5-21.5, , ||||, , 4, , 21.5-28.5, , 8, , |||| |||, , 28.5-35.5, , ||, , 2, , 35.5-42.5, , ||||, , 5, , 42.5-49.5, , |||| ||, , 7, , Total, , 30, , MEASURES OF CENTRAL TENDENCY, One of the most important aspects of describing a distribution is the central value, around which the observations are distributed., , Any mathematical measure which is, , intended to represent the center or central value of a set of observations is known as, measure of central tendency (or ), The single value, which represents the group of values, is termed as a „measure of central, tendency‟ or a measure of location or an average., , BS Publication

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7, Characteristics of a Satisfactory Average:, 1. It should be rigidly defined, 2. It should be easy to understand and easy to calculate, 3. It should be based on all the observations, 4. It should be least affected by fluctuations in sampling, 5. It should be capable of further algebraic treatment, 6. It should not be affected much by the extreme values, 7. It should be located easily, Types of average:, , 1. Arithmetic Mean, 2. Median, 3. Mode, 4. Geometric Mean, 5. Harmonic Mean, , Arithmetic Mean (A.M): It is defined as the sum of the given observations divided by the, number of observations. A.M. is measured with the same units as that of the observations., Ungrouped data:, Direct Method: Let x1, x2, ………,xn be „n‟ observations then the A.M is computed from, the formula:, n, , x  x 2  ........  x n, A.M. = 1, , n, , where, , n, , x, i 1, , i, , x, i 1, , i, , n, , = sum of the given observations, , n = Number of observations, Linear Transformation Method or Deviation Method: When the variable constitutes large, values of observations, computation of arithmetic mean involves more calculations. To, overcome this difficulty, Linear Transformation Method is used., , The value xi, , is, , transformed to di., n, , and, , A.M. = A+, , d, i 1, , i, , n, , where A = Assumed mean which is generally taken as class mid point of middle class, or the class where frequency is large., , BS Publication

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8, di = xi  A = deviations of the ith value of the variable taken from an assumed, mean and n = number of observations, Grouped Data:, Let f1, f2, ………., fn be „n‟ frequencies corresponding to the mid values of the class, intervals x1, x2, ………xn then, n, , f x  f 2 x 2  .............  f n x n, A.M. = 1 1, , f1  f 2  ..........  f n, , , i 1, n, , f i xi, , f, i 1, , n, , , , fx, i, , i 1, , i, , N, , (direct method), , i, , and, , , A.M. =A+ , , ,  n, , fi di , , , f1 d1  f 2 d 2  .............  f n d n , C (indirect method), C  A   i 1, f1  f 2  ..........  f n,  N , , , , , where di = deviation =, , xi  A, ; f = frequency; C = class interval;, c, , x = mid values of classes., Arithmetic mean, when computed for the data of entire population, is represented by the, symbol „  ‟. Where as when it is computed on the basis of sample data, it is represented, as X , which is the estimate of  ., Properties of A.M.:, i), , The algebraic sum of the deviations taken from arithmetic mean is zero, i.e. (x-A.M.) = 0, , ii), , Let x1 be the mean of n1 observations, x 2 be the mean of the n2, observations ……… x k be the mean of nk observations then the mean x, of n = (n1+n2+…..nk) observations is given by, k, , n x  n2 x 2  ......  nk xk, =, x = 1 1, n1  n2  ....  nk, , n x, i 1, k, , i, , i, , n, i 1, , Merits:, 1. It is well defined formula defined, 2. It is easy to understand and easy to calculate, 3. It is based upon all the observations, , BS Publication, , i

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9, 4. It is amenable to further algebraic treatments, provided the sample is randomly, obtained., 5. Of all averages, arithmetic mean is affected least by fluctuations of sampling, Demerits:, 1. Cannot be determined by inspection nor it can be located graphically, 2. Arithmetic mean cannot be obtained if a single observation is missing or lost, 3. Arithmetic mean is affected very much by extreme values, 4. Arithmetic mean may lead to wrong conclusions if the details of the data from which, it is computed are not given, 5. In extremely asymmetrical (skewed) distribution, usually arithmetic mean is not a, suitable measure of location, Examples:, i) Ungrouped data:, If the weights of 7 ear-heads of sorghum are 89, 94, 102, 107, 108, 115 and 126 g. Find, arithmetic mean by direct and deviation methods, Solution:, n, , i) Direct Method:, , A.M. =, , x, i 1, , i, , n, n, , A.M. = A , , ii) Deviation Method:, , Where n = number of observations;, , d, i 1, , i, , n, , xi = given values;, , A = arbitrary mean (assumed value); d = deviation =xi -A, xi, 89, 94, 102, 107, 108, 115, 126,  x  741, , di = xi –A, 89-102= -13, 94-102= -8, 102-102= 0, 107-102= 5, 108-102= 6, 115-102= 13, 126-102= 24,  d  29, , here A = assumed value = 102, n, , i), , A.M. =, ,  xi, i 1, , n, , n, , (ii) A.M. = A , , BS Publication, , d, i 1, , n, , i

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10, , =, ,  27 , = 102   ,  7 , , 741, 7, , = 105.86 g, , = 105.86 g., , ii) Grouped Data:, The following are the 405 soybean plant heights collected from a particular plot. Find the, arithmetic mean of the plants by direct and indirect method:, Plant height, (Cms), No. of, plants( f i ), Solution:, , 812, 6, , 1317, 17, , 1822, 25, , 2327, 86, , 2832, 125, , 3337, 77, , 3842, 55, , 4347, 9, , 4852, 4, , a) Direct Method:, n, , A.M. = , Where, , fx, , i i, , i 1, , N, , xi = mid values of the corresponding classes, N = Total frequency =, , n, , f, i 1, , i, , fi = frequency, b) Deviation Method:,   fi di, A.M. = A  ,  N, , , , C, , , , Where di = deviation ( i.e. d i , , xi  A, ), C, , Length of class interval (C ) = 5; Assumed value, C.I, 8-12, 13-17, 18-22, 23-27, 28-32, 33-37, 38-42, 43-47, 48-52, 53-57, Total, , fi, 6, 17, 25, 86, 125, 77, 55, 9, 4, 1, N = 405, , xi, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, , fixi, 60, 255, 500, 2150, 3750, 2695, 2200, 405, 200, 55, Σfi xi = 12270, , BS Publication, , di , , xi  A, C, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, , (A) = 30, , fi di, -24, -51, -50, -86, 0, 77, 110, 27, 16, 5, Σfidi = 24, , 5357, 1

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11, a) Direct Method:, A.M. =, , 12270,  30.30 cms., 405, , b) Deviation Method:, ,  24 , A.M. = 30  , 5,  405 , = 30.30 cms., MEDIAN, The median is the middle most item that divides the distribution into two equal, parts when the items are arranged in ascending order., Ungrouped data: If the number of observations is odd then median is the middle value, after the values have been arranged in ascending or descending order of magnitude. In, case of even number of observations, there are two middle terms and median is obtained, by taking the arithmetic mean of the middle terms., In case of discrete frequency distribution median is obtained by considering the, cumulative frequencies. The steps for calculating median are given below:, i) Arrange the data in ascending or descending order of magnitude, ii) Find out cumulative frequencies, iii) Apply formula: Median = Size of, , N 1, , where N=, 2, , f, , iv) Now look at the cumulative frequency column and find, that total which is either equal, to, , N 1, or next higher to that and determine the value of the variable corresponding to it,, 2, , which gives the value of median., Continuous frequency distribution:, If the data are given with class intervals then the following procedure is adopted for the, calculation of median., N 1, , where N = f, 2, , i), , find, , ii), , see the (less than) cumulative frequency just greater than, , iii), , the corresponding value of x is median, , BS Publication, , N 1, 2

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12, In the case of continuous frequency distribution, the class corresponding to the cumulative, frequency just greater than, , N 1, is called the median class and the value of median is, 2, , obtained by the following formula:, ,  N 1, ,  2  m, Median = l + , C, f, , , , , Where l is the lower limit of median class, f is the frequency of the median class, m is the cumulative frequency of the class preceding the median class, C is the class length of the median class, N = total frequency, Examples:, Case-i) when the number of observations (n) is odd:, The number of runs scored by 11 players of a cricket team of a school are, 5, 19, 42, 11, 50, 30, 21, 0, 52, 36, 27, To compute the median for the given data, we proceed as follows:, In case of ungrouped data, if the number of observations is odd then median is the middle, value after the values have been arranged in ascending or descending order of magnitude., Let us arrange the values in ascending order:, 0, 5, 11, 19, 21, 27, 30, 36, 42, 50, 52, th, , th, ,  n 1,  11  1 ,  Median = ,  value = ,  value,  2 ,  2 , = 6th value, Now the 6th value in the data is 27., , , Median = 27 runs, , Case-ii) when the number of observations (n) is even:, Find the median of the following heights of plants in Cms:, 6, 10, 4, 3, 9, 11, 22, 18, , BS Publication

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13, In case of even number of observations, there are two middle terms and median is, obtained by taking the arithmetic mean of the middle terms., Let us arrange the given items in ascending order, 3, 4, 6, 9, 10, 11, 18, 22, In this data the number of items n = 8, which is even., n, Median = average of   th and, 2, , n ,   1 th terms., 2 , , = average of 9 and 10, ,  Median = 9.5 Cms., Grouped Data:, Find out the median for the following frequency distribution of 180 sorghum ear-heads., Weight of ear-heads (in g), , 4060, , 6080, , 80100, , 100120, , 120140, , 140160, , 160180, , 180200, , No.of ear-heads, , 6, , 28, , 35, , 45, , 30, , 15, , 12, , 9, ,  N 1, , m, , C ------------------ (1), Solution: Median = l   2, f, , , , , , , Where l is the lower limit of the median class, f is the frequency of the median class, m is the cumulative frequency of the class preceding the median class, C is the class interval of the median class, and N =, ,  f = Total number of observations, , BS Publication

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14, Weight of earheads (in g), , No. of ear-heads, , Cumulative, Frequency (CF), , 40-60, , 6, , 6, , 60-80, , 28, , 34, , 80-100, , 35, , 69 - m, , 100-120, , 45 – f, , 114, (Median class), , 120-140, , 30, , 144, , 140-160, , 15, , 159, , 160-180, , 12, , 171, , 180-200, , 9, , 180, , N=, Here, , f, , =180, , N  1 181, =,  90.5 . Cumulative frequency just greater than 90.5 is 69 and the, 2, 2, , corresponding class is 100-120. The median class is 100-120., N = 121; L =40; f = 24; m = 39 and C = 10, Substituting the above values in equation (1), we get, ,  90.5  69 , Median = 100  , 20, 45 , , = 109.56 g, Merits and Demerits of Median:, Merits:, 1. It is rigidly defined., 2. It is easily understood and is easy to calculate. In some cases it can be located, merely by inspection., 3. It is not at all affected by extreme values., 4. It can be calculated for distributions with open-end classes, Demerits:, 1. In case of even number of observations median cannot be determined exactly. We, merely estimate it by taking the mean of two middle terms., , BS Publication

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15, 2. It is not amenable to algebraic treatment, 3. As compared with mean, it is affected much by fluctuations of sampling., MODE, Mode is the value which occurs most frequently in a set of observations or mode is the, value of the variable which is predominant in the series., In case of discrete frequency distribution mode is the value of x corresponding to, maximum frequency, In case of continuous frequency distribution, mode is obtained from the formula:, ,  ( f  f1 ) , Mode = l + , C, 2, f, , f, , f, 1, 2 , , Where l is the lower limit of modal class, C is class interval of the modal class, f the frequency of the modal class, f1 and f2 are the frequencies of the classes preceding and succeeding, the modal class respectively, If the distribution is moderately asymmetrical, the mean, median and mode obey the, empirical relationship:, Mode = 3 Median – 2 Mean, Example: Find the mode value for the following data:, 27, 28, 30, 33, 31, 35, 34, 33, 40, 41, 55, 46, 31, 33, 36, 33, 41, 33., Solution: As seen from the above data, the item 33 occurred maximum number of times, i.e. 5 times. Hence 33 is considered to be the modal value of the given data., Grouped Data:, Example: The following table gives the marks obtained by 89 students in Statistics. Find, the mode., Marks, , 10-14, , 15-19, , 20-24, , 25-29, , 30-34, , 35-39, , 40-44, , 45-49, , No. of, , 4, , 6, , 10, , 16, , 21, , 18, , 9, , 5, , students, Solution:, , Mode = l , , ( f  f1 ), xC, 2 f  f1  f 2, , Where l = the lower limit of the modal class ; C = length of the modal class, , BS Publication

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16, f = the frequency of the modal class, , f1 = the frequency of the class preceding modal class, , f 2 = the frequency of the class succeeding modal class, Sometimes it so happened that the above formula fails to give the mode. In this case, the, modal value lies in a class other than the one containing maximum frequency. In such, cases we take the help of the following formula;, Mode = l , , f2, xC, f1  f 2, , Where f , c, f1 and f 2 have usual meanings., Marks, , No. of students (f), , 10-14, , 4, , 15-19, , 6, , 20-24, , 10, , 25-29, , 16f1, , 30-34, , 21f, , 35-39, , 18f2, , 40-44, , 9, , 45-49, , 5, , From the above table it is clear that the maximum frequency is 21 and it lies in the class, 30-34., Thus the modal class is 29.5-34.5, Here L = 29.5 , c = 5, f = 21, f1 =16, f 2 = 18, , 21  16, , , Mode = 30  , *5,  2 * 21  16  18 , = 30 + 3.13, = 33.63 m, , BS Publication

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17, Merits and Demerits of Mode:, Merits:, 1. Mode is readily comprehensible and easy to calculate., 2. Mode is not at all affected by extreme values., 3. Mode can be conveniently located even if the frequency distribution has classintervals of unequal magnitude provided the modal class and the classes preceding, and succeeding it are of the same magnitude. Open-end classes also do not pose, any problem in the location of mode, Demerits:, 1. Mode is ill defined. It is not always possible to find a clearly defined mode. In, some cases, we may come across distributions with two modes. Such distributions, are called bi-modal., , If a distribution has more than two modes, it is said to be, , multimodal., 2. It is not based upon all the observations., 3. It is not capable of further mathematical treatment., 4. As compared with mean, mode is affected to a greater extent by fluctuations of, sampling., , Dispersion, Dispersion means scattering of the observations among themselves or from a, central value (Mean/ Median/ Mode) of data. We study the dispersion to have an idea, about the variation., Suppose that we have the distribution of the yields (kg per plot) of two Ground nut, varieties from 5 plots each. The distribution may be as follows:, Variety 1:, , 46, , 48, , 50, , 52, , 54, , Variety 2:, , 30, , 40, , 50, , 60, , 70, , It can be seen that the mean yield for both varieties is 50 k.g. But we can not say that the, performances of the two varieties are same. There is greater uniformity of yields in the, first variety where as there is more variability in the yields of the second variety. The first, variety may be preferred since it is more consistent in yield performance., , BS Publication

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18, , Measures of Dispersion:, These measures give us an idea about the amount of dispersion in a set of, observations. They give the answers in the same units as the units of the original, observations. When the observations are in kilograms, the absolute measure is also in, kilograms. If we have two sets of observations, we cannot always use the absolute, measures to compare their dispersion. The absolute measures which are commonly used, are:, 1. The Range, 2. The Quartile Deviation, 3. The Mean Deviation, 4. The Standard Deviation and Variance, 5. Coefficient of Variation, 6. Standard Error, Standard Deviation: It is defined as the positive square root of the arithmetic mean of the, squares of the deviations of the given values from arithmetic mean. The square of the, standard deviation is called variance., Ungrouped data:, Let x1, x2, …….,xn be n observations then the standard deviation is given by the, formula, n, , S.D. =, ,  ( xi  A.M .) 2, i 1, , n, , n, , where A.M. =, , x, i 1, , n, , i, , ,, , where n = no. of observations., Simplifying the above formula, we have, , or, , S.D. =, , 2, ,  n  , ,   xi , 2, 1 n,  i 1  ,   xi , , n  i 1, n, , , , , , , by linear transformation method, we have, , BS Publication

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19, , =, , 2, ,  n,  , ,   di , 2, 1 n,  i 1  ,   di , , n  i 1, n, , , , , , , where di = xi  A ; A= Assumed value; xi = Given values, Continuous frequency distribution: (grouped data):, Let f1, f2, ……,fn be the „n‟ frequencies corresponding to the mid values of the, classes x1, x2, ……,xn respectively, then the standard deviation is given by, n, ,  f (x, i 1, , S.D. =, ,  A.M .) 2, , i, , where, , N, , n, , f, i 1, , i, , N, , Simplifying the above formula, we have, , S.D. =, , ,  n, ,   f i xi, 1 n, 2, f i xi   i 1, , ,  N, N  i 1, , , , , , , , , , , , , 2, , , , , , , , , , by linear transformation method, we have, , ( f i d i ) 2 , 1 , 2, S.D. = C,  f i d i  N , N , , where di =, , xi  A, ; A = assumed value; and C = class interval, C, , S.D. for population data is represented by the symbol „‟, Ungrouped data:, Example:, Calculate S.D. for the kapas yields (in kg per plot) of a cotton variety recorded from seven, plots 5, 6, 7, 7, 9, 4, 5, i) Direct method:, , S.D., , =, , 2, ,  n  , ,   xi , 2, 1 n,  i 1  ,   xi , , n  i 1, n, , , , , , , BS Publication

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20, ii) Deviation Method:, , S.D., , 2, n, , , , , , , d, , , , 2, i, , 1 n, i 1, , ,   di , , n  i 1, n, , , , , , , =, , Where xi = given values, Assumed value (A) = 7, di = deviation (i.e. di =xi-A), 2, , d i  xi  A, , di, , 2, , xi, , xi, , 5, , 25, , 5-7=-2, , 4, , 6, , 36, , 6-7=-1, , 1, , 7, , 49, , 7-7=0, , 0, , 7, , 49, , 7-7=0, , 0, , 9, , 81, , 9-7=2, , 4, , 4, , 16, , 4-7=-3, , 9, , 5, , 25, , 5-7=-2, , 4, , Σxi = 43, , Σxi2 = 281, , Σdi = -6, , Σ d i = 22, , i) Direct method:, S.D., , 2, , 1, 43 , =, 281 , , 7, 7 , , = 1.55 kg., ii) Deviation Method:, S.D. =, ,  6 2 , 1, 22, , , , 7, 7 , , = 1.55 kg, , BS Publication, , 2

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21, Grouped Data:, Example:, The following are the 381soybean plant heights in Cms collected from a particular plot., Find the Standard deviation of the plants by direct and deviation method:, Plant heights (Cms), , No. of Plants (fi), , 6.8 -7.2, , 9, , 7.3 -7.7, , 10, , 7.8-8.2, , 11, , 8.3-8.7, , 32, , 8.8-9.2, , 42, , 9.3-9.7, , 58, , 9.8-10.2, , 65, , 10.3-10.7, , 55, , 10.8-11.2, , 37, , 11.3-11.7, , 31, , 11.8-12.2, , 24, , 12.3-12.7, , 7, , Solution:, i) Direct method:, n, , A.M. = , , fx, i 1, , i i, , N, , S.D. =, , ;, , where N =, , n, , f, i 1, , i, , 2, ,  n,  ,   f i xi  , , 1 n, 2,  , f i xi   i 1, , , , N i 1, N, , , , , , ii) Deviation Method:, ,  n,   fi di, A.M. = A   i 1,  N, , , , , , C, , , , , BS Publication

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22, 2, ,  n,  ,   fi di  , , 1 n, 2,  , S.D. = C, f i d i   i 1, , , , N i 1, N, , , , , , f i xi, , xi  A, C, , di , , 2, , fi, , xi, , 6.8-7.2, , 9, , 7.0, , 63, , 441, , -5, , -45, , 225, , 7.3-7.7, , 10, , 7.5, , 75, , 562.5, , -4, , -40, , 160, , 7.8-8.2, , 11, , 8.0, , 88, , 704, , -3, , -33, , 99, , 8.3-8.7, , 32, , 8.5, , 272, , 2312, , -2, , -64, , 128, , 8.8-9.2, , 42, , 9.0, , 378, , 3402, , -1, , -42, , 42, , 9.3-9.7, , 58, , 9.5, , 551, , 5234.5, , 0, , 0, , 0, , 9.8-10.2, , 65, , 10, , 650, , 6500, , 1, , 65, , 65, , 10.3-10.7, , 55, , 10.5, , 577.5, , 6063.75, , 2, , 110, , 220, , 10.8-11.2, , 37, , 11.0, , 407, , 4477, , 3, , 111, , 333, , 11.3-11.7, , 31, , 11.5, , 356.5, , 4099.75, , 4, , 124, , 496, , 11.8-12.2, , 24, , 12.0, , 288, , 3456, , 5, , 120, , 600, , 12.3-12.7, , 7, , 12.5, , 87.5, , 1093.75, , 6, , 42, , 252, , Σfixi, =3793.5, , Σfixi2, =38346.25, , Σfidi, =, 348, , Σ fi di =, 2620, , i) Direct method:, A.M. =, , S.D. =, , =, , 3793.5, = 9.96 Cms, 381, 2, , 1 , 3793.5 , 38346.25 , , 381 , 381 , , 1, 38346.25  37770.71, 381, , = 1.5106 = 1.23 Cms., ii) Deviation Method:, ,  348 , A.M. = 9.5+ , 0.5 = 9.96 Cms,  381 , , BS Publication, , fidi, , fi di, , C.I., , N, =381, , fixi, , 2, , 2

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23, 2, , 1 , 348 , = 0.5, 2620 , , 381 , 381 , , S.D., , = 0.5x47.98 = 1.23 Cms., Measures of Relative Dispersion:, These measures are calculated for the comparison of dispersion in two or more, than two sets of observations. These measures are free of the units in which the original, data is measure. If the original is in dollar or kilometers, we do not use these units with, relative measure of dispersion. These are a sort of ratio and are called coefficients., Suppose that the two distributions to be compared are expressed in the same units and, their means are equal or nearly equal. Then their variability can be compared directly by, using their standard deviations. However, if their means are widely different or if they are, expressed in different units of measurement. We can not use the standard deviations as, such for comparing their variability. We have to use the relative measures of dispersion in, such situations., Coefficient of Variation (C.V.), Coefficient of variation is the percentage ratio of standard deviation and the, arithmetic mean. It is usually expressed in percentage. The formula for C.V. is,, C.V. =, , S .D., x100, Mean, , Where S.D. =, , 2, ,  n  , ,   xi , 2, 1 n,  i 1   and,   xi , , n  i 1, n, , , , , , n, , Mean , , x, i 1, , i, , n, , The coefficient of variation will be small if the variation is small of the two groups,, the one with less C.V. said to be more consistent., Note: 1. Standard deviation is absolute measure of dispersion, 2. Coefficient of variation is relative measure of dispersion., , BS Publication

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24, Example: Consider the distribution of the yields (per plot) of two ground nut varieties., For the first variety, the mean and standard deviation are 82 kg and 16 kg respectively., For the second variety, the mean and standard deviation are 55 kg and 8 kg respectively., Then we have, for the first variety, C.V. =, , 16, x100 = 19.5%, 82, , C.V. =, , 8, x100 = 14.5%, 55, , For the second variety, , It is apparent that the variability in second variety is less as compared to that in the first, variety. But in terms of standard deviation the interpretation could be reverse., Example: Below are the scores of two cricketers in 10 innings. Find who is more, „consistent scorer‟ by Indirect method., A, , 204, , 68, , 150, , 30, , 70, , 95, , 60, , 76, , 24, , 19, , B, , 99, , 190, , 130, , 94, , 80, , 89, , 69, , 85, , 65, , 40, , Solution:, , Let the player A = x, And the player B = y, , Coefficient of variation of x= (C.V .) x, , =, , , Where x = A  , , , , Standard deviation of x, , and coefficient of variation of y, , = x , , x, x, , X 100, , d, , x, , n, , ,  where d x  xi  A, , , ,  dx , 1,  d x 2  , n, n, , , = (C.V .) y =, , y, y, , = y , , Where, ,  dy, y = B,  n, , , , , , , X 100, ,  dy , 1, 2,  d y  , n, n, , , Standard deviation of y, , 2, , 2, , , , , , ,  where d y  yi  B, , , , BS Publication

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25, Here A = 150 and B = 190, xi, , yi, , dx=xi –A, , dy=yi –B, , dxi2, , dyi2, , 204, , 99, , 54, , -91, , 2916, , 8281, , 68, , 190, , -82, , 0, , 6724, , 0, , 150, , 130, , 0, , -60, , 0, , 3600, , 30, , 94, , -120, , -96, , 14400, , 9216, , 70, , 80, , -80, , -110, , 6400, , 12100, , 95, , 89, , -55, , -101, , 3025, , 10201, , 60, , 69, , -90, , -121, , 8100, , 14641, , 76, , 85, , -74, , -105, , 5476, , 11025, , 24, , 65, , -126, , -125, , 15876, , 15625, , 19, , 40, , -131, , -150, , 17161, , 22500, , Σ dx = -704, , Σdx2, , Σdy = -959, , Σdy2, , = 80078, , =107189, ,   704 , x = 150  , ,  10 , ,   959 , y = 190  , ,  10 , , = 79.6 runs, , x , =, , = 94.1 runs, ,  704 2 , 1 , 80078, , , , 10 , 10, , , y , , 1, 80078  49561.6, 10, , =, , = 55.24 runs, , (C.V .) x =, , 55.24, X 100, 79.6, , = 69.4%, Coefficient of variation of A, ,  9592 , 1 , 107189, , , , 10 , 10 , 1, 107189  91968.1, 10, , = 39.01 runs, (C.V .) y =, , 39.01, x100, 94.1, , = 41.46%, is greater than coefficient of variation of B and hence we, , conclude that coefficient of player B is more consistent, , BS Publication

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26, NORMAL DISTRIBUTION, The Normal Distribution (N.D.) was first discovered by De-Moivre as the limiting, form of the binomial model in 1733, later independently worked Laplace and Gauss., The Normal distribution is „probably‟ the most important distribution in statistics., It is a probability distribution of a continuous random variable and is often used to model, the distribution of discrete random variable as well as the distribution of other continuous, random variables. The basic from of normal distribution is that of a bell, it has single, mode and is symmetric about its central values., , The flexibility of using normal, , distribution is due to the fact that the curve may be centered over any number on the real, line and it may be flat or peaked to correspond to the amount of dispersion in the values of, random variable., Definition: A random variable X is said to follow a Normal Distribution with parameter , and and 2 if its density function is given by the probability law, , 1, f(x) =, e,  2, , ( x )2, , - < x < ; - <  < ;  > 0, , 2 2, , where  = a mathematical constant equality = 22/7, e = Naperian base equaling 2.7183,  = population mean,  = population standard deviation, x = a given value of the random variable in the range - < x <, Characteristics of Normal distribution and normal curve:, The normal probability curve with mean  and standard deviation  is given by the, equation, f(x) =, , 1, ,  2, , e, , , , ( x )2, 2 2, , ; - < x <, , and has the following properties, i., , The curve is bell shaped and symmetrical, about the mean , , ii., , The height of normal curve is at its maximum at the mean. Hence the mean, and mode of normal distribution coincides. Also the number of observations, below the mean in a normal distribution is equal to the number of observations, about the mean. Hence mean and median of N.D. coincides. Thus, N.D. has, Mean = median = mode, , BS Publication

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27, iii., , As „x‟ increases numerically, f(x) decreases rapidly, the maximum probability, occurring at the point x = , and given by, p[(x)]max =, , 1, ,  2, , iv., , 32, Skewness 1 =, =0, 23, , v., , Kurtosis = 2 =, , vi., , All odd central moments are zero‟s, , 4, = 3 (  i ,  2 ,  3 and 4 are called central moments), 22, , i.e. 1   3   5  ..............  0, vii., , The first and third quartiles are equidistant from the median, , viii., , Linear combination of independent normal variates is also a normal variate, , ix., , The, , points, , of, , inflexion, , of, , the, , 1, , , 1, 2, x, , , , , ,, f, (, x, ), , e, , ,  2 , , , , , x., , If, ,  f ( x)dx  1 then, , , , the area under the normal curve is distributed as follows, i), , - < x < + covers 68.26% of area, , ii), , -2 < x < +2 covers 95.44% of area, , iii), , -3 < x < +3 coves 99.73% of area, , Area under Normal curve, , BS Publication, , curve, , is, , given, , by

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28, The Normal Curve: The graph of the normal distribution depends on two factors - the, mean and the standard deviation. The mean of the distribution determines the location of, the center of the graph, and the standard deviation determines the height and width of the, graph. When the standard deviation is large, the curve is short and wide; when the, standard deviation is small, the curve is tall and narrow. All normal distributions look like, a symmetric, bell-shaped curve, as shown below., , The curve on the left is shorter and wider than the curve on the right, because the curve on, the left has a bigger standard deviation., Standard Normal Distribution: If „X‟ is a normal random variable with Mean  and, standard deviation  , then Z =, , X , , , , is a standard normal variate with zero mean and, , standard deviation = 1., The probability density function of standard normal variate „z‟ is, f(z) =, , 1, 2, , e, , , , z2, 2, , , , and, ,  f ( z )dz, , =1, , , , A graph representing the density function of the Normal probability distribution is also, known as a Normal Curve or a Bell Curve (see Figure below). To draw such a curve, one, needs to specify two parameters, the mean and the standard deviation. The graph below, has a mean of zero and a standard deviation of 1, i.e., (m =0, s =1). A Normal distribution, with a mean of zero and a standard deviation of 1 is also known as the Standard Normal, Distribution., , Standard Normal Distribution, , BS Publication

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29, , Testing of Hypothesis, Introduction: The estimate based on sample values do not equal to the true value in the, population due to inherent variation in the population. The samples drawn will have, different estimates compared to the true value. It has to be verified that whether the, difference between the sample estimate and the population value is due to sampling, fluctuation or real difference. If the difference is due to sampling fluctuation only it can, be safely said that the sample belongs to the population under question and if the, difference is real we have every reason to believe that sample may not belong to the, population under question. The following are a few technical terms in this context., Hypothesis: The assumption made about any unknown characteristics is called hypothesis., It may or may be true., Ex:, , 1., ,  = 2.3;  be the population mean, , 2., ,  = 2.1 ;  be the population standard deviation, , Population follows Normal Distribution. There are two types of hypothesis, namely null, hypothesis and alternative hypothesis., Null Hypothesis:, , Null hypothesis is the statement about the parameters. Such a, , hypothesis, which is usually a hypothesis of no difference is called null hypothesis and is, usually denoted by H0., or, any statistical hypothesis under test is called null hypothesis. It is denoted by H0., 1. H0:  = 0, 2. H0: 1 = 2, Alternative Hypothesis: Any hypothesis, which is complementary to the null hypothesis,, is called an alternative hypothesis, usually denoted by H1., Ex:, , 1. H1:  # 0, 2. H1: 1 # 2, , Parameter: A characteristics of population values is known as parameter. For example,, population mean () and population variance (2)., , BS Publication

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30, In practice, if parameter values are not known and the estimates based on the sample, values are generally used., Statistic: A Characteristics of sample values is called a statistic. For example, sample, mean ( x ), sample variance (s2) where x =, , x1  x2  ..........  xn, n, , and s2 =, , ,  n 2 ,   xi  , n, 1, 2, xi   i 1  , , n  i 1, n , , , , , , Sampling distribution: The distribution of a statistic computed from all possible samples, is known as sampling distribution of that statistic., Standard error: The standard deviation of the sampling distribution of a statistic is, known as its standard error, abbreviated as S.E., S.E.( x ) =, , , ; where  = population standard deviation and n = sample size, n, , Sample: A finite subset of statistical objects in a population is called a sample and the, number of objects in a sample is called the sample size., Population: In a statistical investigation the interest usually lies in the assessment of the, general magnitude and the study of variation with respect to one or more characteristics, relating to objects belonging to a group. This group of objects under study is called, population or universe., Random sampling: If the sampling units in a population are drawn independently with, equal chance, to be included in the sample then the sampling will be called random, sampling. It is also referred as simple random sampling and denoted as SRS. Thus, if the, population consists of „N‟ units the chance of selecting any unit is 1/N. A theoretical, definition of SRS is as follows: Suppose we draw a sample of size „n‟ from a population, ,  , size N; then there are  N  possible samples of size „n‟. If all possible samples have an,  n , equal chance,, , 1,  , N,  n , , of being drawn, then the sampling is said to be simple random, , sampling., Simple Hypothesis: A hypothesis is said to be simple if it completely specifies the, distribution of the population. For instance, in case of normal population with mean , , BS Publication

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31, and standard deviation  , a simple null hypothesis is of the form H0:  =  0 ,  is, known, knowledge about  would be enough to understand the entire distribution. For, such a test, the probability of committing the type-1 error is expressed as exactly  ., Composite Hypothesis: If the hypothesis does not specify the distribution of the, population completely, it is said to be a composite hypothesis. Following are some, examples:, H0 :    0 and  is known, H0 :    0 and  is known, All these are composite because none of them specifies the distribution completely., Hence, for such a test the LOS is specified not as  but as „at most  ‟., Types of Errors: In testing of statistical hypothesis there are four possible types of, decisions, 1. Rejecting H0 when H0 is true, 2. Rejecting H0 when H0 is false, 3. Accepting H0 when H0 is true, 4. Accepting H0 when H0 is false, 1 and 4th possibilities leads to error decisions. Statistician gives specific names to, these concepts namely Type-I error, , and Type-II error respectively., , the above decisions can be arranged in the following table, Rejecting H0, , H0 is true, , H0 is false, , Type-I error, , Correct, , (Wrong decision), Accepting H0, , Correct, , Type-II error, , Type-I error: Rejecting H0 when H0 is true, Type-II error: Accepting H0 when H0 is false, The probabilities of type-I and type-II errors are denoted by  and  respectively., Degrees of freedom: It is defined as the difference between the total number of items and, the total number of constraints., If „n‟ is the total number of items and „k‟ the total number of constraints then the degrees, of freedom (d.f.) is given by d.f. = n-k, Level of significance(LOS): The maximum probability at which we would be willing to, risk a type-I error is known as level of significance or the size of Type-I error is level of, significance. The level of significance usually employed in testing of hypothesis are 5%, , BS Publication

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32, and 1%. The Level of significance is always fixed in advance before collecting the sample, information. LOS 5% means the results obtained will be true is 95% out of 100 cases and, the results may be wrong is 5 out of 100 cases., Critical value: while testing for the difference between the means of two populations, our, concern is whether the observed difference is too large to believe that it has occurred just, by chance. But then the question is how much difference should be treated as too large?, Based on sampling distribution of the means, it is possible to define a cut-off or threshold, value such that if the difference exceeds this value, we say that it is not an occurrence by, chance and hence there is sufficient evidence to claim that the means are different. Such a, value is called the critical value and it is based on the level of significance., Steps involved in test of hypothesis:, 1. The null and alternative hypothesis will be formulated, 2. Test statistic will be constructed, 3. Level of significance will be fixed, 4. The table (critical) values will be found out from the tables for a given level of, significance, 5. The null hypothesis will be rejected at the given level of significance if the value of, test statistic is greater than or equal to the critical value. Otherwise null hypothesis, will be accepted., 6. In the case of rejection the variation in the estimates will be called „significant‟, variation. In the case of acceptance the variation in the estimates will be called „notsignificant‟., , BS Publication

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33, STANDARD NORMAL DEVIATE TESTS, OR, LARGE SAMPLE TESTS, If the sample size n >30 then it is considered as large sample and if the sample size n< 30, then it is considered as small sample., SND Test or One Sample (Z-test), Case-I: Population standard deviation () is known, Assumptions:, 1. Population is normally distributed, 2. The sample is drawn at random, Conditions:, 1. Population standard deviation  is known, 2. Size of the sample is large (say n > 30), Procedure: Let x1, x2, ………x,n be a random sample size of n from a normal population, with mean  and variance  2 ., Let x be the sample mean of sample of size „n‟, Null hypothesis (H0): population mean (µ) is equal to a specified value 0, i.e. H0 :  = 0, Under H0, the test statistic is, Z=, , x  0, , , , ~ N(0,1), , n, i.e the above statistic follows Normal Distribution with mean „0‟ and varaince‟1‟., If the calculated value of Z < table value of Z at 5% level of significance, H0 is accepted, and hence we conclude that there is no significant difference between the population, mean and the one specified in H0 as 0., Case-II: If  is not known, Assumptions:, 1. Population is normally distributed, 2. Sample is drawn from the population should be random, 3. We should know the population mean, , BS Publication

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34, Conditions:, 1. Population standard deviation  is not known, 2. Size of the sample is large (say n > 30), Null hypothesis (H0) :  = 0, under H0, the test statistic, Z=, , x  0, s, , ~ N(0,1), , where s =, , n, ,  xi , 1 , 2, xi , , n, n, , , 2, , , , , , , x = Sample mean; n = sample size, If the calculated value of Z < table value of Z at 5% level of significance, H0 is accepted, and hence we conclude that there is no significant difference between the population mean, and the one specified in H0 otherwise we do not accept H0., The table value of Z at 5% level of significance = 1.96 and table value of Z at 1% level of, significance = 2.58., Two sample Z-Test or Test of significant for difference of means, Case-I: when  is known, Assumptions:, 1. Populations are distributed normally, 2. Samples are drawn independently and at random, Conditions:, 1. Populations standard deviation  is known, 2. Size of samples are large, Procedure: Let x1 be the mean of a random sample of size n1 from a population with mean, , 1 and variance  1 2, Let x 2 be the mean of a random sample of size n2 from another population with mean  2, and variance  2, , 2, , Null hypothesis, Alternative Hypothesis, , H0 : 1 =  2, H1 : 1   2, , i.e. The null hypothesis states that the population means of the two samples are identical., Under the null hypothesis the test statistic becomes, , BS Publication

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35, , Z=, , x1  x 2, 1,  2 , ,  n  n , 2 ,  1, 2, , 2, , ~ N(0,1)  (1), , i.e the above statistic follows Normal Distribution with mean „0‟ and varaince‟1‟., If 12 = 22 = 2 (say) i.e both samples have the same standard deviation then the test, statistic becomes, Z=, , x1  x 2, 1, 1,   ,  n1 n2, , , , , , ~ N(0,1)  (2), , If the calculated value of Z < table value of Z at 5% level of significance, H0 is accepted, otherwise rejected., If H0 is accepted means, there is no significant difference between two population means, of the two samples are identical., Example: The Average panicle length of 60 paddy plants in field No. 1 is 18.5. cms and, that of 70 paddy plants in field No. 2 is 20.3 cms. With common S.D. 1.15 cms. Test, whether there is significant difference between tow paddy fields w.r.t panicle length., Solution:, Null hypothesis: H0: There is no significant difference between the means of two paddy, fields w.r.t. panicle length., H0: 1   2, Under H0, the test statistic becomes, , Z=, , Where, , x1  x 2, 1, 1 ,    ,  n1 n2 , , ~ N (0,1) ------------ (1), , x1  first field sample mean = 18.5 inches, x 2  second field sample mean = 20.3 inches, n1 = first sample size = 60, n2 = second sample size = 70,  = common S.D. = 1.15 inches, , Substitute the given values in equation (1), we get, , BS Publication

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36, , Z=, , 18.5  20.3, 1.15, , 1, 1, , 60 70, , =, , 1.8, = 8.89, 1.15 x0.176, , Calculated value of Z = 5.1, Cal. Value of Z > table value of Z at 5% LOS(1.96), H0 is rejected. This means, there is, highly significant difference between two paddy fields w.r.t. panicle length., Case-II: when  is not known, Assumptions:, 1. Populations are normally distributed, 2. Samples are drawn independently and at random, Conditions:, 1. Population standard deviation  is not known, 2. Size of samples are large, Null hypothesis, , H0 : 1 =  2, , Under the null hypothesis the test statistic becomes, Z=, , Where, , x1  x 2,  s1 2 s 2 2 , , , n n , 2 ,  1, , ~ N(0,1)  (2), , x1 = 1st sample mean ,, , x 2 = 2nd sample mean, , s1 = 1st sample variance,, s 2 = 2nd sample variance, st, n1 = 1 sample size,, n2 = 2nd sample size, If the calculated value of Z < table value of Z at 5% level of significance, H0 is accepted, 2, , 2, , otherwise rejected., Example:, A breeder claims that the number of filled grains per panicle is more in a new variety of, paddy ACM.5 compared to that of an old variety ADT.36. To verify his claim a random, sample of 50 plants of ACM.5 and 60 plants of ADT.36 were selected from the, experimental fields. The following results were obtained:, (ForACM.5), x1  139.4-grains/panicle, s1 = 26.864, n1 = 50, Test whether the claim of the breeder is correct., , (For ADT.36), x 2  112.9 grains/panicle, s2 = 20.1096, n2 = 60, , BS Publication

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37, Sol:, , Null hypothesis H0 : 1   2, , (i.e. the average number of filled grains per panicle is the same for both ACM.5 and, ADT.36), Under H0, the test statistic becomes, , Z=, , x1  x 2, 2, , 2, , s1, s,  2, n1, n2, , ~ N (0,1) ------------ (1), , x1  first variety sample mean = 139.4 grains/panicle, , Where, , x 2  Second variety sample mean = 112.9 grains/panicle, , s1= first sample standard deviation = 26.864, s2= second sample standard deviation = 20.1096, n1 = first sample size = 50, and n2 = second sample size = 60, Substitute the given values in equation (1), we get, 139.4  112.9, , Z=, , =, , (26.864) 2 (20.1096) 2, , 50, 60, , 26.5, 14.4335  6.7399, , = 4.76, Calculated value of Z > Table value of Z at 5% LOS (1.96), H0 is rejected. We conclude, that the number of filled grains per panicle is significantly greater in ACM.5 than in, ADT.36, , BS Publication

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38, SMALL SAMPLE TESTS, The entire large sample theory was based on the application of “normal test”., However, if the sample size „n‟ is small, the distribution of the various statistics, e.g., Z =, , x  0, , , , are far from normality and as such „normal test‟ cannot be applied if „n‟ is small., , n, In such cases exact sample tests, we use t-test pioneered by W.S. Gosset (1908) who wrote, under the pen name of student, and later on developed and extended by Prof. R.A. Fisher., Student’s t-test: Let x1, x2,……….,xn be a random sample of size „n‟ has drawn from a, normal population with mean  and variance 2 then student‟s t – is defined by the, statistic, x  0, t=, s, , n, 2, ,  n  , , xi,  xi , , 1  n 2  i 1  , 2, i 1, where x , and s =,   xi , , n  1  i 1, n, n, , , , , , this test statistic follows a t – distribution with (n-1) degrees of freedom (d.f.). To get the, n, , critical value of t we have to refer the table for t-distribution against (n-1) d.f. and the, specific level of significance. Comparing the calculated value of t with critical value, we, can accept or reject the null hypothesis., The Range of t – distribution is -  to +  ., , One Sample t – test, One sample t-test is a statistical procedure that is used to know the population, mean and the known value of the population mean. In one sample t-test, we know the, population mean. We draw a random sample from the population and then compare the, sample mean with population mean and make a statistical decision as to whether or not the, sample mean is different from the population mean. In one sample t-test, sample size, should be less than 30., Assumptions: 1. Population is normally distributed, 2. Sample is drawn from the population and it should be random, 3. We should know the population mean, Conditions:, , 1. Population S.D.  is not known, 2. Size of the sample is small (<30)., , BS Publication

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39, Procedure: Let : Let x1, x2, …….,xn be a random sample of size „n‟ has drawn from a, normal population with mean  and variance 2., Null hypothesis (H0): population mean (µ) is equal to a specified value 0, i.e. H0:  = 0, Under H0, the test statistic becomes, t=, , x  0, s, n, , and follows student‟s t – distribution with (n-1) degrees of freedom., n, , where x , , x, i 1, , n, , i, , 2, , x , 1 , , 2,  x , , and s =, n 1 , n , , , , 2, , We now compare the calculated value of t with the tabulated value at certain level of, significance, If calculated value of t < table value of t at (n-1) d.f., the null hypothesis is accepted and, hence we conclude that there is no significant difference between the population mean and, the one specified in H0 as 0 ., Example: Based on field experiments, a new variety of greengram is expected to give an, yield of 13 quintals per hectare. The variety was tested on 12 randomly selected farmer, fields. The yields (quintal/hectare) were recorded as 14.3, 12.6, 13.7, 10.9,13.7, 12.0, 11.4,, 12.0, 13.1, 12.6, 13.4 and 13.1. Do the results conform the expectation?, Solution:, Null Hypothesis: H0 :  = 0 = 13, i.e. the results conform the expectation, The test statistic becomes, t=, , x  0, s, , ~t (n-1) d.f., , n, , x, Where x , n, , and s =, , ( xi ) 2 , 1 , 2,  xi  n  is an unbiased estimate of , n  1 , , , BS Publication

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40, Let yield = xi (say), xi2, 204.49, 158.76, 187.69, 118.81, 187.69, 144, 129.96, 144, 171.61, 158.76, 179.56, 171.61, 2, Σx = 1956.94, , xi, 14.3, 12.6, 13.7, 10.9, 13.7, 12, 11.4, 12, 13.1, 12.6, 13.4, 13.1, Σx = 152.8, x, , 152.8, = 12.73, 12, , s=, , 1, (152.8) 2, 1956.94 , 11 , 12, , , , , , = 1.01, t=, , 12.73  13, 1.01, , =, , 0.27, = 0.93 qa/h., 0.29, , 12, t-table value at (n-1) = 11d.f. at 5 percent level of significance is 2.20., Calculated value of t < table value of t, H0 is accepted and we may conclude that the, results conform to the expectation., , t-test for Two Samples, Assumptions: 1. Populations are distributed normally, 2. Samples are drawn independently and at random, Conditions:, , 1. Standard deviations in the populations are same and not known, 2. Size of the sample is small, , Procedure: If two independent samples xi ( i = 1,2,….,n1) and yj ( j = 1,2, …..,n2) of sizes, n1and n2 have been drawn from two normal populations with means 1 and 2, respectively., Null hypothesis H0 : 1 = 2, , BS Publication

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41, The null hypothesis states that the population means of the two groups are identical, so, their difference is zero., Under H0, the test statistics is t =, , x y, 1, 1 , S   ,  n1 n2 , , 2, 2, , , , x, y , 1, , , 2, 2,  x , , =, y , n1  n2  2 , n1, n2 , , , , 2, , Where S =, or, , 2, , (n1  1) s1  (n2  1) s 2, n1  n2  2, , S2 = pooled variance =, , 2, , where s12 and s22 are the variances of the first and second samples respectively., n2, , n1, , and x , , x, i 1, , n1, , i, , and y , , y, j 1, , n2, , j, , ; where x and y are the two sample means., , Which follows Student‟s t – distribution with (n1+n2-2) d.f., If calculated value of t < table value of t with (n1+n2-2) d.f. at specified level of, significance, then the null hypothesis is accepted otherwise rejected., Example: Two verities of potato plants (A and B) yielded tubes are shown in the, following table. Does the mean number of tubes of the variety „A‟ significantly differ, from that of variety B?, Tuber yield, kg/plant, Variety-A, , 2.2, , 2.5, , 1.9, , 2.6, , 2.6, , 2.3, , 1.8, , 2.0, , 2.1 2.4 2.3, , Variety-B, , 2.8, , 2.5, , 2.7, , 3.0, , 3.1, , 2.3, , 2.4, , 3.2, , 2.5 2.9, , Solution:, Hypothesis H0 : 1 =  2, i.e the mean number of tubes of the variety „A‟ significantly differ from the variety „B‟, Statistic, , t=, , x y, 1, 1 , S   ,  n1 n2 , , ~ t (n1  n2  2)d . f, , n1 = 1st sample size;, , n2 = 2nd sample size, , x = Mean of the first sample;, , y = mean of the second sample, , BS Publication

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42, , x=, , x, , y, , and, , n1, , y, n2, , 2, 2, , , , , , x , y , 1, , , , , , , 2, 2,  x , Where S =, , y, ,  , , n1  n2  2 , n1  , n2 , , , , , , , 2, , y, x, 2.20, 2.80, 2.50, 2.50, 1.90, 2.70, 2.60, 3.00, 2.60, 3.10, 2.30, 2.30, 1.80, 2.40, 2.00, 3.20, 2.10, 2.50, 2.40, 2.90, 2.30, Σy = 27.40, Σx = 24.70, 24.70, =, x, 11, , x2, 4.84, 6.25, 3.61, 6.76, 6.76, 5.29, 3.24, 4.00, 4.41, 5.76, 5.29, Σx2 = 56.21, , y, , = 2.25 Kg, Where S 2, , =, , y2, 7.84, 6.25, 7.29, 9.00, 9.61, 5.29, 5.76, 10.24, 6.25, 8.41, Σy2 = 75.94, 27.40, 10, , = 2.74 Kg, , , 1, (24.70) 2  , (27.40) 2 , 56, ., 21, , , 75, ., 94, ,  , , , 11  10  2 , 11  , 10 , , =, , 1, 56.21  55.46  75.94  75.07, 19, , = 0.09 Kg2, S=, Test statistic, , t=, , =, , S 2 = 0.3 Kg., , 2.25  2.74, 1 1, 0.3   ,  11 10 , , 0.49, = 3.77, 0.13, , Calculated value of t = 3.77, Table value of t for 19 d.f. at 5 % level of significance is 2.09, Since the calculated value of t > table value of t, the null hypothesis is rejected and hence, we conclude that the mean number of tubes of the variety „A‟ significantly not differ from, the variety „B‟, , BS Publication

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43, Paired t – test, The paired t-test is generally used when measurements are taken from the same subject, before and after some manipulation such as injection of a drug. For example, you can use, a paired t test to determine the significance of a difference in blood pressure before and, after administration of an experimental pressor substance., Assumptions: 1. Populations are distributed normally, 2. Samples are drawn independently and at random, Conditions:, , 1. Samples are related with each other, 2. Sizes of the samples are small and equal, 3. Standard deviations in the populations are equal and not known, , Hypothesis, , H0:  d  0, , Under H0, the test statistic becomes,, t=, , d, S, ,  t(n-1) d.f., , n, n, , where d , , d, i 1, , n, , i, , 2, ,  n  ,   di  , , 1  n 2  i 1  , 2, and S =,  di  n , n  1  i 1, , , , , , where S2 is variance of the deviations, n = sample size; where di = xi-yi ( i = 1,2,……,n), If calculated value of t < table value of t for (n-1)d.f. at α% level of significance, then the, null hypothesis is accepted and hence we conclude that the two samples may belong to the, same population. Otherwise, the null hypothesis rejected., Example: The average number of seeds set per pod in Lucerne were determined for top, flowers and bottom flowers in ten plants. The values observed were as follows:, Top, 4.2, 5.0, 5.4, 4.3, 4.8, 3.9, 4.2, 3.1, 4.4, 5.8, flowers, Bottom 4.6, 3.5, 4.8, 3.0, 4.1, 4.4, 3.6, 3.8, 3.2, 2.2, flowers, Test whether there is any significant difference between the top and bottom flowers with, respect to average numbers of seeds set per pod., Solution:, Null Hypothesis H0: d = 0, , BS Publication

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44, Under H0 becomes, the test statistic is, , t, , d, , ~ t ( n1) d . f ., , s, n, , d, Where d , n, , 2, , d , 1 , , 2,  d , , and s , n 1 , n, , , 2, , x, 4.2, 5.0, 5.4, 4.3, 4.8, 3.9, 4.2, 3.1, 4.4, 5.8, , y, 4.6, 3.5, 4.8, 3.0, 4.1, 4.4, 3.6, 3.8, 3.2, 2.2, , d, , d=x-y, -0.40, 1.50, 0.60, 1.30, 0.70, -0.50, 0.60, -0.70, 1.20, 3.60, , d, n, , Σd = 7.90, =, , d2, 0.16, 2.25, 0.36, 1.69, 0.49, 0.25, 0.36, 0.49, 1.44, 12.96, Σd2 =, 20.45, , 7.90, 10, , = 0.79, , s2 , , 1, (7.90) 2 , 20, ., 45, , , , 9, 10 , , = 2.27, s=, , s 2  2.27, , = 1.51, , t, , 0.79, 1.51, 10, , = 1.65, Calculated value of t = 1.65, Table value of t for 9 d.f. at 5% level of significance is 2.26, Calculated value of t < table value of t, the null hypothesis is accepted and we conclude, that there is no significant difference between the top and bottom flowers with respect to, average numbers of seeds set per pod., , BS Publication

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45, , F – Test, In agricultural experiments the performance of a treatment is assessed not only by, its mean but also by its variability. Hence, it is of interest to us to compare the variability, of two populations. In testing of hypothesis the equality of variances, the greater variance, is always placed in the Numerator and smaller variance is placed in the denominator., F- test is used to test the equality of two population variances, equality of several, regression coefficients, ANOVA ., F- test was discovered by G.W. Snedecor. The range of F : 0 to ∞, Let x1, x2 ,..............., xn1 and y1, y2 ,....... yn2 be the two independent random samples of sizes, n1 and n2 drawn from two normal populations N ( 1 ,1 ) and, 2, , N (  2 ,  2 ) respectively., 2, , S12 and S22 are the sample variances of the two samples., Null hypothesis H0 :  1   2, 2, , 2, , Under H0, the test statistic becomes, F=, , S1, , 2, , S2, , 2, , where, S12 > S22, , Which follows F-distribution with (n1-1, n2-1)d.f., , Where, , 2, 2, , , x , y , 1 , 1 , , , 2, 2, 2,  x ,  and S2 =,  y , , S , n1  1 , n1 , n2  1 , n2 , , , , , 2, 1, , or, , F=, , S2, , 2, , S1, , 2, , where S22 > S12, , Which follows F-distribution with (n2-1, n1-1)d.f., If calculated value of F < table value of F with (n2-1, n1-1)d.f at specified level of, significance, then the null hypothesis is accepted and hence we conclude that the variances, of the populations are homogeneous otherwise heterogeneous., Example: The heights in meters of red gram plants with two types of irrigation in two, fields are as follows:, Tap water (x), , 3.5, , 4.2, , 2.8, , 5.2, , 1.7, , 2.6, , 3.5, , 4.2, , 5.0, , Saline water (y), , 1.9, , 2.6, , 2.3, , 4.3, , 4.0, , 4.2, , 3.8, , 2.9, , 3.7, , Test whether the variances of the two system of irrigation are homogeneous., Solution:, H0: The variances of the two systems of irrigation are homogeneous., , BS Publication, , 5.2

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46, i.e.  1   2, 2, , 2, , Under H0, the test statistic becomes, 2, , S, 2, 2, F = 1 2 ; ( S1  S2 ), S2, 2, , x  , 1 , , 2, Where S1 = fist sample variance =, x  n , n1  1 , 1, , , 2, ,  y , 1 , and S 2 = second sample variance =, y2 , , n2  1 , n2, , , 2, , 2, , x, , y, , 3.5, 4.2, 2.8, 5.2, 1.7, 2.6, 3.5, 4.2, 5, 5.2, , 1.9, 2.6, 2.3, 4.3, 4, 4.2, 3.8, 2.9, 3.7,  y  29.7, ,  x =37.9, , x2, 12.25, 17.64, 7.84, 27.04, 2.89, 6.76, 12.25, 17.64, 25, , 27.04,  x  156.35, , , , , , y2, 3.61, 6.76, 5.29, 18.49, 16, 17.64, 14.44, 8.41, 13.69, 2,  y  104.33, , 2, , 1, (37.9) 2 ,  = 1.41 mt2, S1 = 156.35 , 9, 10 , 2, , and, , 2, , S2 =, , 1, (29.7) 2 , 104.33 ,  = 0.79 mt2, 8, 9 , 2, , F=, , S1, 2, S2, , =, , 1.41, = 1.78, 0.79, , F calculated value = 1.78, Table value of F0.05 for (n1-1, n2-1) d.f. = 3.23, Calculated value of F < Table value of at 5% level of significance, H0 is accepted and, hence we conclude that the variances of the two systems of irrigation are homogeneous., 1, 1,  F2 or F1 , Fi, F2, , BS Publication

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47, , Chi-square (2 ) test, The various tests of significance studied earlier such that as Z-test, t-test, F-test, were based on the assumption that the samples were drawn from normal population., Under this assumption the various statistics were normally distributed., , Since the, , procedure of testing the significance requires the knowledge about the type of population, or parameters of population from which random samples have been drawn, these tests are, known as parametric tests., But there are many practical situations in which the assumption of any kind about, the distribution of population or its parameter is not possible to make. The alternative, technique where no assumption about the distribution or about parameters of population is, made are known as non-parametric tests., , Chi-square test is an example of the non, , parametric test. Chi-square distribution is a distribution free test., If Xi  N(0,1) then, , x, , 2, i, , ~  2n, , Chi-square distribution was first discovered by Helmert in 1876 and later independently, by Karl Pearson in 1900. The range of chi-square distribution is 0 to ∞., Measuremental data: the data obtained by actual measurement is called measuremental, data. For example, height, weight, age, income, area etc.,, Enumeration data: the data obtained by enumeration or counting is called enumeration, data. For example, number of blue flowers, number of intelligent boys, number of curled, leaves, etc.,, 2 – test is used for enumeration data which generally relate to discrete variable where as, t-test and standard normal deviate tests are used for measure mental data which generally, relate to continuous variable., 2 –test can be used to know whether the given objects are segregating in a theoretical, ratio or whether the two attributes are independent in a contingency table., The expression for 2 –test for goodness of fit, , (Oi  Ei ) 2,  =, Ei, i 1, 2, , n, , where Oi = observed frequencies, Ei = expected frequencies, n = number of cells( or classes), Which follows a chi-square distribution with (n-1) degrees of freedom, , BS Publication

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48, , The null hypothesis H0 = the observed frequencies are in agreement with the expected, frequencies, If the calculated value of 2 < Table value of 2 with (n-1) d.f. at specified level of, significance (), we accept H0 otherwise we do not accept H0., Conditions for the validity of 2 –test:, The validity of 2-test of goodness of fit between theoretical and observed, the, following conditions must be satisfied., i) The sample observations should be independent, ii) Constraints on the cell frequencies, if any, should be linear oi = ei, iii) N, the total frequency should be reasonably large, say greater than 50, iv) If any theoretical (expected) cell frequency is < 5, then for the application of chi-square, test it is pooled with the preceding or succeeding frequency so that the pooled frequency is, more than 5 and finally adjust for the d.f. lost in pooling., Applications of Chi-square Test:, 1. testing the independence of attributes, 2. to test the goodness of fit, 3. testing of linkage in genetic problems, 4. comparison of sample variance with population variance, 5. testing the homogeneity of variances, 6. testing the homogeneity of correlation coefficient, Test for independence of two Attributes of (2x2) Contingency Table:, A characteristic which can not be measured but can only be classified to one of the, different levels of the character under consideration is called an attribute., 2x2 contingency table: When the individuals (objects) are classified into two categories, with respect to each of the two attributes then the table showing frequencies distributed, over 2x2 classes is called 2x2 contingency table., , BS Publication

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49, Suppose the individuals are classified according to two attributes say intelligence (A) and, colour (B). The distribution of frequencies over cells is shown in the following table., A, B, , A1, , A2, , Row totals, , B1, a, B, R1+(a+b), B2, c, D, R2 = (c+d), Column total, C1 = (a+c), C2 = (b+d), N = (R1+R2) or (C1+C2), Where R1 and R2 are the marginal totals of 1st row and 2nd row, C1 and C2 are the marginal totals of 1st column and 2nd column, N = grand total, The null hypothesis H0: the two attributes are independent ( if the colour is not dependent, on intelligent), Based on above H0, the expected frequencies are calculated as follows., R xC, R xC, R xC, R xC, E(a) = 1 1 ;, E(b) = 1 2 ; E(c) = 2 1 ; E(d) = 2 2, N, N, N, N, Where N = a+b+c+d, To test this hypothesis we use the test statistic, n, (O  Ei ) 2, 2 =  i, Ei, i 1, the degrees of freedom for mxn contingency table is (m-1)x(n-1), the degrees of freedom for 2x2 contingency table is (2-1)(2-1) = 1, This method is applied for all rxc contingency tables to get the expected frequencies., The degrees of freedom for rxc contingency table is (r-1)x(c-1), If the calculated value of 2 < table value of 2 at certain level of significance, then H0 is, accepted otherwise we do not accept H0, The alternative formula for calculating 2 in 2x2 contingency table is, 2 =, , (ad  bc) 2 N, R1 xR2 xC1 xC 2, , Example: Examine the following table showing the number of plants having certain, characters, test the hypothesis that the flower colour is independent of the shape of leaf., Flower colour, , Shape of leaf, , Totals, , Flat leaves, , Curled leaves, , White flowers, , 99 (a), , 36 (b), , R1 = 135, , Red flowers, , 20( c), , 5 (d), , R2 = 25, , C1 = 119, , C2 = 41, , N = 160, , Totals, , BS Publication

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50, Solution:, Null hypothesis H0: attributes „flower colour‟ and „shape of leaf‟ are independent of each, other., Under H0 the statistic is, n, , 2 , ,  (o, i 1, , i, ,  ei ) 2, , ei, where oi = observed frequency, and, , ei = expected frequency, , Expected frequencies are calculated as follows., E(a) =, , R1 * C1 135 *119, = 100.40, , N, 160, , E(b) =, , R1 * C2 135 * 41, =34.59, , N, 160, , E( c ) =, , E(d) =, , where R1 and R2 = Row totals, C1 and C2 = column totals, , R2 * C1 25 *119, =18.59, , N, 160, , N= Grand totals, , R2 * C2 25 * 41, =6.406, , N, 160, , oi, , ei, , oi  ei, , (oi- ei)2, , 99, 36, 20, 5, , 100.40, 34.59, 18.59, 6.41, , -1.4, 1.41, 1.41, -1.41, , 1.96, 1.99, 1.99, 1.99, , (o i - e i )2, ei, 0.02, 0.06, 0.11, 0.31, n, ,  (o, i 1, , i, ,  ei ) 2, , ei, n, , Calculated value of  2 , ,  (o, i 1, , i, ,  ei ) 2, , ei, , = 0.49, , Direct Method:, Statistic:  2 =, , N (ad  bc) 2, R1 R2 C1C 2, , here a = 99, b = 36, c = 20 and d = 5 and N = 160, , 2=, , 160(99 * 5  36 * 20) 2, 135 * 25 *119 * 41, , BS Publication, , =0.49

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51, , =, , 160 * 50625, 164666.25, , =, , 18100000, = 0.49, 16466625, , Calculated value of  2 = 0.40, Table value of  2 for (2-1) (2-1) = 1 d.f. is 3.84, Calculated value of  2 < Table value of  2 at 5% LOS for 1 d.f. , Null hypothesis is, accepted and hence we conclude that two characters, flower colour and shape of leaf are, independent of each other., Yates correction for continuity in a 2x2 contingency table:, In a 2x2 contingency table, the number of d.f. is (2-1)(2-1) = 1. If any one of, Expected cell frequency is less than 5, then we use of pooling method for 2 –test results, with `0‟ d.f. (since 1 d.f. is lost in pooling) which is meaningless. In this case we apply a, correction due to Yates, which is usually known a Yates correction for continuity., Yates correction consists of the following steps; (1) add 0.5 to the cell frequency which is, the least, (2) adjust the remaining cell frequencies in such a way that the row and column, totals are not changed. It can be shown that this correction will result in the formula., N, , N  ad  bc  , 2, 2 (corrected) = , R1 R2 C1C 2, , 2, , Example: The following data are observed for hybrids of Datura., Flowers violet, fruits prickly =47, Flowers violet, fruits smooth = 12, Flowers white, fruits prickly = 21, Flowers white, fruits smooth = 3. Using chi-square test, find the association between, colour of flowers and character of fruits., Sol: H0: The two attributes colour of flowers and fruits are independent., We cannot use Yate‟s correction for continuity based on observed values., expected frequency less than 5, we use Yates‟s correction for continuity., The test statistic is, , BS Publication, , If only

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52, N, , N  ad  bc  , 2, = , R1R2C1C2, , 2, , Flowers Violet, , Flowers white, , Total, , Fruits Prickly, , 47(48.34), , 21(19.66), , 68, , Fruits smooth, , 12(10.66), , 3(4.34), , 15, , Total, , 59, , 24, , 83, ,  2 (corrected), , The figures in the brackets are the expected frequencies, ,  2(corrected), , 83 , , 83 (47 * 3)  (21 *12)  , 2, = , 68 *15 * 59 * 24, , 2, , 83141  252  41.5, , 2, , =, =, , 68 *15 * 59 * 24, 400910.75, = 0.28, 1444320, , Calculated value of  2 = 0.28, Table value of  2 for (2-1) (2-1) = 1 d.f. is 3.84, Calculated value of  2 < table value of  2 , H0 is accepted and hence we conclude that, colour of flowers and character of fruits are not associated, CORRELATION, When there are two continuous variables which are concomitant their joint, distribution is known as bivariate normal distribution. If there are more than two such, variables their joint distribution is known as multivariate normal distributions. In case of, bivariate or multivariate normal distributions, we may be interested in discovering and, measuring the magnitude and direction of the relationship between two or more variables., For this purpose we use the statistical tool known as correlation., Definition: If the change in one variable affects a change in the other variable, the two, variables are said to be correlated and the degree of association ship (or extent of the, relationship) is known as correlation., Types of correlation:, a). Positive correlation: If the two variables deviate in the same direction, i.e., if the, increase (or decrease) in one variable results in a corresponding increase (or decrease) in, the other variable, correlation is said to be direct or positive., , BS Publication

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53, Ex: (i) Heights and weights, (ii) Household income and expenditure, (iii) Amount of rainfall and yield of crops, (iv) Prices and supply of commodities, (v) Feed and milk yield of an animal, (vi) Soluble nitrogen and total chlorophyll in the leaves of paddy., b). Negative correlation: If the two variables constantly deviate in the opposite direction, i.e., if increase (or decrease) in one variable results in corresponding, , decrease (or, , increase) in the other variable, correlation is said to be inverse or negative., Ex: (i) Price and demand of a goods, (ii) Volume and pressure of perfect gas, (iii) Sales of woolen garments and the day temperature, (iv) Yield of crop and plant infestation, c) No or Zero Correlation: If there is no relationship between the two variables such that, the value of one variable change and the other variable remain constant is called no or zero, correlation., Figures:, , Methods of studying correlation: 1. Scatter Diagram, Correlation, , 3. Spearman‟s Rank Correlation, , 2. Karl Pearson‟s Coefficient of, , 4. Regression Lines, , 1. Scatter diagram: It is the simplest way of the diagrammatic representation of bivariate, data. Thus for the bivariate distribution (xi,yi); i = j = 1,2,…n, If the values of the, variables X and Y be plotted along the X-axis and Y-axis respectively in the xy-plane, the, diagram of dots so obtained is known as scatter diagram. From the scatter diagram, if the, points are very close to each other, we should expect a fairly good amount of correlation, , BS Publication

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54, between the variables and if the points are widely scattered, a poor correlation is expected., This method, however, is not suitable if the number of observations is fairly large., If the plotted points shows an upward trend of a straight line then we say that both the, variables are positively correlated, , Positive Correlation, When the plotted points shows a downward trend of a straight line then we say that both, the variables are negatively correlated, , Negative Correlation, If the plotted points spread on whole of the graph sheet, then we say that both the variables, are not correlated., , No Correlation, , BS Publication

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55, Karl Pearson’s Coefficient of Correlation: Prof. Karl Pearson, a British Biometrician, suggested a measure of correlation between two variables. It is known as Karl Pearson‟s, coefficient of correlation. It is useful for measuring the degree of linear relationship, between the two variables X and Y. It is usually denoted by rxy or „r‟., i) Direct Method:, , =, , cov( x, y ), , rxy =, , V ( x)V ( y ), , 1 n,  ( xi  x)( yi  y), n i 1, 1 n, 1 n, ( xi  x ) 2, ( yi  y ) 2, , , n i 1, n i 1, , After simplification =, ,  x  x)( y  y),  x  x   y  y , 2, , 2, , or, , r =, , { x, , 2, ,  xy , ( x), , , ( x)( y ), n, 2, , } { y , 2, , n, , ( y ) 2, n, , }, , ii) Deviation method:, , r=, , { dx, , 2, ,  dxdy , ( dx), , n, , ( dx)( dy ), n, 2, , } { dy , 2, , ( dy ) 2, n, , }, , Where x = S.D. of x and y = S.D. of Y, n = number of items;, , dx = x-A, dy = y-B, , A = assumed value of and B = assumed value of y, The correlation coefficient never exceed unity. It always lies between –1 and +1 (i.e. –1,  r  1), If r = +1 then we say that there is a perfect positive correlation between x and y, If r = -1 then we say that there is a perfect negative correlation between x and y, If r = 0 then the two variables x and y are called uncorrelated variables, No unit of measurement., , BS Publication

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56, Test for significance of correlation coefficient, If „r‟ is the observed correlation coefficient in a sample of „n‟ pairs of observations from a, bivariate normal population, then Prof. Fisher proved that under the null hypothesis, H0 :  = 0, the variables x, y follow a bivariate normal distribution. If the population correlation, coefficient of x and y is denoted by , then it is often of interest to test whether  is zero or, different from zero, on the basis of observed correlation coefficient „r‟. Thus if „r‟ is the, sample correlation coefficient based on a sample of „n‟ observations, then the appropriate, test statistic for testing the null hypothesis H0 :  = 0 against the alternative hypothesis H1:,  # 0 is, t=, , r, 1 r2, , (n  2), , t follows Student‟s t – distribution with (n-2) d.f., If calculated value of t > table value of t with (n-2) d.f. at specified level of significance,, then the null hypothesis is rejected. That is, there may be significant correlation between, the two variables. Otherwise, the null hypothesis is accepted., Example: From a paddy field, 12 plants were selected at random. The length of panicles, in cm (x) and the number of grains per panicle (y) of the selected plants were recorded., The results are given in the following table. Calculate correlation coefficient and its, testing., y, , 112, , 131, , 147, , 90, , 110, , 106, , 127, , 145, , 85, , 94, , 142, , 111, , x, , 22.9 23.9, , 24.8, , 21.2, , 22.2, , 22.7, , 23.0, , 24.0, , 20.6, , 21.0, , 24.0, , 23.1, , Solution:, a) Direct Method:, , Correlation coefficient rxy = r =, , { x 2, ,  xy , ( x), , n, , ( x)( y ), n, 2, , } { y 2 , , Where n = number of observations, Testing the correlation coefficient:, Null hypothesis H0: Population correlation coefficient „‟ = 0, Under H0, the test statistic becomes, , BS Publication, , ( y ) 2, n, , }

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57, , t=, , r n2, 1 r2, , x, , y, , x2, , 112, 131, 147, 90, 110, 106, 127, 145, 85, 94, 142, 111,  x  1400, , 22.9, 23.9, 24.8, 21.2, 22.2, 22.7, 23.0, 24.0, 20.6, 21.0, 24.0, 23.1,  y  273.4, , 12544, 17161, 21609, 8100, 12100, 11236, 16129, 21025, 7225, 8836, 20164, 12321,  x 2 =168450, , ~ t( n  2) d.f., y2, 524.41, 571.21, 615.04, 449.44, 492.84, 515.29, 529.00, 576.00, 424.36, 441.00, 576.00, 533.61,  y 2  6248.20, , xy, 2564.8, 3130.9, 3645.6, 1908.0, 2442.0, 2406.2, 2921.0, 3480.0, 1751.0, 1974.0, 3408.0, 2564.1,  xy  32195.6, , (1400) x 273.4), 12, 2, , (1400)  , (273.4) 2 , 168450 ,   6248.20 , , 12, 12, ,  , , 32195.6 , , r =, , =, , 298.9333, = 0.95, 71.5308 x19.2367, , The value of coefficient of determination (r2) = 0.90, t=, , =, , 0.95 12  2, 1  (0.95) 2, 3.0042, 0.3122, , = 9.6, t critical (table) value for 10 d.f. at 5% LOS is 2.23, Since calculated value i.e. 9.6 is > t table value i.e. 2.23, it can be inferred that there exists, significant positive correlation between (x, y)., , BS Publication

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58, b) Indirect Method:, Here A = 127 and B = 24, , dy, , x, , y, , dx  x  A, , dy  y  B, , dx, , 112, 131, 147, 90, 110, 106, 127, 145, 85, 94, 142, 111, , 22.9, 23.9, 24.8, 21.2, 22.2, 22.7, 23, 24, 20.6, 21, 24, 23.1, , -15, 4, 20, -37, -17, -21, 0, 18, -42, -33, 15, -16,  dx =, -124, , -1.1, -0.1, 0.8, -2.8, -1.8, -1.3, -1, 0, -3.4, -3, 0, -0.9, dy =, -14.6, , 225, 16, 400, 1369, 289, 441, 0, 324, 1764, 1089, 225, 256, 2,  dx =, 6398, ,  d  d , d d , n, ( d, ( d ),  d  n  d  n, x, , r=, , 2, , x, , y, , y, , 2, , 2, , 2, , x, , x, , y, , =, , (124)(14.6), 12, 2, (124), (14.6) 2, 6398 , 37 , 12, 12, , =, , 298.9333, 5116.6667 x 19.2367, , =, , 298.9333, = 0.95, 71.5309 x 4.3860, , 449.8 , , BS Publication, , y, , )2, , 2, , 1.21, 0.01, 0.64, 7.84, 3.24, 1.69, 1, 0, 11.56, 9, 0, 0.81, 2,  dd =, 37, , d xd y, , 16.5, -0.4, 16, 103.6, 30.6, 27.3, 0, 0, 142.8, 99, 0, 14.4,  d xd y =, 449.8

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59, REGRESSION, The term `regression‟ literally means “stepping back towards the average”. It was first, used by a British Biometrician Sir Francis Galton., The relationship between the independent and dependent variables may be expressed as a, function. Such functional relationship between two variables is termed as regression., In regression analysis independent variable is also known as regressor or, predictor or explanatory variable while dependent variable is also known as regressed or, explained variable., When only two variables are involved the functional relationship is known as, simple regression. If the relationship between two variables is a straight line, it is known, as simple linear regression, otherwise it is called as simple non-linear regression., Direct Method:, The regression equation of Y on X is given as, Y= a+bX, Where Y= dependent variable;, , X = independent variable and a = intercept, , b = the regression coefficient (or slope) of the line. a and b are also called as, Constants, the constants a and b can be estimated with by applying the `least squares method‟. This, involves minimizing, , n, , e, i, , 2, , i, ,    y  a  bx  . This gives, 2, , i, , byx = b =, , Cov( XY ), =, v( X ), and, ,  XY , X, , ( X )( Y ), , 2, , , , n, ( X ) 2, n, , a = Y  bX, , where b is called the estimate of regression coefficient of Y on X and it measures the, change in Y for a unit change in X., Similarly, the regression equation of X on Y is given as, X = a1+b1Y, where X = dependent variable and Y = independent variable, , bxy = b1 = =, , Cov( XY ), =, v(Y ), ,  XY , Y, , BS Publication, , 2, , ( X )( Y ), , , n, ( Y ) 2, n

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60, and a1 = X - b1 Y, where b1 is known as the estimate of regression coefficient of X on Y and `a‟ is intercept, Deviation Method:, The regression equation of Y on X is, (Y- Y ) = b (X- X ), ,  Y = Y  b( X  X ), The regression equation of X on Y, (X- X ) = b1(Y- Y ), ,  X = X + b1 (Y- Y ), Properties of regression coefficient:, 1. Correlation coefficient is the geometric mean of the two regression coefficients., r =  b.b1, , i.e, , 2. If one of the regression coefficient is greater than unity, the other must be less than, unity., 3. Arithmetic mean of the regression coefficients is greater than the correlation, coefficient `r‟., 4. Regression coefficients are independent of the change of origin but not of scale., 5. Units of „b‟ are same as that of the dependent variable., 6. Regression is only a one-way relationship between y (dependent variable) and x, (independent variable)., 7. The range of b is from -∞ to ∞. - ∞ for negative b and +∞ for positive b., Note:, 1. Both the lines regression pass through the point ( X , Y ). In other words, the, mean values ( X , Y ) can be obtained as the point of intersection of the two regression lines, 2. If r = 0, the two variables are uncorrelated, the lines of regression become, perpendicular to each other, 3. If r =  1, in this case the two lines of regression either coincide or they are parallel to, each other, 4.If the regression coefficients are positive, r is positive and if the regression, coefficients are negative, r is negative, , BS Publication

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61, Distinguish between correlation and regression:, Correlation, , Regression, , 1. Correlation is the relationship between two, , 1. Regression, , is, , or more variables. Where the change in one, , measure, , variable affects a change in other variable, , relationship between two or more, , of, , mathematical, the, , average, , variables. Where one variable is, dependent and other variable is, independent, 2. correlation coefficient measures extent of, relationship between two variables, , 2. regression coefficient estimates, the change in one variable for a, unit change in other related, variable, , 3. correlation is a two way relationship, , 3. regression, , is, , a, , one, , way, , coefficient, , is, , relationship, 4. correlation coefficient is independent of, units of the variables., , 4. regression, expressed, , in, , the, , units, , of, , dependent variable, 5. correlation, , coefficient, , always, , lies, , between –1 and +1, , 5. regression, , coefficient, , lies, , between, - and +, , 6. In correlation both the variables are, random, , 6. In regression one variable will, be in dependent and other will be, dependent, , 7. correlation coefficient calculated by, rxy =, , 8. r =, , bb1, , we can not predict, , cov( x, y ), V ( x)V ( y ), , 7. the regression coefficient of y on, ( x)( y ),  xy , n, x is b =, (, x) 2, , 2, x  n, the regression coefficient of x on y is, ( x)( y ), xy , , n, b1 =, ( y ) 2, 2, y  n, ,  y  y ,  x  x , , 2, , b=r, , BS Publication, , 2, , ;we can predict

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62, Example: The following data give the yield per plant (gm) (Y) and days to flowering (X), of 11 pigeonpea plants. Fit regression equations of Y on X and X on Y. Also estimate Y, when the flowering period is 149., Solution:, The regression equation of Y on X is given by, Y= a+bX, Where Y= dependent variable; X = independent variable and a = intercept, b = the regression coefficient of Y on X, The constants a and b can be estimated with the help of `least squares method‟, , Where b =, , ( X )( Y ), ,  XY , X, , 2, , n, ( X ) 2, , , , ;, , a = Y  bX, , n, , Similarly, the regression equation of X on Y is given by, X = a1 + b1Y, Where X = dependent variable and Y = independent variable, , Where b1 =, ,  XY , Y, , Y, 24.72, 20.25, 38.56, 74.72, 72.75, 78.45, 69.8, 80.4, 160.2, 165.75, 77.56, ,  Y =863.16, , 2, , ( X )( Y ), , , n, ( Y ) 2, n, , Y2, 611.08, 410.06, 1486.87, 5583.08, 5292.56, 6154.40, 4872.04, 6464.16, 25664.04, 27473.06, 6015.55, Y 2 =,  X =1339 90026.91, X, 91, 76, 98, 136, 128, 119, 142, 135, 147, 145, 122, , ; and a1 = X - b1 Y, , X2, XY, 8281, 2249.52, 5776, 1539.00, 9604, 3778.88, 18496, 10161.92, 16384, 9312.00, 14161, 9335.55, 20164, 9911.60, 18225, 10854.00, 21609, 23549.40, 21025, 24033.75, 14884, 9462.32, 2,  X =  XY =, 168609 114187.94, , (1339)(863.16), 11, (1339) 2, 168609 , 11, , 114187.94 , , b=, , BS Publication

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63, , =, , 9117.83, 5616.18, , = 1.62, , and, , X , , X, , Y, , Y, , n, n, , , , =, , 1339,  121.73 and, 11, 863.16, = 78.47, 11, , a = 78.47 – 1.62*121.73 = - 118.74, , The regression equation of Y on X is Y = a+bX = Y = -118.74 + 1.62X, To estimate Y when X= 149, Yˆ = -118.74 + 1.62 * 149 = 122.64, , b1 =, , (1339)(863.16), 9117.83, 11, =, = 0.41, 2, 22295.53, (863.16), 90026.91 , 11, , 114187.94 , , a1 = X - b1 Y = 121.73 – 0.41*78.47 = 89.56, The regression equation of X on Y is X = 89.56 + 0.41Y, Result:, 1.The regression equation of Y on X is Y = a + bX = Y = -118.74 + 1.62X, 2. The regression equation of X on Y is X = 89.56 + 0.41Y, 3. Estimated Yield is 122.64, , BS Publication

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64, , Analysis of Variance (ANOVA), The ANOVA is a powerful statistical tool for tests of significance. The test of, significance based on t-distribution is an adequate procedure only for testing the, significance of the difference between two sample means. In a situation when we have, two or more samples to consider at a time, an alternative procedure is needed for testing, the hypothesis that all the samples have been drawn from the same population. For, example, if three fertilizers are to be compared to find their efficacy, this could be done by, a field experiment, in which each fertilizer is applied to 10 plots and then the 30 plots are, later harvested with the crop yield being calculated for each plot. Now we have 3 groups, of ten figures and we wish to know if there are any differences between these groups. The, answer to this problem is provided by the technique of ANOVA., The term ANOVA was introduced by Prof. R.A. Fisher in 1920‟s to deal with, problem in the analysis of agronomical data. Variation is inherent in nature, The total variation in any set of numerical data is due to a number of causes which, may be classified as i) Assignable causes and ii) chance causes, The variation due to assignable causes can be detected and measured where as the, variation due to chance causes is beyond the control of humans and cannot be traced, separately, ANOVA: The ANOVA is a simple arithmetical process of sorting out the components of, variation in a given data, Types of ANOVA: There are two types i) One way classification and ii) Two way, classification, Assumptions of ANOVA:, 1. The observations are independent, 2. Parent population from which observations are taken is normal, 3. Various treatment and environmental effects are additive in nature, 4. The experimental errors are distributed normally with mean zero and variance 2, , Experimental Designs, In order to verify a hypothesis pertaining to some scientific phenomena we have to, collect data. Such data are obtained by either observation or by experimentation. The, main topics connected with data collection are Theory of Sample Surveys and, Experimental Designs. In sample survey, a researcher makes observations on existing, population and records data without interfering with the process that is being observed. In, , BS Publication

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65, experimentation, on the other hand, the researcher controls or manipulates the, environment of the subjects that constitute the population. The experiments allow a, researcher to study the factors of his interest and show that these factors actually cause, certain effects. Hence, whenever the objective is to study the effects of variables rather, than simply to describe a population, we prefer the data collection through, experimentation., Modern concepts of experimental design are due primarily to R.A. Fisher. He, developed them in the planning of agricultural field experiments. They are now used in, many fields of science., , Basic concepts:, Blocks: In agricultural experiments, most of the times we divide the whole experimental, unit (field) into relatively homogeneous sub-groups or strata. These strata, which are more, uniform amongst themselves than the field as a whole are known as blocks., Treatments: the objects of comparison in an experiment are defined as treatments, For example: i) suppose an Agronomist wishes to know the effect of different, spacings on the yield of a crop, different spacings will be treatments. Each spacing, will be called a treatment., ii) If different of fertilizer are tried in an experiment to test the responses of a crop, to the fertilizer doses, the different doses will be treatments and each dose will be a, treatment., ii) A teacher practices different teaching methods on different groups in his class to, see which yields the best results., iii) A doctor treats a patient with a skin condition with different creams to see, which is most effective., Experimental unit: Experimental unit is the object to which treatment is applied to, record the observations., For example i) In laboratory insects may be kept in groups of five or six. To each group,, different insecticides will be applied to know the efficacy of the insecticides. In this study, different groups of insects will be the experimental unit., ii) If treatments are different varieties, then the objects to which treatments are applied to, make observations will be different plot of land. The plots will be called experimental, units., , BS Publication

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66, Basic principles of experimental designs:, The purpose of designing an experiment is to increase the precision of the experiment. In, order to increase the precision, we try to reduce the experimental error. For reducing the, experimental error, we adopt some techniques. These techniques form the basic principles, of experimental designs. The basic principles of the experimental designs are replication,, randomization and local control., 1. Replication: Repetition of, , treatment to different experimental units is known as, , Replication. In other words, the repetition of treatments under investigation is known as, replication. We have no means of knowing about the variations in the results of a, treatment., , Only when we repeat the treatment several times we can estimate the, , experimental error., A replication is used (i) to secure more accurate estimate of the experimental error, a term, which represents the differences that would be observed if the same treatments were, applied several times to the same experimental units;, (ii) to reduce the experimental error and thereby to increase precision, which is a measure, of the variability of the experimental error., The standard error of treatment mean is, , , . Where  is S.D. of treatment in the, r, , population and `r‟ is the number of replications. As `r‟ increases, the standard error of, mean decreases. Also in the analysis of variance the replication of treatments provides, estimate of experimental error which is essential for the application of F-tes., 2. Randomization: when all the treatments have equal chances of being allocated to, different experimental units it is known as randomization, or, Random allocation of treatments to different experimental units known as randomization., The purpose of randomization is to remove bias and other sources of extraneous variation, which are not controllable., , Another advantage of randomization (accompanied by, , replication) is that it forms the basis of any valid statistical test. Hence the treatments, must be assigned at random to the experimental units. Randomization is usually done by, using tables of random numbers., 3.Local control: It has been observed that all extraneous sources of variation are not, removed by randomization and replication., , This necessitates a refinement in the, , experimental technique. For this purpose, we make use of local control, a term referring, to the grouping of homogeneous experimental units., , BS Publication

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67, The main purpose of the principle of local control is to increase the efficiency of an, experimental design by decreasing the experimental error., Replication, , II, , III, , Random, , Local control, , distribution, Validity of estimate, of experimental error, , Reducing the, experimental error, , Shape of blocks and plots: the shape and size of the blocks will usually depend up on the, shape and size of the plots. In order to control the experimental error it is desirable to, divide the whole experimental area into different subgroups (blocks) such that within each, block there is as much homogeneity as possible but between blocks there is maximum, variation. Further each block is to be divided into as many plots as the number of, treatments. For maximum precision the plots should be rectangular in shape with their, long sides parallel to the direction of the fertility gradient and the blocks should be, arranged one after the other along the fertility gradient as shown in the figure., Fertility gradient, Block-II, , ……….., , No variation in this direction, , Block-I, , ………., , Plots, , Plots, , BS Publication, , Block-K

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68, , COMPLETELY RANDOMIZED DESIGN (CRD), The CRD is the simplest of all the designs. In this design, treatments are allocated, at random to the experimental units over the entire experimental material. In case of field, experiments, the whole field is divided into a required number of plots equal size and then, the treatments are randomized in these plots., , Thus the randomization gives every, , experimental unit an equal probability of receiving the treatment., In field experiments there is generally large variation among experimental plots, due to soil heterogeneity. Hence, CRD is not preferred in field experiments. In laboratory, experiments and green house studies, it is easy to achieve homogeneity of experimental, materials. Therefore, CRD is most useful in such experiments., Layout of CRD: The placement of the treatments on the experimental units along with the, arrangement of experimental units is known as the layout of an experiment., For example, suppose that there are 5 treatments A,B,C,D and E. each with 4 replications,, we need 20 experimental units. Here, since the number of units is 20, a two digit random, number of table will be consulted and a series of 20 random numbers will be taken, excluding those which are greater than 20., , suppose, the random numbers are, , 4,18,2,14,3,7,13,1,6,10,17,20,8,15,11,5,9,12,16,19. After this the plots will be serially, numbered and the treatment A will be allotted to the plots bearing the serial numbers 4,, 18, 2, 14 and so on., 1, , 2, , 3, , 4, , 5, , B, , A, , B, , A, , D, , 6, , 7, , 8, , 9, , 10, , C, , B, , D, , E, , C, , 11, , 12, , 13, , 14, , 15, , D, , E, , B, , A, , D, , 16, , 17, , 18, , 19, , 20, , E, , C, , A, , E, , C, , BS Publication

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70, ANOVA table, Sources of, , D.F., , S.S., , k-1, , Tr.S.S., , M.S., , F-cal.value, , F-table value, , variation, Treatments, , Error, , N-k, , ESS, , Total, , N-1, , TSS, , TMS =, , Tr.S .S ., k 1, , EMS=, , ESS, N k, , Ft =, , TMS, EMS, , F[k-1, N-k] at, ,  % LOS, , -, , If the calculated value of F < table value of F at certain level of significance, we accept H0, and hence we may conclude that there is no significant difference between the treatment, means, If the calculated value of F > table vale of F, H0 is rejected. Then the problem is to know, which of the treatment means are significantly different. For this, we calculate critical, difference (CD), CD = SED x t – table value for error d.f. at 5% LOS., where, , SED = Standard Error of Difference between the Treatments., , SED =, , 2 EMS, r, , SED =, , 1 1, EMS    , where i = 1,2,…..,k and j = 1,2,…..,k (unequal No. replications), r r , j ,  i, , (equal No. of replications), where r is number of replications, , The treatment means are arranged first in descending order of magnitude. If the difference, between the two treatment means is less than CD value, it will he declared as non, significant otherwise significant, Advantages and disadvantages of CRD:, 1. This design is most commonly used in laboratory experiments such as in Ag., Chemistry, plant pathology, and animal experiments where the experimental, material is expected to be homogeneous., 2. This design is useful in pot cultural experiments where the same type of soil is, usually used. However, in greenhouse experiments care has to be taken with, regard to sunshade, accessibility of air along and across the bench before, conducting the experiment., , BS Publication

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71, 3. Any number of replications and treatments can be used., , The number of, , replications may vary from treatment to treatment., 4. The analysis remains simple even if information on some units are missing, 5. This design provides maximum number of degrees of freedom for the estimation of, error than the other designs, 6. The only draw back with this design is that when the experimental material is, heterogeneous, the experimental error would be inflated and consequently the, treatments are less precisely compared. The only way to keep the experimental, error under control is to increase the number of replications thereby increasing the, degrees of freedom for error., Applications: 1. CRD is most useful in laboratory technique and methodological studies., Ex: in physics, chemistry, in chemical and biological experiments, in some, greenhouse studies etc., 2. CRD is also recommended in situations where an appreciable fraction of, units is likely to be destroyed or fail to respond., Case i) CRD with equal repetitions:, Example: In order to find out the yielding abilities of five varieties of sesamum an, experiment was conducted in the greenhouse using a completely randomized design, with four pots per variety. Analyze the data and state your conclusions, Seed yield of sesamum, g/pot, Varieties, 1, , 2, , 3, , 4, , 5, , 8, , 10, , 18, , 12, , 8, , 8, , 12, , 17, , 10, , 11, , 6, , 13, , 13, , 15, , 9, , 10, , 9, , 16, , 11, , 8, , Sol: H0: There is no significant difference between effect varities., i.e. 1   2  3 =  4 =  5, Correction factor (CF) =, , (GT ) 2, N, , Total sum of squares (TSS) =, , {y112 + y122 + y132+ …………+ y452 } – CF, , BS Publication

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72, Variety Sum of squares (VSS) =, , ( v1 ) 2  ( v2 ) 2  ( v3 ) 2  ( v4 ) 2  ( v5 ) 2, r, , - CF, , Error Sum of Square (ESS) = TSS- VSS, Varieties, , Variety, , 1, , 2, , 3, , 4, , 5, , 8, , 10, , 18, , 12, , 8, , 8, , 12, , 17, , 10, , 11, , 6, , 13, , 13, , 15, , 9, , 10, , 9, , 16, , 11, , 8, , 32, , 44, , 64, , 48, , 36, , GT =, , totals, , 224, , Means, , 8, , 11, , CF =, , 16, , 12, , 9, , (224) 2, = 2508.8, 20, , TSS = {(8)2  (10)2  (18)2  ..................  (16)2  (11)2  (8)2  2508.8  207.2, VSS =, , (32)2  (44)2  (64)2  (48)2  (36)2, - 2508.8 = 155.2, 4, , ESS= TSS-VSS = 207.2-155.2 = 52.0, ANOVA TABLE, Sources of, , d.f., , S.S., , M.S., , F-cal value, , F- table value, , Varieties, , 5-1 =4, , 155.20, , 38.80, , 11.19, , F0.05(4,15) =3.06, , Error, , 19-4=15, , 52.0, , 3.47, , variation, , Total, , 20-1=19, , 207.20, , Calculated value of F > Table value of F at 5%LOS, H0 is rejected and hence we conclude, that there is significant difference between variety means., Critical difference (CD): SED x t0.05 for error d.f., Where SED =, , 2 EMS, r, , t-table value for 15 d.f at 5%LOS is 2.13, CD =, , 2 * 3.47, x 2.13 = 2.80, 4, , BS Publication

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73, The varieties means are arranging descending order of magnitude. If the difference, between the verities mean is less than CD value, it will be declared as non significant, otherwise significant, V3, 16, , V4, V2, V5, V1, 12, 11, 9, 8, ________, ________, ________, , The varieties which do not differ significantly have been underlined by a bar., , Coefficient of variation (CV) = Coefficient of variation =, , EMS, x100, X, , Where X = Grand mean, Coefficient of variation =, , 3.47, x100  = 16.6%, 11.2, , Conclusion: Lesser CV% indicates more consistency in the data, Case –ii) CRD with unequal repetitions:, Example: A Completely Randomized Design was conducted with the three treatments A, B and C where treatment A is replicated 6 times and B is replicated 4 times and C is, replicated 5 times. Analyze the data and state your conclusions., A, 16.5, 17.0, 16.0, 12.0, 18.0, 14.0, , B, 15.0, 13.8, 14.0, 17.9, -, , C, 18.2, 24.3, 25.0, 18.9, 21.0, -, , Sol: Null hypothesis H0: There is no significant difference between the effect treatments, i.e. 1   2  3, First treatment A is replicated r1 = 6 times, Second treatment B is replicated r2 = 4 times, Third treatment C is replicated r3 = 5 times, Correction factor (CF) =, , (GT ) 2, N, , Total sum of squares (TSS) = {y112 + y122 + y132+ …………+ y352 } – CF, , BS Publication

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74, , Treatment of Sum of square (Tr.S.S.) =, , ( A) 2, r1, , , , ( B) 2, r2, , , , ( c ) 2, r3, , - CF, , Error Sum of square(ESS) = TSS-Tr.S.S., Standard error of difference = SED =, , 1 1, EMS   , r r , j ,  i, , Where i = 1, 2, ….., k and j = 1,2,….., r, Coefficient of variation =, , 2 EMS, X100;, X, , where X = Grand mean, , Treatments, , Totals, , Means, , A, , 16.5, , 17.0, , 16.0, , 12.0, , 18.0, , 14.0, , ΣA=93.5, , 15.58, , B, , 15.0, , 13.8, , 14.0, , 17.9, , -, , -, , ΣB=60.7, , 15.18, , C, , 18.2, , 24.3, , 25.0, , 18.9, , 21.0, , -, , ΣC=107.4 21.48, GT=261.6, , CF =, , (261.6) 2, = 4562.30, 15, , TSS = (16.5)2 + (17.0)2 + (16.0)2 + (12.0)2 + (18.0)2 + (14.0)2 + (15.0)2 + (13.8)2, + (14.0)2 + (17.9)2 + (18.2)2 + (24.3)2 + (25.0)2 + (18.9)2 + (21.0)2 – 4562.30, = 4758.04 – 4562.30, = 195.74, , (93.5) 2 (60.7) 2 (107.4) 2, , , Tr.S.S. =, - 4562.30, 6, 4, 5, = 4685.11 – 4562.30, = 122.81, ESS, , = TSS-Tr.S.S., = 195.74 – 122.81, = 72.93, AN0VA TABLE, , Sources, , d.f, , S.S., , M.S., , F-cal. Value, , F- table Value, , Treatments, , 3-1=2, , 122.81, , 61.405, , 10.10, , F0.05(2,12) = 3.89, , Error, , 12, , 72.93, , 6.0775, , Total, , 15-1=14, , 195.74, , Calculated value F(Tr) > Table value of F, H0 is rejected and hence we conclude that there, is significant difference between treatment means., , BS Publication

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75, In CRD with unequal number of replications, we have to calculate CD for each pair of, treatment means. As there are 3 treatments, we have to calculate 3C2 = 3 CD values., , 1 1, CD = SEDxt0.05 d.f., EMS   , r r , i, j, , , ----------------------------------------------------------------------------------------------------------Treatment pair, , SED =, , AB, , 1 1, 6.08   = 1.59, 6 4, , =1.59*2.18=3.47, , AC, , 1 1, 6.08   = 1.49, 6 5, , =1.49*2.18=3.25, , 1 1, 6.08   = 1.65,  4 5, , =1.65*2.18=3.61, , BC, Bar Notation:, , The treatment means are arranged according to their ranks, T3, 21.48, , T1, T2, 15.58, 15.18, _________, , i) Those pairs not scored are significant, ii) Those pairs under scored are non-significant, Coefficient of variation =, , EMS, x100, X, where x = Grand mean, , =, , 6.08, x100 =14.14%, 17.44, , Among three treatments, the third treatment i.e. T3 is found to be superior one, RANDOMIZED BLOCK DESIGN (RBD), We have seen that in a completely randomized design no local control measure, was adopted excepting that the experimental units should be homogeneous. Usually,, when experiments require a large number of experimental units, completely randomized, designs cannot ensure precision of the estimates of treatment effects., In agricultural field experiments, usually the experimental materials are not, homogeneous., , In such situations the principle of local control is adopted and the, , experimental material is grouped into homogeneous sub groups., , BS Publication, , The subgroup is

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76, commonly termed as block. Since each block will consist the entire set of treatments a, block is equivalent to a replication., The blocks are formed with units having common characteristics which may, influence the response under study. In agricultural field experiments the soil fertility is an, important character that influences the crop responses. The uniformity trial is used to, identify the soil fertility of a field. If the fertility gradient is found to run in one direction, (say from north to south) then the blocks are formed in the opposite direction (from east to, west)., If the number of experimental units within each group is same as the number of, treatments and if every treatment appears precisely once in each group, then such an, arrangement is called a randomized block design., Layout: Let us consider 5 treatments A, B, C, D and E each replicated 4 times. We divide, the whole experimental area into 4 relatively homogeneous blocks and each block into 5, plots. Treatments are then allocated at random to the plots of a block, fresh randomization, being done for each block. A particular layout as follows., Block-1, , A, , E, , B, , D, , C, , Block-2, , E, , D, , C, , B, , A, , Block-3, , C, , B, , A, , E, , D, , Block-4, , A, , D, , E, , C, , B, , Let us select one digit random numbers in the order of their occurrence in the table leaving, 0 and greater than 5. suppose we get random numbers from 1 to 5 as: 1,3,5,4,2. So in the, First block we allocate treatment A to the 1st plot and B to 3rd plot and so on., Statistical Analysis: The results from RBD can be arranged in two way table according to, the replications (blocks) and treatments; there will be `rk‟ observations in total. The data, can be arranged in the following table., , BS Publication

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78, 2, , 2, , 2, , ( B )  ( B2 )  ...........  ( Br ), 3) Block Sum of Squares (BSS) = 1,  CF, k, r, , B, j 1, , =, , 2, j, , k, ,  CF, , = {y112 + y122 + y132 + ………+ykr2} – CF, , 4) Total Sum of Squares (TSS), , =, , k, , r, ,  y, i 1 j 1, , 5) Error Sum of Square (ESS), , 2, ij, ,  CF, , = TSS -Tr.S.S .- SSB, , ANOVA TABLE, Sources of, , D.F, , S.S, , M.S., , F-Cal.Value, , variation, Treatments, , At 5% LOS, k-1, , Blocks, (Replications), , F-table value, , Tr.S.S., , TMS =, , BSS, , BMS=, , r-1, , Error, , (r-1)(k-1), , Total, , rk-1, , ESS, , EMS =, , Tr.S .S ., k 1, , Ft =, , TMS, EMS, , F[k-1,{(r-1), , BSS, r 1, , Fb=, , BMS, EMS, , F[r-1, {(r-1), , (k-1)}], (k-1)}], , ESS, (r  1)(k  1), , TSS, , If the calculated value of F (Treatments) < table value of F, we accept H0, and hence we, may conclude that there is no significant difference between the treatment means., If calculated value of F (Treatments) > table value of F, we reject H0 and hence we may, conclude that there is significant difference between the treatment means., If the treatments are significantly different, the comparison of the treatments is carried out, on the basis of Critical Difference (C.D.)., , BS Publication

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79, C.D. = SED(Tr) x t(r-1)(k-1) at  level of significance, , 2 xEMS, r, i), F (Blocks) should be not significant, if the planning of experiment is, well manner., ii), Desirable C.V. (%) in field experiment and lab experiment., Advantages and disadvantages of RBD:, Where SED =, , 1. The principle advantage of RBD is that it increases the precision of the experiment., This is due to the reduction of experimental error by adoption of local control., 2. The amount of information obtained in RBD is more as compared to CRD. Hence,, RBD is more efficient than CRD., 3. Flexibility is another advantage of RBD. Any number of replications can be, included in RBD. If large number of homogeneous units are available, large, number of treatments can be included in this design., 4. Since the layout of RBD involves equal replication of treatments, statistical, analysis, , is simple. Even when some observations are missing of certain, , treatments, the data can be analysed by the use of missing plot technique., 5. When the number of treatments is increased, the block size will increase. If the, block size is large it may be difficult to maintain homogeneity within blocks., Consequently, the experimental error will be increased. Hence, RBD may not be, suitable for large number of treatments. But for this disadvantage, the RBD is a, versatile design. It is the most frequently used design in agricultural experiments., 6. The optimum blocks size in field experiments is 21 plots. i.e. we can not compare, treatments which are > 21 in RBD to preserve homogeneity of plots, within a, block., Example: The yields of 6 varieties of a crop in lbs., along with the plan of the experiment,, are given below. The number of blocks is 5, plot of size is 1/20 acre and the varieties have, been represented by A, B, C, D and E and analyze the data and state your conclusions, B-I, B-II, B-III, B-IV, B-V, , B, 12, E, 23, A, 28, F, 75, D, 17, , E, 26, C, 16, B, 9, D, 20, F, 70, , D, 10, F, 56, E, 35, E, 30, A, 20, , C, 15, A, 30, F, 64, C, 14, C, 12, , BS Publication, , A, 26, D, 20, D, 23, B, 7, B, 9, , F, 62, B, 10, C, 14, A, 23, E, 28

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80, Solution:, Null hypothesis H01: There is no significant difference between variety means, , 1   2  3   4 =  5   6, H02: There is no significant difference between block means, , 1  2  3  4  5, (G.T ) 2, =, rk, , Correction factor, , Variety Sum of Squares due to varieties (VSS) =, , v, , i, , r, , b, , 2, ,  CF, , 2, j, ,  CF, , Block Sum of square(BSS), , =, , Total sum of squares (TSS), , = yij2 - CF, , Error Sum of Square (ESS), , = TSS- VSS- BSS, , k, , First rearrange the given data, Blocks, , Varieties, , Block totals, , Means, , A, , B, , C, , D, , E, , F, , B1, , 26, , 12, , 15, , 10, , 26, , 62, , ΣB1 = 151, , 25.17, , B2, , 30, , 10, , 16, , 20, , 23, , 56, , ΣB2 =155, , 25.83, , B3, , 28, , 9, , 14, , 23, , 35, , 64, , ΣB3 = 173, , 28.83, , B4, , 23, , 7, , 14, , 20, , 30, , 75, , ΣB4 = 169, , 28.17, , B5, , 20, , 9, , 12, , 17, , 28, , 70, , ΣB5 = 156, , 26.00, , Variety, , ΣA =, , ΣB =, , ΣC =, , ΣD =, , ΣE =, , ΣF =, , GT = 804, , -, , totals, , 127, , 47, , 71, , 90, , 142, , 327, , Means, , 25.4, , 9.4, , 14.2, , 18, , 28.4, , 65.4, , -, , -, , CF, , (804) 2, =, = 21547.2, 30, , VSS, , (127) 2  (47) 2  (71) 2  (90) 2  (142) 2  (327) 2,  21547.2, =, 5, = 31714.4 – 21547.2, , = 10167.2, , (151)2  (155)2  (173)2  (169)2  (156)2,  21547.2, BSS =, 6, = 21608.67 – 21547.2 = 61.47, TSS = (12)2 + (26)2 + ( 10)2 + ( 15)2 + + ………… + (12)2 + (9)2 + (28)2 - 21547.2, , BS Publication

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81, = 32194 – 21547.2, = 10646.8, ESS, , = TSS – BSS – Tr.S.S., = 10646.8 – 61.47 – 10167.2, = 418.13, ANOVA TABLE, , Sources of, variation, Blocks, , d.f, , S.S., , M.S., , F-cal. Value, , F- table Value, , 5-1=4, , 61.47, , 15.37, , 0.74, , F0.05 (4, 20) =2.87, , Varieties, , 6-1=5, , 10167.2, , 2033.44, , 97.25, , F0.05 (5, 20) = 2.71, , Error, , 29-4-5= 20, , 418.13, , 20.91, , Total, , 30-1-29, , 10646.8, , Calculated value of F (Treatments) > Table value of F, H0 is rejected and hence we, conclude that there is highly significant difference between variety means., Where SEm =, SED =, , EMS, =, r, , 20.91, = 2.04, 5, , 2 * SEm = 1.414 * 2.04 = 2.88, , Critical difference = SED x t-table value for error d.f. at 5% LOS,  CD, , = 2.88 * 2.09, = 6.04, , Coefficient of variation =, Bar Notation:, F, E, A, 65.4, 28.4 25.4, _________, , EMS, X100 =, X, D, 18.0, , C, 14.2, , 20.91, X100 = 17%, 26.8, , B, 9.40, , _______, ________, i), , Those pairs not scored are significant, , ii), , Those pairs underscored are non-significant, , Variety F gives significantly higher yield than all the other varieties; varieties D,C and B, are on par and gives significantly higher yield than variety A., , BS Publication

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82, LATIN SQUARE DESIGN (LSD), When the experimental material is divided into rows and columns and the, treatments are allocated such that each treatment occurs only once in a row and once in a, column, the design is known as latin square design. In this design eliminating fertility, variations consists in an experimental layout which will control variation in two, perpendicular directions, [Latin square designs are normally used in experiments where it is required to, remove the heterogeneity of experimental material in two directions. This design, requires that the number of replications (rows) equal the number of treatments]. In, LSD the number of rows and number of columns are equal. Hence the arrangement will, form a square., Layout of LSD: In this design the number of rows is equal to the number of columns and, it is equal to the number of treatments. Thus in case of `m‟ treatments, there have to be, mxm = m2 experimental units (plots) arranged in a square so that each row as well as each, column contain `m‟ plots. The `m‟ treatments are then allocated at random to these rows, and columns in such a way that every treatment occurs once and only once in each row, and each column such a layout is known as mxm L.S.D and is extensively used in, agricultural experiments. The minimum and maximum number of treatments required for, layout of LSD is 5 to 12., In LSD the treatments are usually denoted by alphabets like A,B,C…etc. For a, latin square with five treatments the arrangement may be as follows, Square –I, , Square – II……………………….. etc.,, , A B C D E, , A B C D E, , B A E C D, , B A D E C, , C D A E B, , C E A B D, , D E B A C, , D C E A B, , E C D B A, , E D B C A, , Statistical analysis: the mathematical model for LSD is given by, Yijk =    i   j   k  ijk, , ( i = j = k = 1,2,……....,m), , where Yijk denote the response from the unit (plot) in the ith row, jth column and, receiving the kth treatment, ,  = general mean effect, , BS Publication

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83, ,  i = ith row effect,  j = jth column effect, ,  k = kth treatment effect;, , ijk, , = error, , component, , we know that total variation = variation due to rows + variation due to columns +, Variation due to treatments + variation due to error, Null hypothesis (H0) = There is no significant difference between Rows, Columns and, Treatment effects., i.e., , i) H01 : 1   2  ....................   m, ii) H02: 1   2  ....................   m and, iii) H03:  1   2  ....................   m, , The steps in the analysis of the data for verifying the null hypothesis are:, Different component variations can be calculated as follows:, 1) C.F =, , (G.T ) 2, m2, m, , 2).Row Sum of Squares (RSS) =, , r, , 2, , i, , i 1, , -C.F, , m, m, , c, 3) Column Sum of Squares (CSS) =, , j 1, , 2, j, , - C.F, , m, m, , 4) Treatment Sum of Squares (Tr..S.S). =, , m, , 5) Total Sum of Squares (TSS) =, , m, , t, k 1, , 2, k, , - C.F, , m, m, ,  y, i 1 j 1 k 1, , 2, ijk, , - C.F, , 6) Error Sum of Squares (ESS) = TSS –RSS-CSS-Tr.S.S., , BS Publication

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84, ANOVA TABLE, Sources, , D.F, , S.S, , M.S., , F-cal. value, , F-table value at, 5%LOS, , Rows, , m-1, , RSS, , Columns, , m-1, , CSS, , Treatments, , m-1, , Tr.S.S., , Error, , (m-1)(m-2), , ESS, , RMS =, , RSS, m 1, , FR =, , RMS, EMS, , FR[m-1, {(m-1)(m-2)}], , CMS=, , CSS, m 1, , FC =, , CMS, EMS, , , , Tr.S .S ., m 1, , FT =, , TMS, EMS, , , , TMS =, , EMS =, Total, , m2-1, , TSS, , ESS, m 1, , -, , If calculate value of F(Tr) < table value of Fat 5%LOS, H0 is accepted and hence we may, conclude that there is no significance difference between treatment effects., If calculate value of F(Tr) > table value of F at 5%LOS, H0 is rejected and hence we may, conclude that there is significance difference between treatments effects., If the treatments are significantly different, the comparison of the treatments is carried out, on the basis of Critical Difference (C.D.)., C.D. = SED (Tr) x t(r-1)(k-1) at  level of significance, Where SED =, , 2 xEMS, , where m = number of rows, m, , If F is significant, the significance of any treatment contrast can be tested by using the CD, value., Advantages of Latin Square Design: 1) With two way grouping or stratification LSD, controls more of the variation than C.R.D. or R.B.D., 2) L.S.D. is an incomplete 3-way layout. Its advantage over complete 3-way layout is that, instead of m3 experimental units only m2 units are needed. Thus a 4x4 L.S.D. results in, saving of 64-16 = 48 observations over a complete 3-way layout., 3) The statistical analysis is simple though slightly complicated than for R.B.D. Even with, missing data the analysis remains relatively simple., 4) More than one factor can be investigated simultaneously., 5) The missing observations can be analysed by using missing plot technique., , BS Publication

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85, Example: An experiment on cotton was conducted to study the effect of foliar application, of urea in combinations with insecticidal sprays in the cotton yield. Five treatments were, tried in a 6x6 Latin Square Design. The layout plan and yield is given below:, T2, T4, T5, 4.9, 6.4, 3.3, T3, T1, T2, 9.3, 4.0, 6.2, T4, T3, T1, 7.0, 15.4, 6.5, T5, T2, T3, 5.3, 7.6, 13.2, T1, T5, T4, 9.3, 6.3, 11.8, Analyze the data and state your conclusions, , T1, 9.5, T5, 5.1, T2, 6.0, T4, 8.6, T3, 15.9, , T3, 11.8, T4, 5.4, T5, 4.6, T1, 4.9, T2, 7.6, , Sol:, Null hypothesis: Rows, Columns and Treatments effects are equal, or, H01 : 1   2  3   4  5 ;, H02 : 1   2   3   4   5, H03:  1   2   3   4   5, CF =, , (GT ) 2, m2, m, , RSS =, , R, , 2, , i, , i 1, ,  CF, , m, m, , C, j 1, , CSS =, , 2, j, ,  CF, , m, m, , T, k 1, , Tr.S.S =, , m, m, , TSS =, , 2, , k, , m, ,  CF, , m, ,  y, i 1 j 1 k 1, , ij, , 2, ,  CF, , ESS = TSS-RSS-CSS-Tr.S.S., , BS Publication

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87, TSS, , 2, , 2, , 2, , 2, , = Y11  Y12  Y13  ..................  Y55  CF, = (4.9)2 + (6.4)2 + (3.3)2 + ……..+ (15.9)2 + (7.6)2 – 1535.07, = 1821.07 – 1535.07 = 286.0, , ESS, , = TSS – RSS – CSS – Tr.S.S, =286.0 - 46.54 - 14.77 - 196.55 = 28.14, ANVOA table, , Sources, , d.f., , S.S., , M.S., , F cal . value, , F table value, , Rows, , 5-1 = 4, , 46.54, , 11.64, , 4.95, , F0.05(4, 12) = 3.26, , Columns, , 5-1 = 4, , 14.77, , 3.69, , 1.57, , F0.05(4, 12) = 3.26, , Treatments, , 5-1 = 4, , 196.55, , 49.14, , 20.91, , F0.05(4, 12) = 3.26, , Error, , 24 – 12 = 12, , 28.14, , 2.35, , Total, , 25-1 = 24, , 286.0, , Calculated value of F (treatments) > Table value of t at 5% LOS, H0 is rejected and hence, effect of foliar application of urea, have significant effect in the yield. To determine, which of the treatment pairs differ significantly we have to calculate the critical difference, (C.D.), , 2.35, = 0.69, 5, , SEm =, , EMS, =, m, , SED =, , 2 * SEm = 1.414 * 0.69 = 0.97, , CD = SED x t – table value for 12 d.f. at 5 % LOS, = 0.98 * 2.18, = 2.11, Coefficient of variation (CV) =, , 2.35, EMS, *100 =, X100 =19.55%, 7.84, X, , BS Publication

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88, Bar Notation:, , T3, 13.12, , i), , T5, T4, T1, T2, 7.84, 6.84, 6.46, 4.92, ______________, _______________, ________________, The pairs not scored are significant, , ii), , The pairs under scored are non significant, , From the bar chart it can be concluded that third treatment i.e. T3 significantly higher than, all the other treatments., References Books:, 1. Statistics for Agricultural Sciences, G. Nageswara Rao, Second Edition, BS, Publications, Hyderabad, 2. A Text book of Agricultural Statistics, R. Rangaswamy, New Age International (P), Limited, publishers, 3. Statistical Methods, K.P. Dhamu and K. Ramamoorthy, AGROBIOS (INDIA), 4. Fundamentals of Mathematical Statistics, S.C. Gupta and V.K. Kapoor, Sultan, Chand & Sons Educational Publications, 5. Fundamentals Applied Statistics, S.C. Gupta and V.K. Kapoor, Sultan Chand &, Sons Educational Publications, 6. Design Resources Server: www.iasri.res.in, , BS Publication