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Statistical Methods for Social Sciences 3(2-1) Lecture 1 Index Number: Concept of Index Number: According to Prof. Secrist, “Index Numbers are a series of numbers by which changes in the magnitudes of a phenomena are measured from time to time or from place to place”. Suppose it is required to measure the general changes in the price of certain commodities. Let the commodities be wheat, milk, eggs etc. We know that wheat is sold in Rs. per maund, milk in Rs. per litre, eggs in Rs. per dozen. Obviously these commodities are not comparable directly. In order to have comparison between them we compute the percentage changes in these commodities from one period to another on the basis of the prices of some selected date. These percentages are known as index number of prices of these commodities. The averages of these percentages would study the general changes in the level of prices. This average is known as wholesale price index number or general index number. Construction of Wholesale Price Index Numbers: The index number construction involves the following steps; 1. Purpose of Index Number 2. Selection of Commodities 3. Selection of Prices 4. Choice of Base Period 5. Choice of Average 6. Choice of Proper Weights 1. Purpose of Index Number: Since all index numbers are not suitable for all purposes, therefore, it is necessary to ascertain the purpose of index number beforehand for which they are going to be constructed. For example, if it is required to study the effect of increased prices on the labour class, cost of living index numbers should be constructed and not the wholesale index numbers which study the variations in general. 2. Selection of Commodities: It is not possible to include all commodities in the construction of index numbers which are brought and sold in the market due to financial and other difficulties; therefore, it is necessary to include only those commodities which are most commonly used for that class of people for whom the index numbers are going to be constructed. The selected commodities should suit the tastes, habits, customs or requirements of that class of people. As regards the number of commodities to be included there is no hard and fast rule for this purpose. Normally the larger the number of commodities lesser will be the chance of error in the average obtained. Considering the economy, and accuracy, the number of commodities should be atleast 23 for sensitive index numbers. The Pakistan wholesale price index number of Board of Economic Enquiry, Punjab Lahore includes 39 commodities. The British Board of Trade wholesale price index number includes 200 commodities. 3. Selection of Prices: After selecting the commodities, price quotation to be taken from the prominent business houses, standard journals or magazines from different places at the same time. The price should not be quoted as so many units per rupee but should be quoted as so many rupees per unit. It is advisable to take into account the wholesale prices rather than retail prices. The price quotations should be taken daily, weekly, fortnightly etc depending upon the purpose of index numbers. 4. Choice of Base Period: When the prices have been collected, the next step is to reduce them into percentages or relatives by selecting a suitable base. It may be either fixed base method or chain base method. i. Fixed Based Method: In this method the base period is fixed and the prices of subsequent years are expressed as relatives of the prices of the base year. Method of calculating the price relatives by the Fixed Base Method: In order to calculate the price relatives by the fixed based method, the price of each year is divided by the price of base year and this ratio is multiplied by 100. Price relative for the current year = (Price for the current year/Price for the base year) X 100 Symbolically it can be written as; Y1 = (P1/Po) x 100 Y2 = (P2/Po) x 100 Y3= (P3/Po) x 100 Yn = (Pn/ Po) x 100 Where Po = Price for the base year P1 = Price for the current year Example 1. Compute the Index numbers by taking 1957 as the base year. Years 1955 1956 1957 1958 1959 1960 1961 Price of wheat 14, 15, 16, 17, 18, 19, 20 Solution: Computation of Index Numbers Year 1955 1956 1957 1958 1959 1960 1961 Price 14 15 16 17 18 19 20 Index Number with 1957 = 100 14/16 x 100 = 87.50 15/16 x 100 = 93.75 16/16 x 100 = 100 17/16 x 100 = 106.25 18/16 x 100 = 112.5 19/16 x 100 = 118.7 20/16 x 100 = 125.0 ii. Chain Base Method: According to this method, the base is not fixed but it changes from year to year. Here the price of previous year is taken as the base period and thus the relatives are computed. If it is required to compute the price relatives from 1955 to 1961, the price of 1955 is taken as 100. Then the price of 1955 is taken as base for the year 1957 and so on. The price relatives computed by this method are known as Link Relatives. Method of Calculating the Link Relatives by the Chain Base Method. In order to calculate link relatives by the chain base method, the price of the current year is divided by the price of preceding year and this ratio is multiplied by 100. Link Relative for the current year = (Price in the current year/ price in the preceding year) x 100 Symbolically it can be written as; L1 = P1/Po x 100 L2 = P2/Po x 100 L3 = P3/Po x 100 Ln = Pn/Po x 100 Where Po = Price in the first year P1 = Price in the 2nd year P3 = Price in the 3rd year and so on. Example 2. Use the following data of industrial production in Pakistan to compare the annual fluctuations in Pakistan Industrial activity by the chain base method. Index Number of Industrial Production in Pakistan Year 1956 1957 1958 1959 1960 1961 1962 1963 1964 Index Number 120 122 116 120 120 137 136 149 156 Solution: In order to construct the index numbers of chain base method, we take the relative for 1956 =100 and then compute the other index. Base year 1956 =100 1956 1957 1958 1959 1960 1961 1962 1963 1964 Index Number 100 =100 122/120 x 100 = 101.66 116/122 x 100 = 95.08 120/116 x 100 = 103.4 120/120 x 100 =100.0 137/120 x 100 = 114.16 136/137 x 100 = 99.27 149/136 x 100 = 109.6 156/149 x 100 = 104.7 5. Choice of Average: After computing the link relatives, their average is taken to get the required index number. Theoretically any average (such as mean, median, mode, G.Mean, H. Mean) can be taken in the construction of the index number but practically mean and geometrical means are suitable because of mean of relatives is reversible and geometrical means of prices is reversible. Moreover, mean gives heavy weight to commodities of high prices and light to those of low prices. The geometric mean gives more weight to small items and less weight to large items. Example 3. From the data given below, compute the index number of prices, taking 1962 as base. Year 1962 1963 1964 1965 Commodity (Price in Rs.) Firewood Soft coke 3.25 2.50 3.44 2.80 3.50 2.00 3.75 2.50 Kerosene oil 0.22 0.22 0.25 0.25 Matches 0.06 0.06 0.06 0.06 Solution: Calculation of index Numbers Yea r Commodity (Price in Rs.) Firewood 196 2 100 196 3 3.44/3.25x100=1 06 196 4 3.50/3.25x100=1 08 196 5 3.75/3.25x100=1 15 Tota l Soft coke 2.50/2.5 0 x100 =100 2.80/2.5 0 x100=11 2 2.0/ 2.50 x 100 =80 2.50/2.5 0 x 100 =125 Kerosen e oil 100 Matche s 100 400 Index Number by Mea Media n n 100 100 100 100 418 105 108 114 100 402 101 104 114 100 454 114 108 Example 4. Construct Chain Indices for the following years, taking 1940 as the base. Year 1940 1941 1942 1943 1944 Commodity Wheat 2.8 3.4 3.6 4.0 4.2 Rice 10.5 10.8 10.6 11.0 11.5 Maize 2.7 3.2 3.5 3.8 4.0 Solution: Calculation of Chain Indices: Year Commodity Wheat 1940 100 Total Average Chain Indices Rice 100 Maize 100 100 100 100 1941 1942 1943 1944 3.4/2.8x100=121 3.6/3.4x100=106 4.0/3.6x100=111 4.2/4.0x100=105 103 98 104 105 119 109 109 105 114 104 108 105 114 104 108 105 100x114/100=114 114x104/100=118 118.6x108/100=128 128x105/100=134.5 6. Choice of Proper Weight: When the average of the relatives is taken we get the required index number. While taking the average we see that all the commodities are treated alike while in actual practice that some commodities are more important than others and as such they need weight in the construction of index numbers. Thus weights are assigned to the commodities depending upon their relative importance. Methods used in weighing the indices of prices: i. Weighted Aggregate Method: According to this method, the current year’s prices are multiplied by the base year quantities. The sum of the products so obtained is then divided by the sum of products of the base year’s prices and base year quantities. Symbolically it can be written as; Index Number for the current year = (Σ p1 qo/ Σ po qo) x 100 Where p1= Price for the current year po = Price for the base year qo = Quantity in the base year Example 3. On analyzing bills of certain food items consumed at a club for the years 1954 and 1955, the following tables are drawn up: Items Meat Fish Eggs Vegetables Fruit 1954 Prices 2.2 2.5 1.1 0.7 0.3 1955 Prices 2.0 2.8 1.3 0.7 0.2 Quantities 100 seers 30 seers 50 doz 100 seers 150 Nos Obtain an index by an appropriate method for indicating the relative change in prices for 1955 over 1954. Solution: Computation of weighted index number for 1955 by using aggregate expenditure method. Base 1955 =100 Items po Meat 2.2 Fish 2.5 Eggs 1.1 Vegetable 0.7 p1 2.0 2.8 1.3 0.7 qo 100 30 50 100 poqo 220 75 55 70 p1qo 200 84 65 70 Fruit 0.3 0.2 150 45 30 Total 465 449 Weighted Index Number = (Σ p1 qo/ Σ po qo) x 100 = 449/465 x 100 =96.56 Weight Average of Relative: In this method, the weights are the values in the base year are calculated on the basis of the aggregates expenditure of the commodities. In order to calculate the aggregate expenditure of the commodities in the base year, the quantities are multiplied with their respective prices. The sum of the products of the price relatives of the current year and the values of the base year is divided by the sum of the weights (values). The resulting figure is required index number for the current year. Symbolically it can be written as Index Number for the current year = Σ IV/ Σ V Where I = Price relative of the current year V = Value (weights) of the base year Example 4. Calculate the weighted index number of the data given in Example 3 by the method of weighted average of relatives. Soulation: Computation of weighted index number for 1955 by the method of weighted average relative Base year 1954=100 Items Prices po p1 Wights qo Meat Fish Eggs Vegetable Fruit Total 2.2 2.5 1.1 0.7 0.3 100 30 50 100 150 2.0 2.8 1.3 0.7 0.2 Values consumed in the base V= poqo 220 75 55 70 45 465 I=p1/qox100 IxV 99.9 112.0 118.0 110.0 66.7 19998 8400 6501 7000 3000.5 44899.5 Weighted index number for the current year (1955)= Σ IV/ Σ V= 44899.5/465 =96.6 Cost of living Index Number: Are especially designed to study the effect of changes in prices on the people as consumers. Uses of Index Numbers: Following are come of uses of Index Numbers 1. Index numbers are used in the department of Commerce, Meteriology, Labour and Industries etc. 2. The Insurance Companies use index numbers for determining the probable times of death or duration of life of those persons who are insured. 3. Index number works like a barometer showing fluctuations in daily life, cost of living, employment, public health etc. 4. Index numbers are useful in the study of changes in price levels over a period of time. 5. Index numbers are of great use in forecasting. There are many forecasting organizations which compile forecasting index numbers. Such index number works well during the period of mild prosperity and depression but fail during a severe depression. 6. Index numbers are of great help for studying physical changes over a period of time. For examples various types of index numbers were compiled such as index of industrial production, index of factory production etc, which study seasonal variations, seasonal trends etc. 7. Index numbers are of great help foe the purpose of comparison among different regions. For example, Government may be interested in compiling the cost of living index numbers of various regions within the country for the purpose of establishing an equitable standard of living. 8. Index numbers are widely used by the economists, social workers and businessmen in order to measure the changes i.e wages, prices, sales, stocks, production and cost of living. Lecture 2 Random Variable The outcome of an experiment need not be a number, for example, the outcome when a coin is tossed can be 'heads' or 'tails'. However, we often want to represent outcomes as numbers. A random variable is a function that associates a unique numerical value with every outcome of an experiment. The value of the random variable will vary from trial to trial as the experiment is repeated. There are two types of random variable - discrete and continuous. A random variable has either an associated probability distribution (discrete random variable) or probability density function (continuous random variable). Examples 1. A coin is tossed ten times. The random variable X is the number of tails that are noted. X can only take the values 0, 1, ..., 10, so X is a discrete random variable. 2. A light bulb is burned until it burns out. The random variable Y is its lifetime in hours. Y can take any positive real value, so Y is a continuous random variable. Expected Value The expected value (or population mean) of a random variable indicates its average or central value. Expected value gives a general impression of the behaviour of some random variable without giving full details of its probability distribution (if it is discrete) or its probability density function (if it is continuous). Two random variables with the same expected value can have very different distributions. There are other useful descriptive measures which affect the shape of the distribution, for example variance. The expected value of a random variable X is symbolised by E(X) or µ. If X is a discrete random variable with possible values x1, x2, x3, ..., xn, and p(xi) denotes P(X = xi), then the expected value of X is defined by: where the elements are summed over all values of the random variable X. If X is a continuous random variable with probability density function f(x), then the expected value of X is defined by: Example Discrete case : When a die is thrown, each of the possible faces 1, 2, 3, 4, 5, 6 (the xi's) has a probability of 1/6 (the p(xi)'s) of showing. The expected value of the face showing is therefore: µ = E(X) = (1 x 1/6) + (2 x 1/6) + (3 x 1/6) + (4 x 1/6) + (5 x 1/6) + (6 x 1/6) = 3.5 Notice that, in this case, E(X) is 3.5, which is not a possible value of X. Variance The (population) variance of a random variable is a non-negative number which gives an idea of how widely spread the values of the random variable are likely to be; the larger the variance, the more scattered the observations on average. Stating the variance gives an impression of how closely concentrated round the expected value the distribution is; it is a measure of the 'spread' of a distribution about its average value. Variance is symbolised by V(X) or Var(X) or The variance of the random variable X is defined to be: where E(X) is the expected value of the random variable X. Notes a. the larger the variance, the further that individual values of the random variable (observations) tend to be from the mean, on average; b. the smaller the variance, the closer that individual values of the random variable (observations) tend to be to the mean, on average; c. taking the square root of the variance gives the standard deviation, i.e.: d. the variance and standard deviation of a random variable are always non-negative. See also sample variance. Probability Distribution The probability distribution of a discrete random variable is a list of probabilities associated with each of its possible values. It is also sometimes called the probability function or the probability mass function. More formally, the probability distribution of a discrete random variable X is a function which gives the probability p(xi) that the random variable equals xi, for each value xi: p(xi) = P(X=xi) It satisfies the following conditions: a. b. Cumulative Distribution Function All random variables (discrete and continuous) have a cumulative distribution function. It is a function giving the probability that the random variable X is less than or equal to x, for every value x. Formally, the cumulative distribution function F(x) is defined to be: for For a discrete random variable, the cumulative distribution function is found by summing up the probabilities as in the example below. For a continuous random variable, the cumulative distribution function is the integral of its probability density function. Example Discrete case : Suppose a random variable X has the following probability distribution p(xi): xi 0 1 2 3 4 5 p(xi) 1/32 5/32 10/32 10/32 5/32 1/32 This is actually a binomial distribution: Bi(5, 0.5) or B(5, 0.5). The cumulative distribution function F(x) is then: xi 0 1 2 3 4 5 F(xi) 1/32 6/32 16/32 26/32 31/32 32/32 F(x) does not change at intermediate values. For example: F(1.3) = F(1) = 6/32 F(2.86) = F(2) = 16/32 Probability Density Function The probability density function of a continuous random variable is a function which can be integrated to obtain the probability that the random variable takes a value in a given interval. More formally, the probability density function, f(x), of a continuous random variable X is the derivative of the cumulative distribution function F(x): Since it follows that: If f(x) is a probability density function then it must obey two conditions: a. that the total probability for all possible values of the continuous random variable X is 1: b. that the probability density function can never be negative: f(x) > 0 for all x. Discrete Random Variable A discrete random variable is one which may take on only a countable number of distinct values such as 0, 1, 2, 3, 4, ... Discrete random variables are usually (but not necessarily) counts. If a random variable can take only a finite number of distinct values, then it must be discrete. Examples of discrete random variables include the number of children in a family, the Friday night attendance at a cinema, the number of patients in a doctor's surgery, the number of defective light bulbs in a box of ten. Continuous Random Variable A continuous random variable is one which takes an infinite number of possible values. Continuous random variables are usually measurements. Examples include height, weight, the amount of sugar in an orange, the time required to run a mile. Independent Random Variables Two random variables X and Y say, are said to be independent if and only if the value of X has no influence on the value of Y and vice versa. The cumulative distribution functions of two independent random variables X and Y are related by F(x,y) = G(x).H(y) where G(x) and H(y) are the marginal distribution functions of X and Y for all pairs (x,y). Knowledge of the value of X does not effect the probability distribution of Y and vice versa. Thus there is no relationship between the values of independent random variables. For continuous independent random variables, their probability density functions are related by f(x,y) = g(x).h(y) where g(x) and h(y) are the marginal density functions of the random variables X and Y respectively, for all pairs (x,y). For discrete independent random variables, their probabilities are related by P(X = xi ; Y = yj) = P(X = xi).P(Y=yj) for each pair (xi,yj). Probability-Probability (P-P) Plot A probability-probability (P-P) plot is used to see if a given set of data follows some specified distribution. It should be approximately linear if the specified distribution is the correct model. The probability-probability (P-P) plot is constructed using the theoretical cumulative distribution function, F(x), of the specified model. The values in the sample of data, in order from smallest to largest, are denoted x(1), x(2), ..., x(n). For i = 1, 2, ....., n, F(x(i)) is plotted against (i-0.5)/n. Quantile-Quantile (QQ) Plot A quantile-quantile (Q-Q) plot is used to see if a given set of data follows some specified distribution. It should be approximately linear if the specified distribution is the correct model. The quantile-quantile (Q-Q) plot is constructed using the theoretical cumulative distribution function, F(x), of the specified model. The values in the sample of data, in order from smallest to largest, are denoted x(1), x(2), ..., x(n). For i = 1, 2, ....., n, x(i) is plotted against F-1((i-0.5)/n). Normal Distribution Normal distributions model (some) continuous random variables. Strictly, a Normal random variable should be capable of assuming any value on the real line, though this requirement is often waived in practice. For example, height at a given age for a given gender in a given racial group is adequately described by a Normal random variable even though heights must be positive. A continuous random variable X, taking all real values in the range is said to follow a Normal distribution with parameters µ and if it has probability density function We write This probability density function (p.d.f.) is a symmetrical, bell-shaped curve, centered at its expected value µ. The variance is . Many distributions arising in practice can be approximated by a Normal distribution. Other random variables may be transformed to normality. The simplest case of the normal distribution, known as the Standard Normal Distribution, has expected value zero and variance one. This is written as N(0,1). Examples Poisson Distribution Poisson distributions model (some) discrete random variables. Typically, a Poisson random variable is a count of the number of events that occur in a certain time interval or spatial area. For example, the number of cars passing a fixed point in a 5 minute interval, or the number of calls received by a switchboard during a given period of time. A discrete random variable X is said to follow a Poisson distribution with parameter m, written X ~ Po(m), if it has probability distribution where x = 0, 1, 2, ..., n m > 0. The following requirements must be met: a. the length of the observation period is fixed in advance; b. the events occur at a constant average rate; c. the number of events occurring in disjoint intervals are statistically independent. The Poisson distribution has expected value E(X) = m and variance V(X) = m; i.e. E(X) = V(X) = m. The Poisson distribution can sometimes be used to approximate the Binomial distribution with parameters n and p. When the number of observations n is large, and the success probability p is small, the Bi(n,p) distribution approaches the Poisson distribution with the parameter given by m = np. This is useful since the computations involved in calculating binomial probabilities are greatly reduced. Examples Binomial Distribution Binomial distributions model (some) discrete random variables. Typically, a binomial random variable is the number of successes in a series of trials, for example, the number of 'heads' occurring when a coin is tossed 50 times. A discrete random variable X is said to follow a Binomial distribution with parameters n and p, written X ~ Bi(n,p) or X ~ B(n,p), if it has probability distribution where x = 0, 1, 2, ......., n n = 1, 2, 3, ....... p = success probability; 0 < p < 1 The trials must meet the following requirements: a. b. c. d. the total number of trials is fixed in advance; there are just two outcomes of each trial; success and failure; the outcomes of all the trials are statistically independent; all the trials have the same probability of success. The Binomial distribution has expected value E(X) = np and variance V(X) = np(1-p). Examples Geometric Distribution Geometric distributions model (some) discrete random variables. Typically, a Geometric random variable is the number of trials required to obtain the first failure, for example, the number of tosses of a coin until the first 'tail' is obtained, or a process where components from a production line are tested, in turn, until the first defective item is found. A discrete random variable X is said to follow a Geometric distribution with parameter p, written X ~ Ge(p), if it has probability distribution P(X=x) = px-1(1-p)x where x = 1, 2, 3, ... p = success probability; 0 < p < 1 The trials must meet the following requirements: a. b. c. d. the total number of trials is potentially infinite; there are just two outcomes of each trial; success and failure; the outcomes of all the trials are statistically independent; all the trials have the same probability of success. The Geometric distribution has expected value E(X)= 1/(1-p) and variance V(X)=p/{(1p)2}. The Geometric distribution is related to the Binomial distribution in that both are based on independent trials in which the probability of success is constant and equal to p. However, a Geometric random variable is the number of trials until the first failure, whereas a Binomial random variable is the number of successes in n trials. Examples Uniform Distribution Uniform distributions model (some) continuous random variables and (some) discrete random variables. The values of a uniform random variable are uniformly distributed over an interval. For example, if buses arrive at a given bus stop every 15 minutes, and you arrive at the bus stop at a random time, the time you wait for the next bus to arrive could be described by a uniform distribution over the interval from 0 to 15. A discrete random variable X is said to follow a Uniform distribution with parameters a and b, written X ~ Un(a,b), if it has probability distribution P(X=x) = 1/(b-a) where x = 1, 2, 3, ......., n. A discrete uniform distribution has equal probability at each of its n values. A continuous random variable X is said to follow a Uniform distribution with parameters a and b, written X ~ Un(a,b), if its probability density function is constant within a finite interval [a,b], and zero outside this interval (with a less than or equal to b). The Uniform distribution has expected value E(X)=(a+b)/2 and variance {(b-a)2}/12. Example Central Limit Theorem The Central Limit Theorem states that whenever a random sample of size n is taken from any distribution with mean µ and variance , then the sample mean will be approximately normally distributed with mean µ and variance /n. The larger the value of the sample size n, the better the approximation to the normal. This is very useful when it comes to inference. For example, it allows us (if the sample size is fairly large) to use hypothesis tests which assume normality even if our data appear non-normal. This is because the tests use the sample mean , which the Central Limit Theorem tells us will be approximately normally distributed. Statistical Inference and Estimation Statistical Inference: The process of drawing inference about a population on the basis of information contained in a sample taken from the population is called statistical inference. Statistical inference is traditionally divided into two main branches: Estimation of parameters and Testing of Hypothesis. Estimation of Parameters: It is a procedure by which we obtain an estimate of the true but unknown value of a population parameter by using the sample observations X1, X2,…..,Xn. For example we may estimate the mean and the variance of population by computing the mean and variance of the sample drawn from the population. Testing of Hypothesis: It is procedure which enables us to decide on the basis of information obtained from sampling whether to accept or reject any specified statement. Estimates and Estimators: An estimate is a numerical value of the unknown parameter obtained by applying a rule or a formula called as estimator, to a sample X1, X2,…, Xn of size n, taken from population. For example, if X1, X2,… Xn is a random sample of size n from a population with mean x , then x = 1/n ∑Xi is an estimator of μ and x , the value of x is an estimate of μ. Kinds of Estimates There are two kinds of estimates as; 1. Point Estimates 2. Interval Estimates 1. Point Estimates: When an estimate of unknown population parameter is expressed by a single value, it is called point estimate. 2. Interval Estimates: An estimate expressed by a range of values within which true value of the population parameters is believed to lie, is referred to as an interval estimate. Suppose we wish to estimate the average height of very large group of student on the basis a sample. If we find sample average height to be 64', then 64' is a point estimate of the unknown population mean. If on the other hand, we state that the true average height is a value between 62' and 66' is an interval estimate. Example: A random sample of n=6 has the elements of 6, 10, 13, 14, 18 and 20. Compute a point estimate of i) The population mean ii) The Population Standard Deviation iii) The Standard Error of Mean. i) The sample mean is Then x = 1/n ∑Xi = (6+10+!3+14+18+20)/6 = 81/6=13.5 Thus point estimate of population mean μ is 13.5 and x is the estimator. ii) The sample standard deviation is S = √ 1/n ∑ (Xi - x )2 = (6-13.5)2 + (10-13.5)2 +(13-13.5)2 +(14-13.5)2 +(18-13.5)2 + (2013.5)2 /6 = 4.68 Thus the point estimate of the population standard deviation σ is 4.68 and S is the estimator iii) When the sample size is less than 5% of the population size, the standard error of mean is S x = S/√n = 4.68/√6 = 1.91 Hence S x is the estimator for σ x and 1.91 is the point estimate of the standard error of mean. Testing of Hypothesis and Significance: Hypothesis Testing: Hypothesis testing is a procedure which enables us to decide on the basis of information obtained from sample data whether to accept or reject a statement. For example, Agriculturists may hypothesize that these farmers are aware about new technology will be the most productive. With statistical techniques, we are able to decide whether or not our theoretical hypothesis is confirmed by the empirical evidence. Null Hypothesis: The null hypothesis (written as HO:) is a statement written in such a way that there is no difference between two items. When we test the null hypothesis, we will determine a P value, which provides a numerical value for the likelihood that the null hypothesis is true. Alternate Hypothesis: If it is unlikely that the null hypothesis is true, then we will reject our null hypothesis in favour of an alternate hypothesis (written as HA), and this states that the two items are not equal. Simple and Composite Hypothesis: In statistical hypothesis completely specifies the distribution is called Simple Hypothesis, otherwise it is called as Composite Hypothesis. Descriptive Statistics: Figure associated with the number of birth, number of employees and other data that average person calls. Characteristic: To describe characteristics of population and sample. Inferential Statistics: It is used to make an inference to a whole population from a sample. For example, when firm tests markets a new product in D.I.Khan, it wishes to make an inference from these sample markets to predict what will happen throughout Pakistan. Characteristic: To generalize from sample to the population. Determining Sample Size: Three factors are required to specify sample size. 1. The variance or heterogeneity of population i.e Standard deviation(S). Only small sample is required if the population is homogeneous. For example, predicting the average age of college students require a smaller sample size than predicting the average age of people visiting the zoo on a given Sunday afternoon. 2. The magnitude of acceptable error i.e E. It indicates how precise the estimate must be. 3. The confidence level i.e Z Sample size (n) = (ZS/E)2 Suppose a survey researcher studying expenditure on major crop wishes to have a 95 % confidence interval level (Table value of which i.e Z = 1.96) and a range of error (E) of less than Rs. 2. The estimate of standard deviation is Rs. 29. n = (ZS/E)2 = [ (1.96 x 29)/ 2]2 = 808 If the range of error (E) is acceptable at Rs. 4, the sample size is reduced. n = (ZS/E)2 = [ (1.96 x 29)/ 4]2 = 202 Thus doubling the range of acceptance error reduces sample size to one quarter of its original size and vise versa. Large Error Small Sample Size Large Figure: 1 Sample Size and Error are inversely related. Confidence Interval: In statistical term, increasing the sample size, decreases the width of the confidence interval at a given confidence level. When the standard deviation of the population is unknown, a confidence interval is calculated by using the following formula X ± Z S/ √ n Standard Error of Mean is E E = Z S/ √n If n increases, E is reduced and vise versa. Level of Significance: Significance level of a test is the probability used as a standard for rejecting a null hypothesis Ho when Ho is assumed to be true. The widely used values of α is called level of significance i.e 1% (0.01), 5% (0.05) or 10% (0.10). Level of Confidence: When H0 is assumed to be true, this probability is equal to some small pre-assigned value usually denoted by α, the equality 1-α is called level of confidence. i.e 99% (0.99), 95% (0.95) or 90% (0.90). Rejection and Acceptance Region: The possible results of an experiment can be divided into two groups; A. Results appearing consistent with the hypothesis. B. Results leading us to reject the hypothesis. The group A is called acceptance region, while group B is called rejection region or critical region. The dividing line in between these two regions is called level of significance (α). All possible values which a test- statistics may assume can be divided into two mutually exclusive groups. One group (A) consisting of values which appear to be consistent with the null hypothesis, and the other (B) having values which are unlikely to occur if Ho is true. For example, If the calculated value of test statistics is higher than its table value, then reject H0 so it is called rejection region or critical region, otherwise accept Ho, then it is called acceptance region. Types of Errors: In theory of hypothesis, two types of errors are committed; A) we reject a hypothesis when it is infact true. B) we accept a hypothesis when it is actually false. The former type i.e rejection of H0, when it is true is called Type I Error and the latter type i.e the acceptance of H0, when is false is called Type II Error may be presented in the following table. True Situation H0 is true H0 is false Accept H0 Correct decision Wrong decision Type II Error Decision Reject H0 Wrong decision Type I Error Correct decision Test Statistics: A test statistics is a procedure or a method by which we verify an established hypothesis or it is function obtained from the sample data on which the decision of rejection or acceptance of H0 is based or a method provides a basis for testing a null hypothesis is known a test statistics. i.e t- test. Z-test, F-test, Chi square test, ANOVA etc. One tailed/ sided and two tailed/ sided: A test of any statistical hypothesis where the alternative hypothesis is one sided such as; H0 : µ = µo H1 : µ > µo or µ < µo The critical region for the H1 : µ > µo has entirely on the right tail of the distribution. α 1-α and the critical region for the H1 : µ < µo has entirely on the left tail of the distribution. α 1-α A test of any statistically hypothesis where alternative hypothesis H1 is two sided as; H0 : µ = µo H1 : µ ≠ µo These constitute on both sides/ tailed of the distribution. α/2 α/2 1- α General Procedure for Testing Hypothesis: The procedure for testing a hypothesis about a population parameter involves the following six steps, 1. State your problem & formulate an appropriate null hypothesis Ho with an alternative hypothesis H1, which to be accepted when Ho is rejected. 2. Decide upon a significance level, α of the test, which is probability of rejecting the null hypothesis if it is true. 3. Choose an appropriate test-statistics, determine & sketch the sampling distribution of the test-statistics, assuming Ho is true. 4. Determine the rejection or critical region in such a way that a probability of rejecting the null hypothesis Ho, if it is true, is equal to the significance level, α the location of the critical region depends upon the form of H1. The significance level will separate the acceptance region from the rejection region. 5. Compute the value of the test-statistics from the sample data in order to decide whether to accept or reject the null hypothesis Ho. 6. Formulate the decision rule as below. a) Reject the null hypothesis Ho, if the computed value of the test-statistics falls in the rejection region & conclude that H1 is true. b) Accept the null hypothesis Ho, otherwise when a hypothesis is rejected, we can give α measure of the strength of the rejection by giving the P-value, the smallest significance level at which the null hypothesis is being rejected. Example A random sample of n = 25 values gives x = 83 can this sample be regarded as drawn from normal population with mean μ = 80 & б =7 Solution i) We formulate our null & alternate hypothesis as Ho: μ = 80 and H1: μ ≠ 80 (two sided). ii) We set the significance level at α = 0.05 iii) Test-statistics is to be used is Z = (x-μ)/б/√n, which under the null hypothesis is a standard normal variable. iv) The critical region for α = 0.05 is for the sample, z ≥ 1.96. z ≥ 1.96 , the hypothesis will be rejected if, v) We calculate the value of Z from the sample data vi) Conclusion: Since our calculated value Z = 2.14 falls in the critical region, so we reject our null hypothesis Ho: μ = 80 & accept H1 : μ ≠ 80. We may conclude that the sample with x = 83 cannot be regarded as drawn from the population with μ = 80 Tests based on Normal Distribution: Two parameters are used in this distribution are µ (population mean) and δ2 (Population variance). Let (x1, x2,-------xn) be a sample from N ~ (µ - δ2) → (Normal Distribution). It is desired to test H0: µ = µo is some predetermined value of µ. Here two cases arise; Case-I δ2 is known: Where δ2 is known, then sample mean (X) is normally distributed with population mean (µ) and population variance (δ2/n), so Z = (X-µ)/ √ δ2/n = (X-µ)/ δ2/√n Where Z is standard normal variance (S) Z = (X-µ)/ δ2/√n (Where H0) Critical region always depends on H1 We know that H0 : µ = µo A. Either H1 : µ≠ µo (two tailed test ) or B1.H1 :µ >µo one tailed test (right hand side) or B2. H1 : µ < µo two tailed test (left hand side) In case│Z1│> Zα/2 Reject Ho otherwise accept Ho Example: A researcher worker is interested in testing a fertilizer effect on wheat production which has average production of 40 kg/acre with δ2 = 25 kgs. He selected at random 16 acres of land which were similar in all respects. The wheat was sown and fertilizers were applied. The yield of 16 plots was observed to be 40, 44, 43, 43, 41, 40, 41, 44, 42, 41, 42, 43, 46, 40, 38, 44. Test this claim that production/ yield of wheat will not be increased due to fertilizer effects. Answer: We formulate hypothesis as H0 : µ = µo (µo = 40 kgs) H1 :µ >µo (one sided right hand test) Level of significance is α is 0.05 Test Statistics to be used is Z = (X-µ)/ δ/√n With given values µo = 40 kgs X= ∑X/n= 672/16= 42, where δ2 =25 Now putting the values, we get Z = (42-40)/ 5/√16 =1.6 Critical region Z > Zα or Z > Z 0.05(n-1) or 1.6 > 1.645 Conclusion: Hence we accept H0, which means that there is no effect of fertilizer on the increase of wheat production. Critical values of Z in a form a table; Level of Significance Two Tailed Test One Tailed Test 0.10 ± 1.645(± Z α/2) ± 1.28(± Z α) 0.05 ± 1.96(± Z α/2) ± 1.645(± Z α) 0.01 ± 2.58(± Z α/2) ± 2.33(± Z α) Example: It is hypothesized that that average diameter of leaves of a certain tree is 20.4 mm with a standard deviation of 2.0 mm. check this supposition; we select a random sample of 16 leaves and found that this mean is 22 mm. Test whether the sample supports this hypothesis. Answer: We formulate our hypothesis as; H0 : µ = µo (µo = 20.4 mm) H1 :µ ≠µo (two side test)l Level of significance α = 0.05 While Test Statistics to be used is Z = (X-µ)/ δ/√n The known values are µo = 20.4 mm, δ = 2.0 mm, n= 16, X = 22.0 mm Hence by putting the values, Z = (22-20.4)/2/√16 = 3.2 Critical region is │Z1│> Zα/2 As 3.2 > 1.96 as Ho is rejected, so this sample does not supports this hypothesis. Note: In case 2, If the δ2 is unknown then the formula will be applied as Z = (X-µ)/ S/√n (under H0), t-Distribution: T-test is used to measure two means. When δ2 is unknown and n> 30, then for a large sample size, Z-test will replace δ2 by S. When n < 30 and population standard deviation is unknown, then t-test instead of Z-test will be used. T-test can be symbolically expressed as; t = (X-µo)/ δ√n with n-1 degree of freedom. Example: Ten students are chosen at random from a normal population and their heights are found to be in inches as 63, 63, 66, 67, 68, 69, 70, 70, 71, 71. In the light of above data, “Is the mean height in the population is 66 inches”?. Answer: We formulate our hypothesis as; H0 : µ = µo (µo = 66 inches) H1 : µ ≠ µo (two tailed test) Level of significance (α) is 0.05 While the test statistics is t = (X-µo)/ δ√n with n-1 df Computations X 63 63 66 67 68 69 70 70 71 71 Total Dx = X-PM (PM =68) -5 -5 -2 -1 0 1 2 2 3 3 ∑Dx = -2 Dx2 25 25 4 1 0 1 4 4 9 9 ∑Dx2 = 82 X = (P.M/1) + ∑DX/n = (68/1) + (-2/10) = 67.8 δ2 = (1/n-1) {∑Dx2 – (∑Dx)2/n} δ2 = 1/10-1 {82 – (-2)2/10 } = 9.066 δ = √9.066 = 3.011 Now putting the values in the formula as t = X - µo/ δ √ n = 67.6- 66/ 3.011 √ 10 with n-1 df t = 1.89 with 9 degree of freedom Here the critical region is │t│> t α/2 (n-1) │1.89│< t 0.025 (9) │1.89│< 2.26 Which proves that H0 is accepted, so we conclude that mean height of population is 66 inches. Testing the equality of two means: Testing equality of two means or the difference of two means, when the population variance are equal (δ1 2 = δ22), but unknown. Suppose we have X1, X2, -------Xn and Y1, Y2, --------Yn are the two independent random small samples with mean X and Y drawn from two normal population with population mean µ1 and µ2 with the same unknown population variance. We wish to test the hypothesis whether two population means are same. t = {(X-Y) – (µ1 - µ2)}/ √ (δ12/n1 + δ22/n1) Since population standard deviation δ is unknown, therefore we take sp as a pooled variance as; sp = √ {1/(n1+n2-2)} [{∑X2 – (∑X)2/n1} + {∑Y2 – (∑Y)2/n2}] Example: In the test, two groups obtained marked as X 9, 11, 13, 11, 15, 9, 12, 14 Y 10, 12, 10, 14, 9, 8, 10 Is there is any difference in their means of their population. Answer: Formulate hypothesis as; H0: µ1 -µ2 = 0 H1 : µ1 -µ2 ≠ 0 Level of significance α is 0.05 Test statistics is t = (X – Y)/ sp√{1/n1 -1/n2} with n1+n2-2 degree of freedom Computations X Y X2 Y2 9 10 81 100 11 12 121 144 13 10 169 100 11 14 121 196 15 9 225 84 9 8 81 64 12 10 144 100 14 196 ∑X = 94 ∑Y= 73 ∑X2 = 1138 ∑Y2 = 785 By putting the values, sp = √ (1/n1+n2-2) [{∑X2 – (∑X)2/n1} + {∑Y2 – (∑Y)2/n2}] =√ (1/8+7-2) [{1138 – (94)2/8} + {785 – (73)2/7}] = 2.097 Hence t = (X – Y)/ sp√{1/n1 -1/n2} = (11.75-10.42)/ 2.097√1/8 + 1/7 = 1.24 Here the critical region is │t│< t α/2 (n1 +n2-2) │1.24│< t 0.025 (8 + 7 - 2) │1.24│< 2.16 which proves that H0 is accepted, so we conclude that there is difference in their population means. Testing Hypothesis about two means with paired observations. Example. Ten young recruits were put through physical training programme. Their weights were recorded before and after the training with the following results; Recruits Weight before Weight after 1 125 136 2 195 201 3 160 158 4 171 184 5 140 145 6 201 195 7 170 175 8 176 190 9 195 190 10 139 145 Use α = 0.05 would you say that that the programme affects the average weights of recruits. Assume the distribution of weights before and after to be approximately normal. Answer: We state our null and alternate hypothesis as; Ho : µD = 0 H1 : µD ≠ 0 Level of significance is α = 0.05 Test statistics under Ho is as t = d/ sd/√n with n-1 degree of freedom Computations Recruits Weights Difference di di2 (after-before) Before After 1 125 136 11 121 2 195 201 6 36 3 160 158 -2 4 4 171 184 13 169 5 140 145 5 25 6 201 195 -6 36 7 170 175 5 25 8 176 190 14 196 9 195 190 -5 25 10 139 145 6 36 ∑di = 47 ∑ di2 = 673 d = ∑di/n = 47/10 = 4.7 sd2 = 1/n-1 [ ∑di2 – (∑di)2/n] = 1/10-1 [ 673- (47)2/10] sd = 7.09 Now by putting the values in the formula; t = d/ sd/√ n = 4.7/ 7.09/√10 = 2.09 Critical region is │ t │< t α/2 (n-1) │ 2.09│< t 0.025 (10-1) 2.09 < 2.262 which proves that H0 is accepted, so that training programme affects the average weights of recruits. Note. When the values are independent, the test statistics will be as t = {(X-Y)-∆}/sp √1/n1 + 1/n2 with n1+ n2-1 degree of freedom Note: Testing the significance of coefficient correlation r by t-test Test statistics will be; t = {r√n-2}/ √1-r2 with n-2 df Chi-Square Test : A test of goodness if fit is a technique by which we test the hypothesis whether the sample distribution is in agreement with theoretical (hypothetical) distribution. Symbolically, it can be expressed as; X2 = ∑(oi2 – ei2)/ei Where X2 = chi- square, oi = Observed value, ei = Expected values Procedure: 1. State the null hypothesis H0, which is usually sample distribution agrees with the theoretical (hypothetical) distribution. 2. Level of significance α = 0.05 3. Test statistics is X2 = ∑(oi2 – ei2)/ei 4. Critical region X2cal = X20.05(r-1) (c-1) degree of freedom Example. The following table shows the academic conditions of 100 people sex. Is there no relationship between sex and academic conditions? Academic Sex Total condition Male Female Strong 30 10 40 Poor 20 40 60 Total 50 50 100 Answer: We formulate the hypothesis as; H0: There is no relationship between sex and academic condition. H1: There is relationship between sex and academic condition. Level of significance α = 0.05 Test statistics is X2 = ∑(oi2 – ei2)/ei with (r-1) (c-1) degree of freedom Computations Academic Sex Total condition Male Female Strong 30 10 40 Poor 20 40 60 Total 50 50 100 e11 = {40 x 50}/100 = 20 e12 = {40 x 50}/100 = 20 e13 = {60 x 50}/100 = 30 e14 = {60 x 50}/100 = 30 Oij eij Oij - eij (Oij –eij)2 (Oij –eij)2/ei 30 20 10 100 5 10 20 -10 100 5 20 30 -10 100 3.33 40 30 10 100 3.33 Total 16.66 2 2 2 X = ∑(oi – ei )/ei with (r-1) (c-1) degree of freedom By putting the values, X2 = 16.66 Critical region: X2cal > X20.05(r-1) (c-1) degree of freedom 16.66 > 3.84 Hence Ho is rejected, which proves that there is relationship between sex and academic conditions. Example: Genetic theory states that children having one parameter of blood type-M and other parameter of blood type -N, will always be one of three types as M, MN, N. The proportion of three types on average will be as 1:2:1. The report says that out of 300 children having one M percent and one N percent 30% were found to be M type, 45% of MN type and remainder of type N. Test the hypothesis whether the traits of genetic theory is not consistent with the report. Answer: We formulate the hypothesis as; H0: The genetic theory is not consistent with the report or the fit is not good. H1: The genetic theory is consistent with the report or the fit is good. Level of significance α = 0.05 Test statistics will be used X2 = ∑(oi2 – ei2)/ei with (n-1) degree of freedom Computations: O1 = (30x300)/100 = 90 O2 = (45x300)/100 = 135 O3 = (25x300)/100 = 75 e1 = (1x300)/4 = 75 e2 = (2x300)/4 = 150 e3 = (1x300)/4 = 75 Oi ei Oi - ei (oi –ei)2 (oi –ei)2/ ei 90 75 15 225 3 135 150 -15 -15 1.5 75 75 0 0 0 ∑(oi2 – ei2)/ei= 4.5 X2 = ∑(oi2 – ei2)/ei with (n-1) degree of freedom X2 = 4.5 Critical region: X2cal < X20.05(3-1) degree of freedom 4.5 < 5.99 Hence Ho is accepted, which proves that the genetic theory is not consistent with the report or the fit is not good. Analysis of Variance: (ANOVA): In simple word, the analysis of variance is defined as, “It is statistical device for partitioning the total variations into separate components that measures the different sources of variation. We use the following terms in making the analysis of variance table; 1. Source of variation: A component of an experiment for which we calculate the sum of squares and mean squares. 2. Degree of freedom: For a given set of conditions, the number of degree of freedom is the total number of observations minus one restriction imparts the aggregate data. 3. Sum of squares: It is sum of squares of squares of deviations for each item from its mean. i.e E (X-X)2. 4. Mean square: When sum of squares divided by respective degree of freedom. It is also known a estimate of variance. (s2). 5. F ratio: The ratio of treatment estimate of variance to error estimate of variance is called F ratio. 6. F tabulated: F ∞ (n1, n2) Analysis of variance technique is applied into different criterion of classification i.e One way classification or one criterion or category classification or two way classification or two criterion or categories classification. ANOVA (TWO WAY WITHOUT INTERACTION) or One Way ANOVA: ANOVA for Randomized Block Design or One Way ANOVA : To test for statistical significance in a randomized block design, the linear model of individual observation is; Yij = µ + αj + ßi + εij Where Yij = Individual observation on the dependent variable. µ = grand mean αj = jth treatment effect ßi = ith block effect εij = random error or residual The statistical objective is to determine if significance differences among treatment means and block means exist. This will be done by calculating an F ratio for each source of effects. Example: To illustrate the analysis of a Latin Square Design, let us return to the experiment in which the letters A, B, C and D represents four varieties of wheat, the rows represents four different fertilizers and the columns accounts for the four varieties of wheat measured in kg per plot. It is assumed that variance of variations do not interact. Using a 0.05 level of significance, test the hypothesis that; a) H/o: There is no difference in the average yields of wheat when different kinds of fertilizers are used. b) H//o: There is no difference in the average yields of wheat due to different years. c) H///o: There is no difference in the average yields of the four varieties of wheat. Table: Yields of wheat in kg per plot Fertilizer Year Treatment 1978 1979 A B T1 70 75 D A T2 66 59 C D T3 59 66 B C T4 41 57 1980 C 68 B 55 A 39 D 39 Solutions: Table: Yields of wheat in kg per plot Fertilizer Year Treatment 1978 1979 1980 1981 A B C D T1 70 75 68 81 D A B C T2 66 59 55 63 C D A B T3 59 66 39 42 B C D A T4 41 57 39 55 Total 236 257 201 241 1 a) H/o: α1 = α2 = α3 = α4 =0 b) H//o: β1 = β 2 = β3 = β3 = 0 c) H//o: TA = TB = TC = TD = 0 2 a) H/1: At least one of the αi is not equal to zero b) H//1: At least one of the βi is not equal to zero c) H///1: At least one of the TK is not equal to zero 1981 D 81 C 63 B 42 A 55 Total 294 243 206 192 935 α = 0.05 Critical region a) f1 > 4.76 b) f2 > 4.76 c) f3 > 4.76 From table, we find the row, column and treatment totals to be; T1 = 294, T2= 243, T3= 206, T4= 192 T.1= 236, T.2= 257, T.3= 201, T.4= 241 T..A=223, T..B=213, T..C= 247, T..D= 252 Hence SST = 702 + 752 + --------+ 552 – 9352/16 = 2500 SSR = (2942 + 2432 + 2062 + 1922)/4 - 9352/16 = 1557 SSC = (2362 + 2572 + 2012 + 2412)/4 - 9352/16 = 418 SSTR = (2232 + 2132 + 2472 + 2522)/4 - 9352/16 = 264 SSE = 2500-1557-418-264 = 261 3 4 5 Two way Analysis of variance (ANOVA) without interaction Table Source of Sum of Degree of Mean square variance squares freedom SSR=1557 r-1 =3 S21 = SSR/(r-1) =519.00 Rows means SSC= 418 c-1= 3 S22 = SSC/(c-1) =139.33 Columns means S23 = SSTR/r-1 =88.00 Treatment SSTA= 264 r-1 = 3 Error SSE= 261 (c-1)(r-2)=6 Total SST = 2500 15 Computed f f1 = S21/ S24 = 11.93 f2 = S22/ S24 = 3.20 f3 = S23/ S24 = 2.02 S24 = SSE/(c-1)(r-2) = 43.5 Decisions: a) Reject H/o and conclude that a difference in the average yields of wheat exists when different kinds of fertilizers are used. As f1c= 11.93 while f10.05 (3,6) =4.76 since f1c> f10.05 (3,6) b) Accept H//o and conclude that there is no difference in the average yields due to different years. As f2c= 3.20 while f10.05 (3,6) = 4.76 since f1c< f10.05 (3,6) c) Accept H///o and conclude that there is no difference in the average yields of the four varieties of wheat. As f3c= 2.02 while f10.05 (3,6) = 4.76 since f1c< f10.05 (3,6) ANOVA (TWO WAY WITH INTERACTION) or Two Way ANOVA or Factorial Design: There is considerable similarity between the factorial design and the one way analysis of variance. The sum of squares for each of the treatment factors (rows and columns) is similar to the between- groups sum of squares in the single factor model- that is , each treatment sum of squares is calculated by taking the deviation of the treatment means from the grand mean. In a two factor experimental design, the linear model for an individual observation is; Yijk = µ + αj + ßi + Iij + εijk Where Yijk = Individual observation on the dependent variable. µ = grand mean αj = jth effect of factor A- column treatment ßi = ith effect of factor B- row treatment Iij= Interaction effect of factors A and B εijk = random error or residual Example: Use a 0.05 level of significance to test the following hypothesis, a) H/o: There is no difference in the average yield of wheat when different kinds of fertilizers are used. b) H//o: There is no difference in the average yield of three varieties of wheat. c) H///o: There is no interaction between the different kinds of fertilizers and the different varieties of wheat. Fertilizer Treatment T1 T2 T3 T4 Varieties of wheat V1 64 66 70 65 63 53 59 68 65 58 41 46 V2 72 81 64 57 43 52 66 71 59 57 61 53 V3 74 51 65 47 58 67 58 39 42 52 59 38 Solutions: Fertilizer Treatment T1 T2 T3 T4 Total Varieties of wheat V1 V2 200 217 186 152 192 196 145 171 723 736 Total V3 190 172 139 150 651 1. a) H/o: α1 = α2 = α3 = α4 =0 b) H//o: β1 = β 2 = β3 = 0 c) H///o: (α β) 11 = (α β) 12 = --------- = (α β) 43 =0 2. a) H/1: at least one of the αi is not equal to zero. b) H//1: at least one of the βj is not equal to zero. c) H///1: at least one of the (αβ)ij is not equal to zero. 3. α = 0.05 607 510 527 466 2110 4. Critical region: a) f1 > 3.01, b) f2 > 3.40 c) f3 > 2.51 5. Computations: SST = Total sum of squares = 642 + 662 + -----+ 382 – 21102/ 36 = 3779 SSR = Row sum of squares = (6072 + 5102 + 5272 + 4662 )/ 9 - 21102/ 36 = 1157 SSC = Column sum of squares = (7232 + 7362 + 6512)/12 - 21102/ 36 = 350 SS (RC) = Sum of squares for interaction of rows and columns = (2002 + 1862 + ----+ 1502)/ 3 - (6072 + 5102 + 5272 + 4662 )/ 9 - (7232 + 7352 + 6512)/12 + 21102/ 36 = (2002 + 1862 + ----+ 1502)/ 3 – 124826 -124019 + 123669 = 771 SSE = Error sum of squares = SST –SSR –SSC-SS(RC) = 3779-1157-350-771= 1501 Two way Analysis of variance (ANOVA) with interaction Table Source of Sum of Degree of Mean square variance squares freedom 1157 r-1 =3 S21 = SSR/(r-1) =385.66 Rows means S22 Columns means Interaction 350 c-1= 2 771 (r-1) (c-1)=6 S23 Error 1501 rc (n-1)=24 Total Decisions: S24 Computed f f1 = S21/ S24 = 6.17 = SSC/(c-1) =175.00 f2 = S22/ S24 = 2.80 = SSR(RC)/ (r-1) (c-1) f3 = S23/ S24 =128.50 = 2.05 = SSE/r c (n-1) = 62.54 3779 rcn-1 = 35 / a) Reject H o and conclude that a difference in the average yield of wheat exists when different kinds of fertilizers are used. As f1c= 6.17 while f10.05 (3,24) =3.01 since f1c> f10.05 (3,24) b) Accept H//o and conclude that there is no difference in the average yield of three varieties of wheat. As f2c= 2.80 while f20.05 (2, 24) =3.40 since f1c< f10.05 (3,24) c) Accept H///o and conclude that there is no interaction between the different kinds of fertilizers and different varieties of wheat As f3c= 2.05 while f10.05 (6, 24) =2.51 since f1c < f10.05 (3,24)