1. Tests of Hypothesis [Motivational Example]. It is claimed that the average grade of all 12 year old children in a country in a particular aptitude test is 60%. A random sample of n= 49 students gives a mean x = 55% with a standard deviation s = 2%. Is the sample finding consistent with the claim? We regard the original claim as a null hypothesis (H 0 ) which is tentatively accepted as TRUE: H 0 :  If the null hypothesis is true, the test statistic t = x -   s  n is a random variable with a N(0, 1) distribution. Thus 55 - 60 = - 35 / 2 = - 17.5 2/  49 is a random value from n(0, 1). rejection regions But this lies outside the 95% confidence interval (falls in the rejection region), so either (i) The null hypothesis is incorrect or(ii) An event with a probability of at most 0.05 has occurred. Consequently, we reject the null hypothesis, knowing that there is a probability of 0.05 that we are acting in error. In technical terms, we say that we are rejecting the null hypothesis at the 0.05 level of significance. The alternative to rejecting H 0, is to declare the test to be inconclusive. By this we mean that there is some tentative evidence to support the view that H 0 is approximately correct. n(0,1) 0.95 1.96-1.96 n(0,1)

2. Modifications Based on the properties of the Normal, student t and other distributions, we can generalise these ideas. If the sample size n < 25, we should use a t n-1 distribution. We can also vary the level of significance of the test and we can apply the tests to proportionate sampling environments. Example. 40% of a random sample of 1000 people in a country indicate satisfaction with government policy. Test at the.01 level of significance if this consistent with the claim that 45% of the people support government policy? Here,H 0 : P = 0.45 p = 0.40,n = 1000 so  p (1-p) / n = 0.015 test statistic = (0.40 - 0.45) / 0.015 = - 3.33 99% critical value = 2.58 so H 0 is rejected at the.01 level of significance. One-Tailed Tests If the null hypothesis is of the form H 0 : P  then arbitrary large values of p are acceptable, so that the rejection region for the test statistic lies in the left hand tail only. Example. 40% of a random sample of 1000 people in a country indicate satisfaction with government policy. Test at the.05 level of significance if this consistent with the claim that at least 45% of the people support government policy? Here the critical value is -1.64, so the the null hypothesis H 0 : P  is rejected at the.05 level of significance N(0,1) -1.64 0.95 Rejection region

3. Testing Differences between Means Suppose that x 1 x 2 … x m is a random sample with mean x and standard deviation s 1 drawn from a distribution with mean  1 and y 1 y 2 … y n is a random sample with mean y and standard deviation s 1 drawn from a distribution with mean  2. Suppose that we wish to test the null hypothesis that both samples are drawn from the same parent population (i.e.) H 0 :  1 =  2. The pooled estimate of the parent variance is  2 = { (m - 1) s 1 2 + (n - 1) s 2 2 } / ( m + n - 2) and the variance of x - y, being the variance of the difference of two independent random variables, is  ’ 2 =  2 / m +  2 / n. This allows us to construct the test statistic, which under H 0 has a t m+n-2 distribution. Example. A random sample of size m = 25 has mean x = 2.5 and standard deviation s 1 = 2, while a second sample of size n = 41 has mean y = 2.8 and standard deviation s 2 = 1. Test at the.05 level of significance if the means of the parent populations are identical. Here H 0 :  1 =  2 x - y = - 0.3 and  2 = {24(4) + 38(1)} / 64 = 2.0938 so the test statistic is - 0.3 /  The.05 critical value for N(0, 1) is  so the test is inconclusive 

4. Paired Tests If the sample values ( x i, y i ) are paired, such as the marks of students in two examinations, then let d i = x i - y i be their differences and treat these values as the elements of a sample to generate a test statistic for the hypothesis H 0 :  1 =  2. The test statisticd / s d /  nhas a t n-1 distribution if H 0 is true. Example. In a random sample of 100 students in a national examination their examination mark in English is subtracted subtracted from their continuous assessment mark, giving a mean of 5 and a standard deviation of 2. Test at the.01 level of significance if the true mean mark for both components is the same. Heren = 100, d = 5, s d /  n = 2/10 = 0.2 so the test statistic is 5 / 0.2 = 10. the 0.1 critical value for a N(0, 1) distribution is 2.58, so H 0 is rejected at the.01 level of significance. Tests for the Variance. For Normally distributed random variables, given H 0 :  2 = k, a constant, then (n-1) s 2 / khas a  2 n - 1 distribution. Example. A random sample of size 30 drawn from a normal distribution has variance s 2 = 5. Test at the.05 level of significance if this is consistent with H 0 :  2 = 2. Test statistic = (29) 5 /2 = 72.5, while the.05 critical value for  2 29 is 45.72, so H 0 is rejected at the.05 level of significance.