Question

A researcher is interested in determining whether or not review sessions affect exam performance. The independent variable, a review session, is administered to a sample of students (n=9) in an attempt to determine if this has an effect on the dependent variable, exam performance. Based on information gathered in previous semesters, I know that the population mean for a given exam is 24. The sample mean is 25, with a sample standard deviation (s) of 4. 1. State the Hypothesis 2. Set the Criterion 3. Calculate the estimated standard error of the mean 4. Calculate the t-statistic for the sample 5. Will you reject or fail to reject the null hypothesis? 6. Calculate confidence interval if appropriate

Answer 1

Solution: Here concept of one sample t test is used.

solution i- The noll and alternative kypotheses are as follows: Hoi M= 24 ie There is no significant affect of review sessions on exam performance. H: M & 24 ie There is significant affect of review sessions on exam performance. 2) The decision criterian is as follows: If It critical t-value (two-tailed) at a significance level, then we reject the null hypothesis aut & significance level. If Its a critical t-valve (two-tailed) at & significance level and then we fail to reject the nell hypothesis at a significance level . A and (n-1) af and (n-1) df

3) Estimated standard error of the mean is given SE(K) = where s is sample standard deviation a is sample size SE(5) = = = 1-3333 Standard error of the mean = 1.3333 . 47 To test the hypothesis we shall use one - sample t-test. The test statistic is given by t = x - Mo ~ t (mai) in Sun ~t was where M s n u is is is is sample nean specified value sample standard sample size of population mean deviation We have, 1 x i t = 25 Mo= 24 S- 4 and n=9 25-24 -= = 0.75 4/59 1 t-statistic is 0.75. 0.75 4/3 The value of

57 In our 0.05 In our question significance level (&) is not geven. So we shall use most commonly used a = 0.05. Now, at x = 0.05 and (n-1) = (3-1) = 8 degrees of freedom the critical t-valve (two-tailed) is given by t 10.05, p = 2.3060 (two-tailed) and we have t = 0.15 7 (0.75 < 2.3060) t < critical t-value It < critical t-value at &=0.05 and 8 degrees of freedom, therefore we shall be fail to reject the noll hypothesis at 0.05 significance level. 6 The 95% confidence interval for population mean is. geven by Since s = 1.3333 we have, x=25, + 10.05, 8) = 2.3060 95% confidence interval is 25 + 12.3060 x 1.3333) (25 + 3.0746) (25 - 3.0746, 25 + 3.0746) (21.9254, 28.0746)