Exhibit 10.7.A stats professor at a large university hypothesizes that students who take statistics in the morning typically do better than those who take it in the afternoon.He takes independently random samples,each of size 36,consisting of students who took a morning and an afternoon class,and compares the scores of each group on a common final exam.He finds that the morning group scored an average of 74 with a standard deviation of 8,while the evening group scored an average of 68 with a standard deviation of 10.The population standard deviation of scores is unknown,but assumed to be equal for morning and evening classes.Let and represent the population mean final exam scores of statistics' courses offered in the morning and the afternoon,respectively Refer to Exhibit 10.7.Compute the appropriate test statistic to analyze the claim at the 1% significance level.

Question

Exhibit 10.7.A stats professor at a large university hypothesizes that students who take statistics in the morning typically do better than those who take it in the afternoon.He takes independently random samples,each of size 36,consisting of students who took a morning and an afternoon class,and compares the scores of each group on a common final exam.He finds that the morning group scored an average of 74 with a standard deviation of 8,while the evening group scored an average of 68 with a standard deviation of 10.The population standard deviation of scores is unknown,but assumed to be equal for morning and evening classes.Let  and  represent the population mean final exam scores of statistics' courses offered in the morning and the afternoon,respectively Refer to Exhibit 10.7.Compute the appropriate test statistic to analyze the claim at the 1% significance level.

Quizplus · Accepted Answer

We can use a two-sample t-test to analyze the claim. The test statistic is calculated as:

t = (x̄1 - x̄2) / (s_pool * √(1/n1 + 1/n2))

where x̄1 and x̄2 are the sample means, s_pool is the pooled standard deviation, and n1 and n2 are the sample sizes.

Here, x̄1 = 74, x̄2 = 68, s1 = 8, s2 = 10, n1 = 36, and n2 = 36. To calculate the pooled standard deviation, we first need to calculate the degrees of freedom:

df = (s1^2/n1 + s2^2/n2)^2 / [(s1^2/n1)^2/(n1-1) + (s2^2/n2)^2/(n2-1)]

= (8^2/36 + 10^2/36)^2 / [(8^2/36)^2/35 + (10^2/36)^2/35]

= 70.53

Using a t-table with df = 70, and a significance level of 0.01 (two-tailed), the critical value is 2.660.

Plugging in all the values into the formula, we get:

t = (74 - 68) / (sqrt(((8^2/36)+(10^2/36))/70.53)*sqrt((1/36)+(1/36)))

= 3.00

Since our calculated t-value (3.00) is greater than the critical value (2.660), we can reject the null hypothesis at the 1% significance level. Therefore, we have evidence to support the claim that students who take statistics in the morning typically do better than those who take it in the afternoon.