A statistics 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 a random sample of 36 students who took a morning class and, independently, another random sample of 36 students who took an afternoon class. 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 is assumed to be equal for morning and evening classes. Let µ₁ and µ₂ represent the population mean final exam scores of statistics' courses offered in the morning and the afternoon, respectively. Compute the appropriate test statistic to analyze the claim at the 1% significance level. A) t₇₀ = −2.811 B) t₇₀ = 2.811 C) z = −2.811 D) z = 2.811

Question

Quizplus · Accepted Answer

Since we are assuming that the population standard deviation is equal for both morning and evening classes, we can use a pooled t-test. The formula for the test statistic is:
t = (x̄1 - x̄2) / s_pooled * sqrt(1/n1 + 1/n2)
where x̄1 and x̄2 are the sample means, n1 and n2 are the sample sizes, and s_pooled is the pooled standard deviation, given by:
s_pooled = sqrt(((n1-1)*s1^2 + (n2-1)*s2^2) / (n1+n2-2))
Plugging in the values from the problem, we get:
t = (74 - 68) / sqrt(((35*8^2) + (35*10^2)) / 70) * sqrt(1/36 + 1/36) = 2.811
Since this is a two-tailed test at the 1% significance level, the critical value is ±2.576. Since our calculated t-value of 2.811 falls outside this range, we reject the null hypothesis and conclude that there is evidence to support the professor's hypothesis that morning students do better than afternoon students.

A Statistics Professor at a Large University Hypothesizes That Students