31. The time needed for college students to complete a certain paper-and-pencil maze follows a normal distribution with a mean of 30 seconds and a standard deviation of 3 seconds. You wish to see if the mean time m is changed by vigorous exercise, so you have a group of nine college students exercise vigorously for 30 minutes and then complete the maze. Assume that s remains unchanged at 3 seconds. The hypotheses you decide to test are H0: m = 30 versus Ha: m ¹ 30. Suppose it takes nine students an average of xbar = 32.05 seconds to complete the maze. At the 1% significance level what can you conclude?

Suppose it takes the nine students an average of = 32.05 seconds to complete the maze. At the 1% significance level, what can you conclude?

A)

H0 should be rejected because the P-value is less than 0.01.

B)

H0 should not be rejected because the P-value is greater than 0.01.

C)

Ha should be rejected because the P-value is less than 0.01.

D)

Ha should not be rejected because the P-value is greater than 0.01.

First, note that we are testing a sample mean against a hypothesized mean. And it is a two-tailed test because the null hypothesis contains a strict equality =.

Get the t stat:

Note that we put in first the t stat, then the degrees of freedom (n-1) and then the number of tails (here two).

Excel gives the p value of 0.075 (rounded). We are asked to test at 1%, which means alpha is 0.01. p is clearly > 0.01 so we fail to reject the null hypothesis. B is the correct answer.

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