An average computer mouse inspector can inspect 45 mice per hour with a population standard deviation of 14 mice per hour. The 44 computer mice inspectors at a particular factory can only inspect 38 mice per hour. Does the company have reason to believe that these inspectors are slower than average at α = 0.10?
A) Yes, because the test value -3.32 falls in the noncritical region.
B) No, because the test value -1.91 falls in the critical region.
C) Yes, because the test value -3.32 falls in the critical region.
D) No, because the test value -1.91 falls in the noncritical region.

Question

Quizplus · Accepted Answer

We need to conduct a one-sample t-test since we have a sample of 44 and we don't know the population standard deviation. We will use a two-tailed test at α=0.10.

The null hypothesis is that the mean inspection rate of the inspectors is equal to the population mean of 45 mice per hour, and the alternative hypothesis is that the mean inspection rate is less than 45 mice per hour.

Our test statistic is:

t = (sample mean - population mean)/(sample standard deviation/sqrt(sample size))

t = (38 - 45)/(14/sqrt(44)) = -3.32

Using a t-table with 43 degrees of freedom (44-1), the critical value for a two-tailed test at α=0.10 is ±1.676.

Since our calculated t-value (-3.32) is less than the critical value (-1.676), we reject the null hypothesis and conclude that there is sufficient evidence to suggest that the inspectors in this factory are slower than average. The test value -3.32 falls in the critical region. Thus, the answer is C.

An Average Computer Mouse Inspector Can Inspect 45 Mice Per