SHORT ANSWER.  Write the word or phrase that best completes each statement or answers the question.
A professor at a large university believes that students take an average of 15 credit hours per term. A random sample of 24 students in her class of 250 students reported the following number of credit hours that they were taking

1) Does this sample indicate that students are taking more credit hours than the professor believes? Test an appropriate hypothesis and state your conclusion.	1)  _____________

2) Find a 95% confidence interval for the number of credit hours taken by the students in the professor's class. Interpret your interval.	2)  _____________

1)  μ = 15. The mean number of credit hours taken by all of the professor's students is 15 credit hours.
 μ > 15. The mean number of credit hours taken by all of the professor's students is greater than 15 credit hours.
* Randomization condition Students from the class were randomly sampled. 
* 10% condition The sample is less than 10% of the class population. 
* Nearly Normal condition The histogram of credit hours is unimodal and reasonably symmetric.

Under these conditions, the sampling distribution of the mean can be modeled by Student's t with df = n — 1 = 24 — 1 = 23. Use a one-sample t-test for the mean.
We know n = 24,  = 16.6, and s = 2.22. So SE() =  = 0.453.
t =  =  = 3.532.
The P-value is P( > 3.532) = 0.0009.

The P-value of 0.0009 is low, so reject the null hypothesis. There is strong evidence that the professor's students are taking more than 15 credit hours, on average.
2) With the conditions satisfied (from Problem 1), we can find a t-interval for mean credit hours.
We know n = 24,  = 16.6, and s = 2.22. So SE() =  = 0.453.
Our confidence interval has the form  ± . We have  = 2.069. Our 95% confidence interval is then 16.6 ± 2.069  = 16.6 ± 0.94, or 15.66 to 17.54.
We are 95% confident that the interval 15.66 to 17.54 contains the true mean number of credit hours that students in the professor's class are taking.