According to a 2014 research study of national student engagement in the U. S., the average college student spends 17 hours per week studying. A professor believes that students at her college study less than 17 hours per week. The professor distributes a survey to a random sample of 80 students enrolled at the college. From her survey data the professor calculates that the mean number of hours per week spent studying for her sample is 15.6 hours per week with a standard deviation of 4.5 hours per week. The professor chooses a 5% level of significance. What can she conclude from her data?

Step-by-step explanation:

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

µ = 17

For the alternative hypothesis,

µ < 17

This is a left tailed test.

Since the population standard deviation is not given, the distribution is a student's t.

Since n = 80

Degrees of freedom, df = n - 1 = 80 - 1 = 79

t = (x - µ) / (s/√n)

Where

x = sample mean = 15.6

µ = population mean = 17

s = samples standard deviation = 4.5

t = (15.6 - 17) / (4.5/√80) = - 2.78

We would determine the p value using the t test calculator. It becomes

p = 0.0034

Since alpha, 0.05 > than the p value, 0.0034, then we would reject the null hypothesis. Therefore, At a 5% level of significance, the sample data showed significant evidence that students at her college study less than 17 hours per week.