A random sample of 12 second-year university students enrolled in a business statistics course wasdrawn. At the course's completion, each student was asked how many hours he or she spent doinghomework in statistics. The data are listed here. It is known that the population standard deviation is ? = 8.0. The instructor has recommended that students devote 3 hours per week for the duration ofthe 12-week semester, for a total of 36 hours. Test to determine whether there is evidence that the average student spent less than the recommended amount of time. Use a 5% significance level.31 40 26 30 36 38 29 40 38 30 35 381. what are the hypotheses?2. what is the rejection region?3. what is the test statistic?4. what is the p-value?5. the conclusion of the hypothesis testing is:

There is enough evidence to support the claim that the population mean of the students at this college is less than the recommended number of 8.4 hours.          

Step-by-step explanation:We are given the following in the question:  

Population mean, μ = 8.4 hours

Sample mean,  = 7.72 hours

Sample size, n = 237

Alpha, α = 0.01

Sample standard deviation, s = 1.02 hours

First, we design the null and the alternate hypothesis

We use one-tailed t test to perform this hypothesis.

Formula:

Putting all the values, we have

Now,

Since, the calculated test statistic is less than the critical value, we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

Conclusion:

Thus, there is enough evidence to support the claim that the population mean of the students at this college is less than the recommended number of 8.4 hour

the answer

b. they are parallel and will never intersect.

table 3 represents a direct variation function. : )