Question

A class survey in a large class for first-year college students asked, "About how many minutes do you study on a typical weeknight?" The mean response of the 255 students was x¯¯¯ = 148 minutes. Suppose that we know that the studey time follows a Normal distribution with standard deviation σ = 65 minutes in the population of all first-year students at this university. Regard these students as an SRS from the population of all first-year students at this university. Does the study give good evidence that students claim to study more than 2 hours per night on the average? (a) State null and alternative hypotheses in terms of the mean study time in minutes for the population. (b) What is the value of the test statistic z? (c) Can you conclude that students do claim to study more than two hours per weeknight on the average?

Answer 1

There is enough evidence to support the claim that students study more than two hours per weeknight on the average.

Explanation:

Answer 2

There is enough evidence to support the claim that students study more than two hours per weeknight on the average.

Explanation:

We are given the following in the question:

Population mean, μ = 2 hours = 120 minutes

Sample mean,
`\bar{x}` = 148 minutes = 2.467 hours

Sample size, n = 255

Alpha, α = 0.05

Population standard deviation, σ = 65 minutes

First, we design the null and the alternate hypothesis

`H_(0): \mu = 120\text{ minutes}\\H_A: \mu > 120\text{ minutes}`

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

Formula:

`z_(stat) = \displaystyle\frac{\bar{x} - \mu}{(\sigma)/(√(n)) }`

Putting all the values, we have

`z_(stat) = \displaystyle(148 - 120)/((65)/(√(255)) ) = 6.8788`

Now, we calculate the p-value from the normal standard table.

P-value = 0.00001

Since the p-value is smaller than the significance level, we fail accept the null hypothesis and reject it, We accept the alternate hypothesis.

Conclusion:

Thus, there is enough evidence to support the claim that students study more than two hours per weeknight on the average.