Question

A group of 54 college students from a certain liberal arts college were randomly sampled and asked about the number of alcoholic drinks they have in a typical week. The purpose of this study was to compare the drinking habits of the students at the college to the drinking habits of college students in general. In particular, the dean of students, who initiated this study, would like to check whether the mean number of alcoholic drinks that students at his college in a typical week differs from the mean of U.S. college students in general, which is estimated to be 4.73.
The group of 51 students in the study reported an average of 4.35 drinks per with a standard deviation of 3.88 drinks.

Find the p-value for the hypothesis test.

The p-value should be rounded to 4-decimal places.

Answer 1

Answer:

P-value =  0.4846

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 4.73

Sample mean, $\bar{x}$ = 4.35

Sample size, n = 51

Alpha, α = 0.05

Sample standard deviation, s = 3.88

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 4.73\\H_A: \mu \neq 4.73$

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

Formula:

$z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Putting all the values, we have

$z_{stat} = \displaystyle\frac{4.35 - 4.73}{\frac{3.88}{\sqrt{51}} } = -0.699$

Now, $z_{critical} \text{ at 0.05 level of significance } = \pm 1.96$

We can calculate the p-value with the help of standard normal table.

P-value =  0.4846

Since the p-value is higher than the significance level, we fail to reject the null hypothesis and accept it.

We conclude that this college has same drinking habit as the college students in general.