A college claims that the proportion, p. of students who commute more than fifteen miles to school is less than 20%. A researcher wants to test this. A random sample of 275 students at this college is selected, and it is found that 37 commute more than fifteen miles to school. Is there enough evidence to support the college's claim at the 0.05 level of significance?

Perform a one-tailed test. Then complete the parts below.

Carry your intermediate computations to three or more decimal places. (If necessary, consult a list of formulas.)

(8) State the null hypothesis H0 and the alternative hypothesis H1

H0

H1

(B) Determine the type of test statistic to use.


(c) Find the value of the best statistic. (Round to three or more decimal places)

(d) Find the p-value. (Round to three or more decimal places.)



(e) Is there enough evidence to support the claim that the proportion of students who commute more than fifteen miles to school is less than 20%?

Yes

No

Let's go through the steps to solve this hypothesis testing problem.

### (a) State the null hypothesis \(H_0\) and the alternative hypothesis \(H_1\):

The null hypothesis \(H_0\) is that the true proportion of students who commute more than fifteen miles to school is 20% (or greater). The alternative hypothesis \(H_1\) is that the true proportion is less than 20%.

- \(H_0: p \geq 0.20\)
- \(H_1: p < 0.20\)

### (b) Determine the type of test statistic to use:

Since we are testing a hypothesis about a population proportion, we should use the **z-test** statistic.

### (c) Find the value of the test statistic:

First, calculate the sample proportion (\(\hat{p}\)):
\[
\hat{p} = \frac{37}{275} \approx 0.1345
\]

Next, calculate the standard error (SE) for the proportion:
\[
SE = \sqrt{\frac{p_0(1 - p_0)}{n}} = \sqrt{\frac{0.20(1 - 0.20)}{275}} = \sqrt{\frac{0.16}{275}} \approx 0.0241
\]

Then, calculate the z-test statistic:
\[
z = \frac{\hat{p} - p_0}{SE} = \frac{0.1345 - 0.20}{0.0241} \approx -2.718
\]

### (d) Find the p-value:

The p-value is the probability that the test statistic is less than or equal to the observed value under the null hypothesis. For a z-test, we can find this using a standard normal distribution table or a calculator:

\[
\text{p-value} = P(Z \leq -2.718) \approx 0.0033
\]

### (e) Is there enough evidence to support the claim?

To determine if there's enough evidence, we compare the p-value to the significance level \(\alpha = 0.05\). Since the p-value (0.0033) is less than \(\alpha = 0.05\), we reject the null hypothesis.

**Conclusion:** Yes, there is enough evidence to support the college's claim that the proportion of students who commute more than fifteen miles to school is less than 20%.

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Do you want more details or have any questions?

Here are 5 questions related to this problem:
1. What would change if the significance level was 0.01 instead of 0.05?
2. How would you interpret the results if the p-value was greater than 0.05?
3. Can you explain why we use a z-test for this hypothesis test?
4. What are the implications of rejecting the null hypothesis in this context?
5. How does increasing the sample size affect the z-test statistic?

**Tip:** Always ensure that the sample size is sufficiently large when performing a z-test for proportions to satisfy the conditions for normal approximation.

Learn how to perform hypothesis testing on the proportion of students commuting over fifteen miles to school. Includes step-by-step calculation of the z-test statistic, determination of p-value, and interpretation of results.

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