Question

an education researcher claims that 60% of college students work year-round. in a random sample of 500 college students, 300 say they work year-round. at α = 0.01, is there enough evidence to reject the researcher’s claim? complete parts (a) through (d) below.

b. the rejection region is z >

c. the rejection region is < z <

d. the rejection region is z <

(c) find the standardized test statistic z.
z =
(round to two decimal places as needed.)

Explanation:

Step1: Identify the type of test

This is a two - tailed test for a proportion since we are testing the claim that \(p = 0.6\) (60% of college students work year - round). The significance level \(\alpha=0.01\), so \(\alpha/2 = 0.005\) and \(1-\alpha/2=0.995\).

Step2: Recall the formula for the z - test statistic for a proportion

The formula for the z - test statistic for a proportion is \(z=\frac{\hat{p}-p}{\sqrt{\frac{p(1 - p)}{n}}}\), where \(\hat{p}=\frac{x}{n}\), \(x\) is the number of successes, \(n\) is the sample size, and \(p\) is the hypothesized population proportion.

Step3: Calculate \(\hat{p}\)

We know that \(n = 500\) and \(x = 300\). So \(\hat{p}=\frac{300}{500}=0.6\). Wait, no, wait. Wait, the researcher claims \(p = 0.6\), and our sample proportion \(\hat{p}=\frac{300}{500} = 0.6\)? Wait, that can't be right. Wait, maybe I made a mistake. Wait, no, let's recalculate. \(x = 300\), \(n=500\), so \(\hat{p}=\frac{300}{500}=0.6\). Then \(p = 0.6\), \(1 - p=0.4\), \(n = 500\).

Step4: Substitute into the formula

\(z=\frac{\hat{p}-p}{\sqrt{\frac{p(1 - p)}{n}}}=\frac{0.6 - 0.6}{\sqrt{\frac{0.6\times0.4}{500}}}=\frac{0}{\sqrt{\frac{0.24}{500}}}=\frac{0}{\sqrt{0.00048}}\approx0.00\)

Wait, but that seems odd. Wait, maybe there is a typo in the problem? Or maybe I misread the numbers. Wait, if \(x = 300\) and \(n = 500\), then \(\hat{p}=0.6\), which is equal to the hypothesized \(p = 0.6\). So the z - statistic is 0.00.

Answer:

\(z\approx\boxed{0.00}\)