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 \\(\alpha = 0.01\\), is there enough evidence to reject the researcher’s claim? complete parts (a) through (d) below.

c. the rejection region is \\(\square < z < \square\\).
d. the rejection region is \\(z < \square\\).
(c) find the standardized test statistic \\(z\\).
\\(z = 0.00\\) (round to two decimal places as needed.)
(d) decide whether to reject or fail to reject the null hypothesis and interpret the decision in the context of the original claim.
\\(\square\\) the null hypothesis. there \\(\square\\) enough evidence to \\(\square\\) the researchers claim

Explanation:

Step1: Identify Hypotheses

Null hypothesis \( H_0: p = 0.6 \), Alternative \( H_a: p
eq 0.6 \) (two - tailed test). Sample proportion \( \hat{p}=\frac{300}{500}=0.6 \).

Step2: Calculate Standard Error

Standard error \( SE=\sqrt{\frac{p(1 - p)}{n}}=\sqrt{\frac{0.6\times(1 - 0.6)}{500}}=\sqrt{\frac{0.6\times0.4}{500}}=\sqrt{\frac{0.24}{500}}=\sqrt{0.00048}\approx0.0219 \)

Step3: Compute z - statistic

\( z=\frac{\hat{p}-p}{SE}=\frac{0.6 - 0.6}{0.0219}=\frac{0}{0.0219}=0.00 \)

Step4: Determine Critical Values

For \( \alpha = 0.01 \) (two - tailed), critical \( z \)-values are \( \pm2.576 \). Rejection region: \( z < - 2.576 \) or \( z>2.576 \)

Step5: Decision

Since \( z = 0.00 \) is not in the rejection region (\( - 2.576<0.00<2.576 \)), we fail to reject \( H_0 \).

Answer:

(c) The standardized test statistic \( z=\boldsymbol{0.00} \)
(d) Fail to reject the null hypothesis. There is not enough evidence to reject the researcher's claim.