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an education researcher claims that 60% of college students work year -…

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 researchers claim? complete parts (a) through (d) below. \\(z_0 = - 4.39,4.39\\) (round to two decimal places as needed. use a comma to separate answers as needed.) identify the rejection region(s). select the correct choice below and fill in the answer box(es) to complete your choice (round to two decimal places as needed.) \\(\bigcirc\\) a. the rejection regions are \\(z < \square\\) and \\(z > \square\\). \\(\bigcirc\\) b. the rejection region is \\(z > \square\\) \\(\bigcirc\\) c. the rejection region is \\(\square < z < \square\\). \\(\bigcirc\\) d. the rejection region is \\(z < \square\\).

Explanation:

Step1: Identify Test Type

This is a two - tailed hypothesis test for a proportion (since we are testing if the proportion is different from 60%), with $\alpha = 0.01$.

Step2: Find Critical Z - Values

For a two - tailed test with $\alpha=0.01$, we split the $\alpha$ into two tails. Each tail has an area of $\frac{\alpha}{2}=\frac{0.01}{2} = 0.005$.
We need to find the $z$ - scores such that $P(Z < - z_{\alpha/2})=0.005$ and $P(Z>z_{\alpha/2}) = 0.005$.
Using the standard normal distribution table or a calculator, the $z$ - score corresponding to an area of $0.005$ in the left tail is $z=- 2.58$ (because $P(Z < - 2.58)\approx0.005$) and the $z$ - score corresponding to an area of $0.005$ in the right tail is $z = 2.58$ (because $P(Z>2.58)\approx0.005$).

Step3: Determine Rejection Region

In a two - tailed test, the rejection regions are $z < - z_{\alpha/2}$ and $z>z_{\alpha/2}$. So the rejection regions are $z < - 2.58$ and $z>2.58$. So the correct choice is A, and the values to fill in are - 2.58 and 2.58.

Answer:

A. The rejection regions are $z < - 2.58$ and $z>2.58$.