Question

an education researcher claims that at most 8% of working college students are employed as teacher assistants. in a random sample of 300 working college students, 9% are employed as teachers assistants. at α = 0.05, is there enough evidence to reject the researcher’s claim? complete parts (a) through (b) find the critical value(s) and identify the rejection region(s). identify the critical value(s) for this test. z₀ = (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 (round to two decimal places as needed.) a. the rejection region is z > b. the rejection regions are z < and z > c. the rejection region is z < d. the rejection region is < z <

Explanation:

Step1: Determine Test Type

This is a two - tailed test (since the claim is "at most" and we are checking for deviation in either direction) with \(\alpha = 0.05\). For a two - tailed test, we split \(\alpha\) into two equal parts: \(\alpha/2=0.025\) for each tail.

Step2: Find Critical Z - Values

We need to find the \(z\) - values such that the area in the two tails is \(0.05\) (total) or \(0.025\) in each tail. Looking at the standard normal distribution table, the \(z\) - value corresponding to an area of \(0.975\) (since \(1 - 0.025=0.975\)) to the left of the \(z\) - value is \(z = 1.96\) and the \(z\) - value corresponding to an area of \(0.025\) to the left of the \(z\) - value is \(z=- 1.96\).

Step3: Identify Rejection Regions

For a two - tailed test, the rejection regions are \(z < - z_{\alpha/2}\) and \(z>z_{\alpha/2}\). Since \(z_{\alpha/2} = 1.96\), the rejection regions are \(z < - 1.96\) and \(z>1.96\).

Answer:

The critical values are \(z_0=- 1.96,1.96\) (rounded to two decimal places). The rejection regions are \(z < - 1.96\) and \(z>1.96\), so the correct option for rejection regions is B with \(z < - 1.96\) and \(z>1.96\).