Question

for the following information, determine whether a normal sampling distribution can be used, where p is the population proportion, α is the level of significance, \hat{p} is the sample proportion, and n is the sample size. if it can be used, test the claim.

claim: p≥0.35; α=0.06. sample statistics: \hat{p}=0.30, n=120

if a normal sampling distribution can be used, identify the rejection region(s). select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.

a. the rejection region is z > \square.
(round to two decimal places as needed.)

b. the rejection region is z < \square.
(round to two decimal places as needed.)

c. the rejection region is \square < z < \square.
(round to two decimal places as needed.)

d. the rejection regions are z < \square and z > \square.
(round to two decimal places as needed.)

Explanation:

Step1: Check Normal Sampling Condition

To use a normal sampling distribution for a proportion, we need \( np \geq 5 \) and \( n(1 - p) \geq 5 \). Here, the claim is \( p \geq 0.35 \), so we use \( p = 0.35 \) for the check.
\( np = 120 \times 0.35 = 42 \geq 5 \)
\( n(1 - p)=120 \times (1 - 0.35)=120 \times 0.65 = 78 \geq 5 \)
So, normal sampling distribution can be used.

Step2: Identify Hypotheses and Test Type

The claim is \( p \geq 0.35 \), so the null hypothesis \( H_0: p = 0.35 \) (or \( p \geq 0.35 \)) and the alternative hypothesis \( H_a: p < 0.35 \) (left - tailed test).

Step3: Find Critical Value

For a left - tailed test with \( \alpha = 0.06 \), we find the z - score such that \( P(Z < z_{\alpha})=\alpha = 0.06 \). Using the standard normal table or calculator, the z - score corresponding to a cumulative probability of 0.06 is approximately \( z=-1.55 \) (since \( P(Z < - 1.55)\approx0.06 \)).

Step4: Determine Rejection Region

In a left - tailed test, the rejection region is \( z < \) critical value. So the rejection region is \( z < - 1.55 \).

Answer:

B. The rejection region is \( z < - 1.55 \)