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 standardized test statistic z. select the correct choice below and, if necessary, fill in the answer box to complete your choice.

○ a. z = \square
(round to two decimal places as needed.)

○ b. a normal sampling distribution cannot be used.

if a normal sampling distribution can be used, decide whether to reject or fail to reject the null hypothesis and interpret the decision. choose the correct answer below.

○ a. reject the null hypothesis. there is enough evidence to reject the claim.

○ b. fail to reject the null hypothesis. there is enough evidence to reject the claim.

○ c. fail to reject the null hypothesis. there is not enough evidence to reject the claim.

○ d. reject the null hypothesis. there is not enough evidence to reject the claim.

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 \( p = 0.35 \) (the hypothesized proportion).
\( np = 120 \times 0.35 = 42 \), \( n(1 - p)=120 \times (1 - 0.35)=120 \times 0.65 = 78 \). Both \( 42 \geq 5 \) and \( 78 \geq 5 \), so normal sampling is valid.

Step2: Calculate Test Statistic \( z \)

The formula for the z - statistic for a proportion is \( z=\frac{\hat{p}-p}{\sqrt{\frac{p(1 - p)}{n}}} \).
Substitute \( \hat{p}=0.30 \), \( p = 0.35 \), \( n = 120 \):
First, calculate the standard error \( \sqrt{\frac{0.35\times(1 - 0.35)}{120}}=\sqrt{\frac{0.35\times0.65}{120}}=\sqrt{\frac{0.2275}{120}}\approx\sqrt{0.001896}\approx0.0435 \).
Then, \( z=\frac{0.30 - 0.35}{0.0435}=\frac{- 0.05}{0.0435}\approx - 1.15 \) (rounded to two decimal places).

Step3: Determine Critical Value and Decision

The significance level \( \alpha = 0.06 \). For a left - tailed test (since the alternative hypothesis \( H_1:p < 0.35 \) as the claim is \( p\geq0.35 \) and the sample proportion is less than \( p \)), the critical value \( z_{\alpha}=- 1.555 \) (from standard normal table, \( z_{0.06}\approx - 1.56 \) for left - tailed).
Since the test statistic \( z=-1.15 > - 1.56 \) (the critical value), we fail to reject the null hypothesis. This means there is not enough evidence to reject the claim \( p\geq0.35 \).

Answer:

For the first part (test statistic): A. \( z=-1.15 \)
For the second part (decision): C. Fail to reject the null hypothesis. There is not enough evidence to reject the claim.