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 > 0.64; α = 0.04. sample statistics: \hat{p} = 0.72, n = 275

let q = 1 - p and let \hat{q} = 1 - \hat{p}. a normal sampling distribution \boxed{} be used here, since \boxed{} \boxed{} 5 and \boxed{} \boxed{} 5.

if a normal sampling distribution can be used, identify the hypotheses for testing the claim. select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.

\bigcirc a. h_0: p > \boxed{}, h_a: p \leq \boxed{} (round to two decimal places as needed.)
\bigcirc b. h_0: p \leq \boxed{}, h_a: p > \boxed{} (round to two decimal places as needed.)
\bigcirc c. h_0: p = \boxed{}, h_a: p \
eq \boxed{} (round to two decimal places as needed.)
\bigcirc d. h_0: p \
eq \boxed{}, h_a: p = \boxed{} (round to two decimal places as needed.)
\bigcirc e. h_0: p < \boxed{}, h_a: p \geq \boxed{} (round to two decimal places as needed.)
\bigcirc f. h_0: p \geq \boxed{}, h_a: p < \boxed{} (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 \( nq \geq 5 \), where \( q = 1 - p \). Here, the claim is \( p > 0.64 \), so we use \( p = 0.64 \) (from the null hypothesis related to the claim). Calculate \( np \) and \( nq \):
\( np = 275 \times 0.64 = 176 \), \( nq = 275 \times (1 - 0.64) = 275 \times 0.36 = 99 \). Both \( 176 \geq 5 \) and \( 99 \geq 5 \), so a normal sampling distribution can be used.

Step2: Identify Hypotheses

The claim is \( p > 0.64 \), so the alternative hypothesis \( H_a \) is \( p > 0.64 \). The null hypothesis \( H_0 \) is the complement of the claim's direction, so \( H_0: p \leq 0.64 \). This matches option B.

Answer:

Part 1 (Normal Sampling):

A normal sampling distribution \(\boldsymbol{\text{can}}\) be used here, since \( np = 275 \times 0.64 = 176 \geq 5 \) and \( nq = 275 \times 0.36 = 99 \geq 5 \).

Part 2 (Hypotheses):

B. \( H_0: p \leq 0.64 \), \( H_a: p > 0.64 \)