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.\
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claim: p ≥ 0.24; α = 0.08. sample statistics: \hat{p} = 0.20, n = 110\
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a normal sampling distribution cannot be used.\
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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.\
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a. z = \
(round to two decimal places as needed.)\
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b. a normal sampling distribution cannot be used.

Explanation:

Step1: Check Normal Conditions

To use a normal sampling distribution for a proportion, we need \( np \geq 5 \) and \( n(1 - p) \geq 5 \). Here, \( p = 0.24 \), \( n = 110 \).
Calculate \( np \): \( 110 \times 0.24 = 26.4 \geq 5 \).
Calculate \( n(1 - p) \): \( 110 \times (1 - 0.24) = 110 \times 0.76 = 83.6 \geq 5 \).
So, normal sampling distribution can be used.

Step2: Calculate Test Statistic

The formula for the standardized test statistic \( z \) for a proportion is:

$$ z = \frac{\hat{p} - p}{\sqrt{\frac{p(1 - p)}{n}}} $$

Substitute \( \hat{p} = 0.20 \), \( p = 0.24 \), \( n = 110 \):
First, calculate the denominator:

$$ \sqrt{\frac{0.24 \times 0.76}{110}} = \sqrt{\frac{0.1824}{110}} \approx \sqrt{0.001658} \approx 0.0407 $$

Then, the numerator: \( 0.20 - 0.24 = -0.04 \)
Now, \( z = \frac{-0.04}{0.0407} \approx -0.98 \)

Answer:

A. \( z \approx -0.98 \)