Questions: Test the hypothesis using the P -value approach. Be sure to verify the requirements of the test. H0: p=0.59 versus H1: p<0.59 n=150, x=81, α=0.01 Is n p0(1-p0) ≥ 10 ? No Yes

Solution Steps

To conduct the hypothesis test, we first verify the requirement for the sample size. We calculate \( n p_0 (1 - p_0) \):

\[ n p_0 (1 - p_0) = 150 \times 0.59 \times (1 - 0.59) = 150 \times 0.59 \times 0.41 = 36.285 \]

Since \( 36.285 \geq 10 \), the requirement is satisfied.

We set up our hypotheses as follows:

• Null Hypothesis: \( H_0: p = 0.59 \)
• Alternative Hypothesis: \( H_1: p < 0.59 \)

The test statistic \( Z \) is calculated using the formula:

\[ Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \]

Where:

• \( \hat{p} = \frac{x}{n} = \frac{81}{150} = 0.54 \)
• \( p_0 = 0.59 \)
• \( n = 150 \)

Substituting the values:

\[ Z = \frac{0.54 - 0.59}{\sqrt{\frac{0.59 \times 0.41}{150}}} = \frac{-0.05}{\sqrt{\frac{0.2419}{150}}} = \frac{-0.05}{0.0401} \approx -1.2451 \]

The P-value associated with the test statistic \( Z = -1.2451 \) is calculated to be \( 0.1066 \).

For a significance level of \( \alpha = 0.01 \) in a left-tailed test, the critical value from the Z-table is approximately \( -2.3263 \). Thus, the critical region is defined as:

\[ Z < -2.3263 \]

We compare the P-value with the significance level:

• P-value: \( 0.1066 \)
• Significance level: \( \alpha = 0.01 \)

Since \( 0.1066 > 0.01 \), we fail to reject the null hypothesis.

Final Answer