To determine whether the proportion found in Population 2 exceeds the proportion in Population 1 using a significance level of $\alpha = 0.05$, we perform a two-proportion z-test. Here’s the step-by-step process to compute the test statistic $z$:

Step 1: Given Information

• Sample Proportions: $\hat{p}_1 = 0.29, \quad \hat{p}_2 = 0.48$
• Sample Sizes: $n_1 = 443, \quad n_2 = 654$
• Null Hypothesis: $H_0: p_1 = p_2$
• Alternative Hypothesis: $H_a: p_2 > p_1$

Step 2: Pooled Proportion

Under the null hypothesis, we assume the proportions are equal, so we compute the pooled proportion: $\hat{p} = \frac{x_1 + x_2}{n_1 + n_2}$ where $x_1 = n_1 \hat{p}_1$ and $x_2 = n_2 \hat{p}_2$.

$x_1 = 443 \cdot 0.29 = 128.47 \quad \text{(round to 128)}$ $x_2 = 654 \cdot 0.48 = 313.92 \quad \text{(round to 314)}$

$\hat{p} = \frac{128 + 314}{443 + 654} = \frac{442}{1097} \approx 0.4028$

Step 3: Standard Error (SE)

The standard error for the difference in proportions is: $SE = \sqrt{\hat{p}(1 - \hat{p}) \left( \frac{1}{n_1} + \frac{1}{n_2} \right)}$

$SE = \sqrt{0.4028(1 - 0.4028) \left( \frac{1}{443} + \frac{1}{654} \right)}$

$SE = \sqrt{0.4028 \cdot 0.5972 \left( 0.002258 + 0.001529 \right)}$

$SE = \sqrt{0.4028 \cdot 0.5972 \cdot 0.003787} \approx \sqrt{0.0009086} \approx 0.03014$

Step 4: Test Statistic ($z$)

The test statistic is: $z = \frac{\hat{p}_2 - \hat{p}_1}{SE}$

$z = \frac{0.48 - 0.29}{0.03014} = \frac{0.19}{0.03014} \approx 6.31$

Final Answer:

The test statistic is: $z \approx 6.31$

Let me know if you want further steps or explanations!

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Related Questions:

1. What is the critical z-value for a one-tailed test at $\alpha = 0.05$?
2. How does the test statistic compare to the critical value?
3. What is the conclusion of the hypothesis test?
4. How would the test change if the sample sizes were smaller?
5. What assumptions must hold for this test to be valid?

Tip:

Always double-check whether you're performing a one-tailed or two-tailed test, as this affects the critical region and conclusion.