Part 5: Constructing a 99% Confidence Interval for $p_1 - p_2$

We are asked to construct a 99% confidence interval for the difference between the proportions of successful challenges by men ($p_1$) and women ($p_2$).

Step 1: Define Proportions and Samples

Let:

• $p_1$ = proportion of successful challenges by men
• $p_2$ = proportion of successful challenges by women

Given data### Part 5: Constructing a 99% Confidence Interval for $p_1 - p_2$

We are asked to construct a 99% confidence interval for the difference between the proportions of successful challenges by men ($p_1$) and women ($p_2$).

Step 1: Define Proportions and Samples

Let:

• $p_1$ = proportion of successful challenges by men
• $p_2$ = proportion of successful challenges by women

Given

• Men: $n_1 = 1386$, successful challenges $x_1 = 426$
• Women: $n_2 = 754$, successful challenges $x_2 = 227$

Step 2: Calculate Sample Proportions

$\hat{p}_1 = \frac{x_1}{n_1} = \frac{426}{1386} \approx 0.3073$ $\hat{p}_2 = \frac{x_2}{n_2} = \frac{227}{754} \approx 0.3013$

Step 3: Calculate the Standard Error

The standard error for the difference of proportions ($p_1 - p_2$) is calculated as: $SE = \sqrt{\frac{\hat{p}_1(1 - \hat{p}_1)}{n_1} + \frac{\hat{p}_2(1 - \hat{p}_2)}{n_2}}$

Substitute the values: $SE = \sqrt{\frac{0.3073 \times (1 - 0.3073)}{1386} + \frac{0.3013 \times (1 - 0.3013)}{754}}$

Calculating each part: $SE \approx \sqrt{\frac{0.3073 \times 0.6927}{1386} + \frac{0.3013 \times 0.6987}{754}}$ $SE \approx \sqrt{\frac{0.2127}{1386} + \frac{0.2105}{754}}$ $SE \approx \sqrt{0.0001535 + 0.0002792}$ $SE \approx \sqrt{0.0004327}$ $SE \approx 0.0208$

Step 4: Find the Critical Value

For a 99% confidence level, the critical z-value is approximately 2.576.

Step 5: Construct the Confidence Interval

The confidence interval for $p_1 - p_2$ is: $(\hat{p}_1 - \hat{p}_2) \pm z \times SE$ $= (0.3073 - 0.3013) \pm 2.576 \times 0.0208$ $= 0.0060 \pm 0.0536$

Step 6: Calculate the Interval

$0.0060 - 0.0536 < p_1 - p_2 < 0.0060 + 0.0536$ $-0.0476 < p_1 - p_2 < 0.0596$

Conclusion for Part 5

The 99% confidence interval for $p_1 - p_2$ is: $-0.048 < p_1 - p_2 < 0.060$

Since this interval includes zero, it supports the conclusion from the hypothesis test that there is no significant difference in the success rates of challenges by men and women.

Would you like additional clarification on any part of this solution?

Follow-up Questions

1. What does it mean for an interval to contain zero in hypothesis testing?
2. How does the confidence interval support the hypothesis test's outcome?
3. Why is a 99% confidence interval wider than a 95% interval?
4. How would changing the significance level to 0.05 affect the test?
5. Can you explain why the critical z-value is 2.576 for a 99% confidence level?

Tip

In hypothesis testing, always ensure your test statistic and confidence interval calculations match the required significance level to accurately interpret results.