Question

identify the test statistic.\\( z = 1.18 \\)\\( \text{(round to two decimal places as needed )} \\)\
identify the p - value.\\( \text{p - value} = 0.238 \\)\\( \text{(round to three decimal places as needed )} \\)\
what is the conclusion based on the hypothesis test?\
the p - value is greater than the significance level of \\( \alpha = 0.01 \\), so fail to reject the null hypothesis. there is not sufficient evidence to warrant rejection of the claim that women and men have equal success in challenging calls.\
b. test the claim by constructing an appropriate confidence interval\
the 99% confidence interval is \\( \square < (p_1 - p_2) < \square \\).\\( \text{(round to three decimal places as needed )} \\)

Explanation:

Step1: Recall Confidence Interval Formula

For the difference between two proportions \((p_1 - p_2)\), the confidence interval formula is \((\hat{p}_1 - \hat{p}_2) \pm z_{\alpha/2} \sqrt{\frac{\hat{p}_1(1 - \hat{p}_1)}{n_1} + \frac{\hat{p}_2(1 - \hat{p}_2)}{n_2}}\). But since we might have missing data (like sample proportions or sample sizes), wait—actually, maybe from the z - test, we can use the fact that for a 99% confidence interval, \(z_{\alpha/2}=2.576\) (since \(\alpha = 0.01\), \(\alpha/2=0.005\), and \(z_{0.005}=2.576\)). Wait, but maybe the problem is about two - proportion z - test, and we need to find the confidence interval for \((p_1 - p_2)\). Let's assume we have the test statistic \(z = 1.18\), but maybe we need to use the formula for the confidence interval. Wait, perhaps the original problem (before the image) had sample proportions and sample sizes. But since the test statistic is \(z = 1.18\) and P - value is 0.238, for a 99% confidence interval, we can use the formula:

The margin of error \(E=z_{\alpha/2}\sqrt{\frac{\hat{p}_1(1 - \hat{p}_1)}{n_1}+\frac{\hat{p}_2(1 - \hat{p}_2)}{n_2}}\), but if we assume that this is a two - proportion test and we can use the fact that the point estimate is \(\hat{p}_1-\hat{p}_2\), and the standard error \(SE=\sqrt{\frac{\hat{p}(1 - \hat{p})}{n_1}+\frac{\hat{p}(1 - \hat{p})}{n_2}}\) (if \(H_0:p_1 = p_2=p\) is assumed). But maybe in the original problem, we have \(\hat{p}_1\) and \(\hat{p}_2\) or other values. Wait, since the test statistic \(z=\frac{(\hat{p}_1-\hat{p}_2)-0}{SE}=1.18\), so \(SE=\frac{(\hat{p}_1-\hat{p}_2)}{1.18}\). For a 99% confidence interval, the interval is \((\hat{p}_1 - \hat{p}_2)\pm z_{\alpha/2}\times SE\). But since we don't have the sample proportions, maybe there was a typo or missing data. Wait, perhaps the user intended to have a problem where we calculate the confidence interval. Alternatively, maybe the test is about two proportions, and we can use the formula for the confidence interval for the difference between two proportions.

Wait, let's assume that we have the following (maybe from a standard problem): Suppose we have two samples, and we calculated the test statistic \(z = 1.18\), and we want to construct a 99% confidence interval for \((p_1 - p_2)\). The formula for the confidence interval for the difference between two proportions is:

\((\hat{p}_1-\hat{p}_2)-z_{\alpha/2}\sqrt{\frac{\hat{p}_1(1 - \hat{p}_1)}{n_1}+\frac{\hat{p}_2(1 - \hat{p}_2)}{n_2}}<(p_1 - p_2)<(\hat{p}_1-\hat{p}_2)+z_{\alpha/2}\sqrt{\frac{\hat{p}_1(1 - \hat{p}_1)}{n_1}+\frac{\hat{p}_2(1 - \hat{p}_2)}{n_2}}\)

But since we know that the test statistic \(z=\frac{(\hat{p}_1 - \hat{p}_2)}{SE}=1.18\), so \(\hat{p}_1-\hat{p}_2 = 1.18\times SE\). The margin of error \(E = z_{\alpha/2}\times SE\), where \(z_{\alpha/2}\) for 99% confidence is 2.576.

If we assume that \(SE\) is the same as in the z - test, then the confidence interval is \((\hat{p}_1 - \hat{p}_2)-2.576\times SE<(p_1 - p_2)<(\hat{p}_1 - \hat{p}_2)+2.576\times SE\)

Since \(\hat{p}_1-\hat{p}_2=1.18\times SE\), substitute:

\(1.18\times SE-2.576\times SE<(p_1 - p_2)<1.18\times SE + 2.576\times SE\)

\((1.18 - 2.576)SE<(p_1 - p_2)<(1.18 + 2.576)SE\)

\((- 1.396)SE<(p_1 - p_2)<(3.756)SE\)

But this is not helpful. Alternatively, maybe the original problem had sample sizes and proportions. For example, suppose \(n_1 = n_2=100\), \(\hat{p}_1 = 0.5\), \(\hat{p}_2 = 0.45\), then \(SE=\sqrt{\frac{0.5\times0.5}{100}+\frac{0.45\times0.55}{100}}=\sqrt{\frac{0.25}{100}+\frac{0.2475}{100}}=\sqrt{\frac{0.4975}{100}}=\sqrt{0.004975}\approx0.0705\)

Then the margin of error \(E = 2.576\times0.0705\approx0.1816\)

\(\hat{p}_1-\hat{p}_2=0.05\)

So the confidence interval is \(0.05 - 0.1816<(p_1 - p_2)<0.05 + 0.1816\), i.e., \(- 0.1316<(p_1 - p_2)<0.2316\)

But since we don't have the actual sample data, maybe the problem is a follow - up to a two - proportion z - test. Let's assume that the point estimate \(\hat{p}_1-\hat{p}_2\) and the standard error are such that when we calculate the 99% confidence interval, we get values. Alternatively, maybe the user made a mistake in the problem presentation.

Wait, another approach: The formula for the confidence interval for the difference between two proportions is:

\(CI=(\hat{p}_1 - \hat{p}_2)\pm z_{\alpha/2}\sqrt{\frac{\hat{p}_1(1 - \hat{p}_1)}{n_1}+\frac{\hat{p}_2(1 - \hat{p}_2)}{n_2}}\)

For a 99% confidence interval, \(z_{\alpha/2}=2.576\)

If we assume that in the hypothesis test, \(H_0:p_1 = p_2\), and we calculated the test statistic \(z = 1.18\), so \(\hat{p}_1-\hat{p}_2=z\times SE\), where \(SE=\sqrt{\frac{\hat{p}(1 - \hat{p})}{n_1}+\frac{\hat{p}(1 - \hat{p})}{n_2}}\) (pooled proportion)

But without the sample data, we can't calculate the exact interval. However, if we assume that this is a problem where we use the test statistic and the standard error, and we know that the margin of error for 99% confidence is \(2.576\times SE\), and since \(z = 1.18=\frac{\hat{p}_1-\hat{p}_2}{SE}\), then \(SE=\frac{\hat{p}_1-\hat{p}_2}{1.18}\)

So the confidence interval is \((\hat{p}_1 - \hat{p}_2)\pm2.576\times\frac{\hat{p}_1 - \hat{p}_2}{1.18}\)

\(=(\hat{p}_1 - \hat{p}_2)(1\pm\frac{2.576}{1.18})\)

\(=(\hat{p}_1 - \hat{p}_2)(1\pm2.183)\)

If \(\hat{p}_1-\hat{p}_2\) is positive (from \(z = 1.18>0\)), then the lower bound is \((\hat{p}_1 - \hat{p}_2)(1 - 2.183)=(\hat{p}_1 - \hat{p}_2)(- 1.183)\) and the upper bound is \((\hat{p}_1 - \hat{p}_2)(1 + 2.183)=(\hat{p}_1 - \hat{p}_2)(3.183)\)

But this is still not helpful. Maybe the original problem had specific values. For example, if we assume that the two - proportion z - test has \(\hat{p}_1 = 0.6\), \(n_1 = 100\), \(\hat{p}_2 = 0.5\), \(n_2 = 100\)

Then \(SE=\sqrt{\frac{0.6\times0.4}{100}+\frac{0.5\times0.5}{100}}=\sqrt{\frac{0.24 + 0.25}{100}}=\sqrt{\frac{0.49}{100}} = 0.07\)

Test statistic \(z=\frac{0.6 - 0.5}{0.07}=\frac{0.1}{0.07}\approx1.428\) (not 1.18, but close). For \(z = 1.18\), \(SE=\frac{0.1}{1.18}\approx0.0847\)

Then margin of error \(E = 2.576\times0.0847\approx0.218\)

Confidence interval: \(0.1\pm0.218\), i.e., \(- 0.118<(p_1 - p_2)<0.318\)

Rounding to three decimal places, \(- 0.118<(p_1 - p_2)<0.318\)

But since we don't have the actual data, maybe the intended answer is based on a standard problem. Let's assume that the 99% confidence interval for \((p_1 - p_2)\) is calculated as follows:

The formula for the confidence interval for the difference between two proportions is:

\(CI=(\hat{p}_1 - \hat{p}_2)\pm z_{\alpha/2}\sqrt{\frac{\hat{p}_1(1 - \hat{p}_1)}{n_1}+\frac{\hat{p}_2(1 - \hat{p}_2)}{n_2}}\)

For a 99% confidence level, \(z_{\alpha/2}=2.576\)

Suppose in the original problem, \(\hat{p}_1-\hat{p}_2 = 0.05\) (a made - up point estimate) and \(SE = 0.1\) (made - up standard error), then the margin of error \(E=2.576\times0.1 = 0.2576\)

Confidence interval: \(0.05-0.2576=- 0.2076\) and \(0.05 + 0.2576 = 0.3076\), rounding to three decimal places, \(- 0.208<(p_1 - p_2)<0.308\)

But since the test statistic is \(z = 1.18\), and if we assume that the point estimate \(\hat{p}_1-\hat{p}_2=z\times SE\), and for 99% confidence, the interval is \((\hat{p}_1 - \hat{p}_2)\pm2.576\times SE\)

Let \(x=\hat{p}_1 - \hat{p}_2\), then \(SE=\frac{x}{1.18}\)

So the interval is \(x\pm2.576\times\frac{x}{1.18}=x(1\pm\frac{2.576}{1.18})=x(1\pm2.183)\)

If \(x = 0.1\) (for example), then the interval is \(0.1\times(1 - 2.183)=- 0.1183\) and \(0.1\times(1 + 2.183)=0.3183\), which rounds to \(- 0.118<(p_1 - p_2)<0.318\)

Answer:

\(-0.118 < (p_1 - p_2) < 0.318\) (Note: This answer is based on assumed values due to incomplete data in the provided problem. If you have the actual sample proportions and sample sizes, you can calculate the exact confidence interval using the formula for the difference between two proportions' confidence interval.)