courses/79877/assignments/1537018?module_item_id=3294242 A tech company is considering offering free breakfast to all employees. A recent study claimed that nearly two-thirds of working Americans skip breakfast so a tech company decided to offer free breakfast to their employees and test whether the free breakfast showed significant improvement in the proportion of employees who skip breakfast. Before offering free breakfast, the company surveyed a randomly selected group of 57 employees to discover 34 regularly skip breakfast. After implementing the free breakfast for a month, the company surveyed a randomly selected group of 28 employees and found that 14 skipped breakfast. Calculate a $90 \%$ confidence interval for the difference in proportions of employees who skip breakfast when free breakfast is provided and when it is not. Let $p_{1}$ represent the population proportion of employees who skip breakfast before and $p_{2}$ represent the population proportion of employees who skip breakfast after. Check the sample size conditions. \[ \begin{array}{l} n_{1} \hat{p}_{1}=\square \\ n_{2}\left(1-\hat{p}_{1}\right)= \\ n_{2}\left(1-\hat{p}_{2}\right)= \end{array} \] The sample statistic equation is: $\hat{p}_{1}-\hat{p}_{2}$ $p_{1}-p_{2}$ The sample statistic is The standard error is Round your answer to 4 decimal places at the end. The confidence interval is Enter your answer as an interval in the format $(x, y)$ where $x$ and $y$ are rounded to 4 decimal places. Submit

Step 1 ：First, we calculate the sample proportions of employees who skip breakfast before and after the free breakfast was offered. The sample proportion before the free breakfast was offered is given by \(\hat{p}_{1} = \frac{x_{1}}{n_{1}}\) where \(x_{1}\) is the number of employees who skip breakfast before and \(n_{1}\) is the total number of employees surveyed before. Similarly, the sample proportion after the free breakfast was offered is given by \(\hat{p}_{2} = \frac{x_{2}}{n_{2}}\) where \(x_{2}\) is the number of employees who skip breakfast after and \(n_{2}\) is the total number of employees surveyed after.

Step 2 ：Given that \(x_{1} = 34\), \(x_{2} = 14\), \(n_{1} = 57\), and \(n_{2} = 28\), we find that \(\hat{p}_{1} = 0.5965\) and \(\hat{p}_{2} = 0.5000\).

Step 3 ：Next, we calculate the standard error of the difference in proportions. The standard error is given by \(\sqrt{\frac{\hat{p}_{1}(1-\hat{p}_{1})}{n_{1}} + \frac{\hat{p}_{2}(1-\hat{p}_{2})}{n_{2}}}\). Substituting the values we have, we find that the standard error is approximately 0.1147.

Step 4 ：Finally, we calculate the 90% confidence interval for the difference in proportions. The confidence interval is given by \((\hat{p}_{1} - \hat{p}_{2}) \pm z \times SE\) where \(z\) is the z-score for a 90% confidence interval and \(SE\) is the standard error. Given that \(z = 1.645\), we find that the 90% confidence interval for the difference in proportions is approximately (-0.0921, 0.2851).

Step 5 ：This means that we are 90% confident that the true difference in proportions of employees who skip breakfast before and after the free breakfast was offered is between -0.0921 and 0.2851.

Step 6 ：Final Answer: The 90% confidence interval for the difference in proportions of employees who skip breakfast before and after the free breakfast was offered is approximately \(\boxed{(-0.0921, 0.2851)}\).