Suppose that an accounting firm does a study to determine the time needed to complete one person's tax forms. It randomly surveys 50 people. The sample mean is 24.6 hours. There is a known standard deviation of 8 hours. The population distribution is assumed to be normal. Round answers to 3 decimal places. a. Find the following: i. $\bar{x}=$ ii. $\sigma=$ iii. $n=$ b. Construct a $90 \%$ confidence interval for the population mean time to complete tax forms. i. State the confidence interval. $\mathrm{Cl}:$ ii. Calculate the error bound. EBM: c. Suppose that the firm decided that it needed to be at least $99 \%$ confident of the population mean length of time to within one hour. How would the number of people the firm surveys change? Select an answer Question Help: 㞔 VIDEO Submit

Step 1 ：Given that the sample mean (\(\bar{x}\)) is 24.6, the standard deviation (\(\sigma\)) is 8, and the sample size (\(n\)) is 50.

Step 2 ：To construct a 90% confidence interval for the population mean time to complete tax forms, we use the formula \(\bar{x} \pm z*\frac{\sigma}{\sqrt{n}}\), where \(z\) is the z-score corresponding to the desired level of confidence. For a 90% confidence level, the z-score is approximately 1.645.

Step 3 ：Substituting the given values into the formula, we get \(24.6 \pm 1.645*\frac{8}{\sqrt{50}}\).

Step 4 ：Solving this gives us the 90% confidence interval for the population mean time to complete tax forms as \(\boxed{(22.739, 26.461)}\).

Step 5 ：The error bound is the difference between the upper limit of the confidence interval and the sample mean, which is \(\boxed{1.861}\).

Step 6 ：If the firm wants to be 99% confident of the population mean length of time to within one hour, we use the formula for the sample size needed for a given margin of error, which is \(n = \left(\frac{z*\sigma}{E}\right)^2\), where \(E\) is the desired margin of error. For a 99% confidence level, the z-score is approximately 2.576, and the desired margin of error is 1.

Step 7 ：Substituting these values into the formula, we get \(n = \left(\frac{2.576*8}{1}\right)^2\).

Step 8 ：Solving this gives us the new sample size as approximately \(\boxed{425}\) people.