2. Find the \( 90 \% \) confidence interval if there are 85 employees taken the same work with a mean of 180 and standard deviation of 40. 3. Find the sample size to estimate a nooulation mean within

To calculate the \(90\%\) confidence interval for the mean, you can use the formula: \[ \text{Confidence Interval} = \text{Mean} \pm \left( z \times \frac{\sigma}{\sqrt{n}} \right) \] Here, \(z\) for \(90\%\) confidence is \(1.645\), \(\sigma\) is the standard deviation of \(40\), and \(n\) is the sample size \(85\). Plugging in the numbers gives: \[ \text{Margin of Error} = 1.645 \times \frac{40}{\sqrt{85}} \approx 11.29 \] Thus, the \(90\%\) confidence interval is \(180 \pm 11.29\), which results in approximately \((168.71, 191.29)\). To determine the sample size needed to estimate the population mean within a specific margin of error, you can use the formula: \[ n = \left( \frac{z \times \sigma}{E} \right)^2 \] Where \(E\) is the desired margin of error. For example, if you want a margin of error of \(5\), and using \(z = 1.645\) and \(\sigma = 40\), you can calculate: \[ n = \left( \frac{1.645 \times 40}{5} \right)^2 \approx 42.56 \] Since you can't have a fraction of a sample, you'd round up to \(43\) as the required sample size.