A small pilot study was conducted prior to planning for a larger research study. Data were collected from students at one college concerning how many hours per week they work In the sample, the mean was 26 hours with a standard deviation of 8 hours. Researchers want to construct a \( 95 \% \) confidence interval to estimate the mean number of hours that students at this college work, with a margin of error of \( \pm 2 \) hours. What is the minimum sample size that they should obtain? Use \( z^{*} \) multiplier of 1.960 .

To determine the minimum sample size needed, we can use the formula for the margin of error in estimating a population mean: \[ E = z^* \left(\frac{\sigma}{\sqrt{n}}\right) \] where: - \(E\) is the margin of error (±2 hours), - \(z^*\) is the z-value for a 95% confidence interval (1.960), - \(\sigma\) is the population standard deviation (8 hours), - \(n\) is the sample size. Setting \(E = 2\), we rearrange the formula to solve for \(n\): \[ 2 = 1.960 \left(\frac{8}{\sqrt{n}}\right) \] Now, isolating \(\sqrt{n}\): \[ \sqrt{n} = \frac{1.960 \times 8}{2} \] Calculating the right side: \[ \sqrt{n} = \frac{15.68}{2} = 7.84 \] Now, squaring both sides to find \(n\): \[ n = (7.84)^2 \approx 61.47 \] Since the sample size must be a whole number, we round up to the next whole number: **Minimum sample size \(n = 62\)**.