Internet Viewing A researcher wishes to estimate the average number of minutes per day a person spends on the Internet. How large a sample must she select if she wishes to be \( 95 \% \) confident that the population mean is within 12 minutes of the sample mean? Assume the population standard deviation is 44 minutes. Round your final answer up to the next whole number. The researcher needs a sample of at least

To determine the sample size needed for the researcher to achieve the desired confidence level and margin of error, you can use the formula for sample size in estimating a population mean: \[ n = \left( \frac{Z \cdot \sigma}{E} \right)^2 \] where: - \( n \) is the sample size, - \( Z \) is the Z-value corresponding to the specified confidence level (for 95% confidence, \( Z \approx 1.96 \)), - \( \sigma \) is the population standard deviation (44 minutes), and - \( E \) is the margin of error (12 minutes). Plugging in the values: \[ n = \left( \frac{1.96 \cdot 44}{12} \right)^2 \] Calculating it step-by-step: 1. Calculate the numerator: \( 1.96 \cdot 44 \approx 86.24 \). 2. Then divide by the margin of error: \( \frac{86.24}{12} \approx 7.1867 \). 3. Finally, square the result: \( (7.1867)^2 \approx 51.5 \). Rounding up to the next whole number, the researcher needs a sample of at least 52 people. The sample size required is therefore \( 52 \).