The population standard deviation for the weights of boxes of breakfast cereal is 0.6 ounces. If we want to be 90% confident that the sample mean is within 1 ounce of the true population mean, what is the minimum sample size that can be taken?\begin{tabular}{|c|c|c|c|c|}\hline z0.10 & z0.05 & z0.025 & z0.01 & z0.005 \\\hline 1.282 & 1.645 & 1.960 & 2.326 & 2.576 \\\hline\end{tabular}Use the table above for the z-score, and be sure to round up to the next integer.Provide your answer below:
Q. The population standard deviation for the weights of boxes of breakfast cereal is 0.6 ounces. If we want to be 90% confident that the sample mean is within 1 ounce of the true population mean, what is the minimum sample size that can be taken?\begin{tabular}{|c|c|c|c|c|}\hline z0.10 & z0.05 & z0.025 & z0.01 & z0.005 \\\hline 1.282 & 1.645 & 1.960 & 2.326 & 2.576 \\\hline\end{tabular}Use the table above for the z-score, and be sure to round up to the next integer.Provide your answer below:
Identify Formula: Identify the formula to calculate the minimum sample size for a given confidence level and margin of error. The formula is:n=(Z⋅σ/E)2where n is the sample size, Z is the z-score corresponding to the desired confidence level, σ is the population standard deviation, and E is the margin of error.
Determine Z-Score: Determine the z-score that corresponds to a 90% confidence level. Looking at the provided z-score table, the z-score that corresponds to a 90% confidence level (which leaves 10% in the tails, or 5% in each tail) is 1.645.
Plug Values: Plug the values into the formula. We have σ=0.6 ounces, E=1 ounce, and Z=1.645. Now we can calculate the sample size:n=(1.645×0.6/1)2
Perform Calculation: Perform the calculation: n=(0.987)2n=0.974169
Round Up: Since we cannot have a fraction of a sample, we need to round up to the next whole number. The minimum sample size is therefore 1 when rounded up.