A study of a local high school tried to determine the mean amount of money that each student had saved. The study surveyed a random sample of 78 students in the high school and found a mean savings of 4800 dollars with a standard deviation of 1000 dollars. At the $95 \%$ confidence level, use the normal distribution/empirical rule to estimate the margin of error for the mean, rounding to the nearest whole number. (Do not write \pm ). Answer: Submit Answer attempt 1 out of 2

Step 1 ：The margin of error for a confidence interval can be calculated using the formula: \[E = Z \times \frac{\sigma}{\sqrt{n}}\] where: E is the margin of error, Z is the Z-score, which corresponds to the desired confidence level (for a 95% confidence level, Z = 1.96), \(\sigma\) is the standard deviation of the population, and n is the size of the sample.

Step 2 ：In this case, we know that \(\sigma = 1000\), \(n = 78\), and \(Z = 1.96\). We can substitute these values into the formula to find the margin of error.

Step 3 ：Calculate the margin of error: \[E = 1.96 \times \frac{1000}{\sqrt{78}}\]

Step 4 ：The calculated margin of error is approximately 221.92629869234074.

Step 5 ：Round the margin of error to the nearest whole number: \[\text{round}(E) = 222\]

Step 6 ：Final Answer: The margin of error for the mean at the 95% confidence level, rounded to the nearest whole number, is \(\boxed{222}\).