The national average for the number of students per teacher for all U.S. public schools is 15.9. A random sample of 12 school districts from a moderately populated area showed the mean number of students per teacher was 19.2 with a standard deviation. of 2.1 . Find a $95 \%$ confidence interval for the true mean number of students per teacher. In your work give the Margin of Error, the confidence interval and the interpretation of the interval. How does your estimate compare with the national average? Edit View Insert Format Tools Table
Solution
Step 1 ：Given in the problem: \(\bar{x} = 19.2\) (sample mean), \(s = 2.1\) (sample standard deviation), and \(n = 12\) (sample size).
Step 2 ：For a 95% confidence interval, the Z-score is approximately 1.96.
Step 3 ：Calculate the margin of error (E) using the formula: \(E = Z * (s/\sqrt{n})\).
Step 4 ：Substitute the given values into the formula: \(E = 1.96 * (2.1/\sqrt{12})\).
Step 5 ：Calculate E: \(E = 1.96 * 0.6055 = 1.1868\).
Step 6 ：Calculate the confidence interval using the formula: \(Confidence\ Interval = \bar{x} ± E\).
Step 7 ：Substitute the values into the formula: \(Confidence\ Interval = 19.2 ± 1.1868\).
Step 8 ：Calculate the confidence interval: \(Confidence\ Interval = (18.0132, 20.3868)\).
Step 9 ：\(\boxed{We\ are\ 95\%\ confident\ that\ the\ true\ mean\ number\ of\ students\ per\ teacher\ in\ this\ moderately\ populated\ area\ is\ between\ 18.0132\ and\ 20.3868.}\)
