Question

In a random sample of 17 ​people, the mean commute time to work was 32.2 minutes and the standard deviation was 7.3 minutes. Assume the population is normally distributed and use a​ t-distribution to construct a 99​% confidence interval for the population mean mu. What is the margin of error of mu​? Interpret the results.

Answer 1

solution? Given that, X-32.2 s= 7.3 n= 17 dF=n-1 =17-1=16 At qq / confidence level the tis L-1-99 % a=1-0.99 -0.01 a - 0.0 -0.005 t212df = tooos; 16 = 2.921 Margin of error = E= €212; dfxs E- 2.921x7.3 Viz E=5.17 Margin of error=E=5.17