Q:

Use the standard normal distribution or the​ t-distribution to construct a 99​% confidence interval for the population mean. Justify your decision. If neither distribution can be​ used, explain why. Interpret the results. In a random sample of 42 ​people, the mean body mass index​ (BMI) was 28.3 and the standard deviation was 6.09.

Accepted Solution

A:
Answer:(25.732,30.868)Step-by-step explanation:Given that in a random sample of 42 ​people, the mean body mass index​ (BMI) was 28.3 and the standard deviation was 6.09.Since only sample std deviation is known we can use only t distributionStd error = [tex]\frac{s}{\sqrt{n} } =\frac{6.09}{\sqrt{42} } \\=0.9397[/tex][tex]df = 42-1 =41[/tex]t critical for 99% two tailed [tex]= 2.733[/tex]Margin of error[tex]= 2.733*0.9397=2.568[/tex]Confidence interval lower bound = [tex]28.3-2.568=25.732[/tex]Upper bound = [tex]28.3+2.568=30.868[/tex]