Question

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.

Answer 1

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 distribution

Std error = $\frac{s}{\sqrt{n} } =\frac{6.09}{\sqrt{42} } \\=0.9397$

$df = 42-1 =41$

t critical for 99% two tailed $= 2.733$

Margin of error$= 2.733*0.9397=2.568$

Confidence interval lower bound = $28.3-2.568=25.732$

Upper bound = $28.3+2.568=30.868$