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

i think its uh

Step-by-step explanation: carrot

Answer 2

(25.732,30.868)

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 =
`(s)/(√(n) ) =(6.09)/(√(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`