Question

A study is run comparing HDL cholesterol levels between men who exercise regularly and those who do not. The data are shown below.
Generate a 95% confidence interval for the difference in mean HDL levels between men who exercise regularly and those who do not.

Answer 1

Answer:

Hello your question is incomplete attached below is the missing part of the question

Regular Exercise           N                             Mean                          Std Dev

Yes                                35                             48.5                                12.5

No                                  120                           56.9                                11.9

answer : ( -12.968, -3.832 )

Step-by-step explanation:

To generate a 95% confidence interval

first step :

Assume population variance is equal we will calculate the number of degrees of freedom

df = n1 + n2 - 2 = 35 + 120 -2 = 153

second step:

Given : critical value = 0.05 , df = 153

hence Tc = 1.976

assuming the population variance is equal calculate pooled std as ( attached below )

Sp = 12.036

third step :

calculate the standard error

Se = $Sp\sqrt{\frac{1}{n1} +\frac{1}{n2} }$

n1 = 35

n2 = 120

Sp = 12.036

hence Se = 2.312

final step

Compute the 95% confidence interval = ( -12.968, -3.832 )

attached below is the detailed solution