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.

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 =

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

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐

you forgot an image

step-by-step explanation: