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denaw5779
22.01.2021 •
Mathematics
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.
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Ответ:
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} }](/tpl/images/1056/9036/dc5be.png)
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
Ответ: