A study is run comparing HDL cholesterol levels between men

A study is run comparing HDL cholesterol levels between men who exercise regularly and those who do not. The data are shown below.

Regular Exercise

N

Mean

Std Dev

Yes

40

48.5

12.5

No

125

56.9

11.9

What is the 95% confidence interval for the difference in the mean HDL levels between men who exercise regularly and those who do not (show your work please)?

A)-12.93 to -3.87

B)12.93 to -3.87

C)-12.70 to -4.10

D)12.70 to -4.10

Solution

CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
Where,
x1 = Mean of Sample 1, x2 = Mean of sample2
sd1 = SD of Sample 1, sd2 = SD of sample2
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x1)=48.5
Standard deviation( sd1 )=12.5
Sample Size(n1)=40
Mean(x2)=56.9
Standard deviation( sd2 )=11.9
Sample Size(n1)=125
CI = [ ( 48.5-56.9) ±t a/2 * Sqrt( 156.25/40+141.61/125)]
= [ (-8.4) ± t a/2 * Sqrt( 5.04) ]
= [ (-8.4) ± 2.023 * Sqrt( 5.04) ]
= [-12.93 , -3.87]

[ANSWER] [-12.93 , -3.87]