Question

Find the 99% confidence interval for the difference between two means based on this information about two samples. Assume independent samples from normal populations. (Use conservative degrees of freedom.) (Give your answers correct to two decimal places.)

Sample	Number	Mean	Std. Dev.
1	25	30	32
2	12	25	24

Lower Limit
Upper Limit

Answer 1

Solution:

Confidence interval = (X1bar – X2bar) ± t*sqrt[(S12 / n1)+(S22 / n2)]

We are given

Confidence level = 99%

X1bar = 30

X2bar = 25

S1 = 32

S2 = 24

n1 = 25

n2 = 12

df = Min(25 – 1, 12 – 1) = Min(24, 11) = 11

Critical value = 3.1058

(by using t-table or excel)

(X1bar – X2bar) = 30 – 25 = 5

Confidence interval = (X1bar – X2bar) ± t*sqrt[(S12 / n1)+(S22 / n2)]

Confidence interval = 5 ± 3.1058*sqrt[(32^2 / 25)+(24^2/ 12)]

Confidence interval = 5 ± 3.1058*9.4319

Confidence interval = 5 ± 29.29359502

Lower limit = 5 - 29.2936 = -24.2936

Upper limit = 5 + 29.2936 = 34.2936

Confidence interval = (-24.2936, 34.2936)

Lower limit = -24.29

Upper limit = 34.29