Find a 90% confidence interval for the difference μ1−μ2 of the means (rounded to 4 decimal places)

Question

Two random samples are selected from two independent populations. A summary of the samples sizes, sample means, and sample standard deviations is given below:

n1=20,

n2=40,

x¯1=58.8

,x¯2=71.7

,s1=5.7

s2=10.5

Answer 1

Given: n₁=20; xbar₁=58.8; s₁=5.7

Given: n₂=40; xbar₂=71.7; s₂=10.5

By "Student's" t-Distribution: Compute Pooled Variance σ-cap² equal to [(20−1)5.7² + (40−1)10.5²] ÷ [20+40−2] or 84.77689655. The Standard Error Of The Difference Between Means based on use of

σ-cap² is σ-cap_{(xbar1-xbar2)} or √[(84.77689655/20)+(84.77689655/40)] or 2.521560477.

The 90% Confidence Interval is then given by (58.8−71.7) ± (Two-Tailed t with 58 Degrees Of Freedom At (1−0.90) or 0.10 Level Of Significance = 1.6716 by Online Calculator for T-Values) × (The Standard Error Of The Difference Between Means = 2.521560477), yielding -17.11504049 and -8.684959507 and equivalent to -17.1150 and -8.6850.

It can then be stated with 90% confidence that xbar₂ is somewhere between 8.6850 and 17.1150 more than xbar₁.