Question

A sample of 173 students using Method 1 produces a testing average of 88. A sample of 127 students using Method 2 produces a testing average of 53.5. Assume the standard deviation is known to be 12.22 for Method 1 and 13.59 for Method 2. Determine the 95% confidence interval for the true difference between testing averages for students Method 2.

Answer 1

43 < μ₁ - μ₂ < 49

Explanation:

Sample 1:

Mean x₁ = 88

Sample size n₁ = 173

Sample standard deviation s₁ = 12,22

Sample 2:

Mean x₂ = 53,5

Sample size n₂ = 127

Sample standard deviation s₂ = 13,59

CI 95 % α = 5% α = 0,05 α/2 = 0,025

We need to find:

( x₁ - x₂ ) - tα/2,v *√ (s₁²/n₁ ) + (s₂²/n₂) < μ₁ - μ₂ < ( x₁ - x₂ ) + tα/2,v *√ (s₁²/n₁ ) + (s₂²/n₂)

v = degree of fredom

v = [ ( s₁²/n₁ + s₂²/n₂)² / (s₁²/n₁)² /n₁-1 + (s₂²/n₂)²/n₂-1

v = [ (0,86 + 1,45 ) / 0,0043 + 0,017

v = 2,31 / 0,0213

v = 108

Then t 0,025, 108 from t table is: We will take v = 100

t = 1,984

Now

√ (s₁²/n₁ ) + (s₂²/n₂) = √ 0,86 + 1,43

√ 2,3 = 1,51

Then: CI:

(173 - 127 ) - 1,984*1,51 < μ₁ - μ₂ < ( 173 - 127 ) + 1,984*1,51

46 - 3 < μ₁ - μ₂ < 46 + 3

43 < μ₁ - μ₂ < 49