John wants to compare two means. His sample statistics were $\bar { x } _ { 1 } = 22.7 , s _ { 1 } ^ { 2 } = 5.4 , n _ { 1 } = 9$ and $\bar { x } _ { 2 } = 20.5 , s _ { 2 } ^ { 2 } = 3.6 , n _ { 2 } = 9$ . Assuming equal variances, the 95 percent confidence interval for the difference of the population means is approximately:
A)[2.44, 6.19].
B)[1.17, 5.08].
C)[0.08, 4.32].
D)[-0.09, 3.19].

Question

Quizplus · Accepted Answer

The pooled standard deviation is calculated, and then the standard error of the difference in means is found. Using the t-distribution with 16 degrees of freedom, the confidence interval is calculated as [0.08, 4.32].

John Wants to Compare Two Means xˉ1=22.7,s12=5.4,n1=9\bar { x } _ { 1 } = 22.7 , s _ { 1 } ^ { 2 } = 5.4 , n _ { 1 } = 9xˉ1​=22.7,s12​=5.4,n1​=9

John Wants to Compare Two Means $\bar { x } _ { 1 } = 22.7 , s _ { 1 } ^ { 2 } = 5.4 , n _ { 1 } = 9$