Question

Samples from two independent, normally-distributed populations produced the following results.

Population 1 Population 2
Sample size 7 9
Sample mean 15.9 12.6
Sample standard deviation 10.2 13.4

Calculate the 95% confidence interval for the difference between population means μ1-μ2

a. 1.889
b. 8.6
c. 1.128
d. 1.286

Answer 1

Answer: The 95% confidence interval for the difference between population means μ1-μ2 is
`(-9.36,\ 15.96)` .

Explanation:

Given : Samples from two independent, normally-distributed populations produced the following results.

Population 1 Population 2

Sample size 7 9

Sample mean 15.9 12.6

Sample standard deviation 10.2 13.4

The confidence interval for the difference between population means μ1-μ2 is given by :-

`\overline{X_1}-\overline{X_2}\pm t^*\sqrt{(s_1^2)/(n_1)+(s_2^2)/(n_2)}`

, where
`n_1` = sample size from population 1

`n_1` = sample size from population 2

`\overline{X_1}-\overline{X_2}` = Difference between sample mean of two population

`s_1`= Sample standard deviation of population 1.

`s_2`= Sample standard deviation of population 2.

t* = Critical value for
`df=n_1+n_2-2` and significance
`\alpha/2`.

As per given :

`n_1=7`
`n_2=9`

df = 7+9-2=14

`\overline{X_1}-\overline{X_2}=15.9-12.6=3.3`

`s_1=10.2`
`s_2=13.4`

`\alpha=1-0.95=0.05`

Critical t-value :
`t_(df, \alpha/2)=t_(14, 0.025)=2.145`

So , the 95% confidence interval for the difference between population means μ1-μ2 would be

`3.3\pm (2.145)\sqrt{(10.2^2)/(7)+(13.4^2)/(9)}`

`3.3\pm (2.145)√(14.86+19.95)`

`3.3\pm (2.145)√(34.81)`

`3.3\pm (2.145)(5.9)`

`3.3\pm 12.66`

`(3.3-12.66,\ 3.3+12.66)=(-9.36,\ 15.96)`

Hence, the 95% confidence interval for the difference between population means μ1-μ2 :
`(-9.36,\ 15.96)`