Question

In order to estimate the difference between the average hourly wages of employees of two branches of a department store, the following data have been gathered. Downtown Store North Mall Store Sample size 25 20 Sample mean $9 $8 Sample standard deviation $2 $1 A 95% interval estimate for the difference between the two population means is a. .071 to 1.929. b. 1.09 to 4.078. c. 1.078 to 2.922. d. .226 to 1.774.

Answer 1

`0.071,1.928`

Explanation:

Downtown Store North Mall Store

Sample size n 25 20

Sample mean
`\bar{x}` $9 $8

Sample standard deviation s $2 $1

`n_1=25\\n_2=20`

`\bar{x_1}=9\\ \bar{x_2}=8`

`s_1=2\\s_2=1`

`x_1-x_2=9-8=1`

Standard error of difference of means =
`\sqrt{(s_1^2)/(n_1)+(s_2^2)/(n_2)}`

Standard error of difference of means =
`\sqrt{(2^2)/(25)+(1^2)/(20)}`

Standard error of difference of means =
`0.458`

Degree of freedom =
`\frac{\sqrt{((s_1^2)/(n_1)+(s_2^2)/(n_2)})^2}{(((s_1^2)/(n_1))^2)/(n_1-1)+(((s_2^2)/(n_2))^2)/(n_2-1)}`

Degree of freedom =
`\frac{\sqrt{((2^2)/(25)+(1^2)/(20)})^2}{(((2^2)/(25))^2)/(25-1)+(((1^2)/(20))^2)/(20-1)}`

Degree of freedom =36

So, z value at 95% confidence interval and 36 degree of freedom = 2.0280

Confidence interval =
`(x_1-x_2)-z * SE(x_1-x_2),(x_1-x_2)+z * SE(x_1-x_2)`

Confidence interval =
`1-(2.0280)* 0.458,1+(2.0280)* 0.458`

Confidence interval =
`0.071,1.928`

Hence Option A is true

Confidence interval is
`0.071,1.928`