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 Refer to Exhibit 10-7. A 95% interval estimate for the difference between the two population means is

Answer 1

`(9-8) -2.02 \sqrt{(2^2)/(25) +(1^2)/(20)}= 0.0743`

`(9-8) +2.02 \sqrt{(2^2)/(25) +(1^2)/(20)}= 1.926`

And we are 9% confidence that the true mean for the difference of the population means is given by:

`0.0743 \leq \mu_1 -\mu_2 \leq 1.926`

Explanation:

For this problem we have the following data given:

`\bar X_1 = 9` represent the sample mean for one of the departments

`\bar X_2 = 8` represent the sample mean for the other department

`n_1 = 25` represent the sample size for the first group

`n_2 = 20` represent the sample size for the second group

`s_1 = 2` represent the deviation for the first group

`s_2 =1` represent the deviation for the second group

Confidence interval

The confidence interval for the difference in the true means is given by:

`(\bar X_1 -\bar X_2) \pm t_(\alpha/2) \sqrt{(s^2_1)/(n_1)+(s^2_2)/(n_2)}`

The confidence given is 95% or 9.5, then the significance level is
`\alpha=0.05` and
`\alpha/2 =0.025`. The degrees of freedom are given by:

`df=n_1 +n_2 -2= 20+25-2= 43`

And the critical value for this case is
`t_(\alpha/2)=2.02`

And replacing we got:

`(9-8) -2.02 \sqrt{(2^2)/(25) +(1^2)/(20)}= 0.0743`

`(9-8) +2.02 \sqrt{(2^2)/(25) +(1^2)/(20)}= 1.926`

And we are 9% confidence that the true mean for the difference of the population means is given by:

`0.0743 \leq \mu_1 -\mu_2 \leq 1.926`