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

Answer:

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

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

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

$0.0743 \leq \mu_1 -\mu_2 \leq 1.926$

Step-by-step 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:

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

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:

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

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

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

$0.0743 \leq \mu_1 -\mu_2 \leq 1.926$