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

1.09 to 4.08

0.07 to 1.9

1.08 to 2.92

1.9 to 2.08

Answer 1

Answer: 0.07 to 1.9

Step-by-step explanation:

Let $\mu_1$ and $\mu_2$ be the population mean of hourly wages of employees of two branches of a department store (Downtown Store and  North Mall Store)

As per given , we have

$n_1=25\\\\ n_2=20$

$\overline{x}_1=\ 9\\\\\overline{x}_2=\ 8$

$s_1=\ 2\\\\s_1=\ 1$

Degree of freedom : $\dfrac{(\dfrac{s_1^2}{n_1}+\dfrac{s _2^2}{n_2})^2}{\dfrac{(\dfrac{s_1^2}{n_1})^2}{n_1-1}+\dfrac{(\dfrac{s_2^2}{n_2})^2}{n_2-1}}$

$=\dfrac{(\dfrac{2^2}{25}+\dfrac{1^2}{20})^2}{\dfrac{(\dfrac{2^2}{25})^2}{24}+\dfrac{(\dfrac{1^2}{20})^2}{19}} =36$

Critical value for 95% confidence interval :

$t_{\alpha/2, df}=t_{0.025,\ 36}=2.028$

Then, 95% interval estimate for the difference between the two population means will be

$\overline{x}_1-\overline{x}_2\pm z_{\alpha/2}\sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s _2^2}{n_2}}$

$9-8\pm (2.028)\sqrt{\dfrac{2^2}{25}+\dfrac{1^2}{20}}$

$1\pm 0.93=(1-0.93,\ 1+0.93 ) = (0.07,\ 1.93)$

Hence, A 95% interval estimate for the difference between the two population means is $(0.07, 1.93)$ .

i.e. 0.07 to 1.9