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

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 :
`(((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))`

`=(((2^2)/(25)+(1^2)/(20))^2)/((((2^2)/(25))^2)/(24)+(((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{(s_1^2)/(n_1)+(s _2^2)/(n_2)}`

`9-8\pm (2.028)\sqrt{(2^2)/(25)+(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