Question

The mean weekly earnings of a sample of 30 construction workers was $759, with a standard deviation of $73, and the mean weekly earnings of a sample of 28 manufacturing workers was $658, with a standard deviation of $65. Construct a 85% confidence interval for the difference between the mean weekly earnings for construction workers and the mean weekly earnings for manufacturing workers. Round to the second decimal place.

Answer 1

`(759-658) -1.46 \sqrt{(73^2)/(30) +(65^2)/(28)}= 74.54`

`(759-658) +1.46 \sqrt{(73^2)/(30) +(65^2)/(28)}= 127.46`

And the confidence interval for the difference of the two means is given by (74.54, 127.46)

Explanation:

Information given:

`\bar X_1 = 759` the sample mean for construction workers

`s_1 =73` the sample standard deviation for construction workers

`n_1 =30` sample size of construction workers

`\bar X_2 = 658` the sample mean for manufacturing workers

`s_2 =65` the sample standard deviation for manufacturing workers

`n_2 =28` sample size of construction workers

Confidence interval

The confidence interval for the difference of means are given by:

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

We need to find the degrees of freedom given by:

`df = n_1 +n_2 -2= 30+28-2=56`

The confidence is 0.85 and the significance level would be
`1-0.85=0.15` and
`\alpha/2 = 0.075`. We need to find a critical value in the t distribution who accumulates 0.075 of the area on each tail and we got:

`t_(\alpha/2)= \pm 1.46`

And then we can replace and we got:

`(759-658) -1.46 \sqrt{(73^2)/(30) +(65^2)/(28)}= 74.54`

`(759-658) +1.46 \sqrt{(73^2)/(30) +(65^2)/(28)}= 127.46`

And the confidence interval for the difference of the two means is given by (74.54, 127.46)