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

Answer:

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

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

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

Step-by-step 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:

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

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:

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

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

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