Question

The General Social Survey described included random samples from two groups: US residents with a college degree and US residents without a college degree. For the 505 sampled US residents with a college degree, the average number of hours worked each week was 41.8 hours with a standard deviation of 15.1 hours. For those 667 without a degree, the mean was 39.4 hours with a standard deviation of 15.1 hours.
Required:

Conduct a hypothesis test to check for a difference in the average number of hours worked for the two groups.

Answer 1

Answer:

$t=\frac{(41.8-39.4)-0}{\sqrt{\frac{15.1^2}{505}+\frac{15.1^2}{667}}}}=2.695$  

Since is a bilateral test the p value would be:  

$p_v =2*P(t_{1170}>2.695)=0.007$  

We have a very low p value so then we have enough evidence to conclude that the two population means are significantly different at a significance level of 1% for example.

Step-by-step explanation:

Data given

$\bar X_{1}=41.8$ represent the mean for sample 1  college degree

$\bar X_{2}=39.4$ represent the mean for sample 2  non college degree

$s_{1}=15.1$ represent the sample standard deviation for 1  

$s_{f}=15.1$ represent the sample standard deviation for 2  

$n_{1}=505$ sample size for the group 2  

$n_{2}=667$ sample size for the group 2  

t would represent the statistic (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the difference in the population means, the system of hypothesis would be:  

Null hypothesis:$\mu_{1}-\mu_{2}=0$  

Alternative hypothesis:$\mu_{1} - \mu_{2}\neq 0$  

We don't have the population standard deviation's, so for this case is better apply a t test to compare means, and the statistic is given by:  

$t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}$ (1)  

And the degrees of freedom are given by $df=n_1 +n_2 -2=505+667-2=1170$  

Statistic

With the info given we can replace in formula (1) like this:  

$t=\frac{(41.8-39.4)-0}{\sqrt{\frac{15.1^2}{505}+\frac{15.1^2}{667}}}}=2.695$  

P value  

Since is a bilateral test the p value would be:  

$p_v =2*P(t_{1170}>2.695)=0.007$  

We have a very low p value so then we have enough evidence to conclude that the two population means are significantly different at a significance level of 1% for example.