Question

In a survey of 100 U.S. residents with a high school diploma as their highest educational degree (Group 1) had an average yearly income was $35,621. Another 120 U.S. residents with a GED (Group 2) had an average yearly income of $33,498. The population standard deviation for both populations is known to be $4,310. At a 0.01 level of significance, can it be concluded that U.S. residents with a high school diploma make significantly more than those with a GED

Answer 1

Answer:

$z=\frac{(35621-33498)-0}{\sqrt{\frac{4310^2}{100}+\frac{4310^2}{120}}}}=3.637$  

$p_v =P(z>3.637)=0.000138$  

Comparing the p value with the significance level $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and the mean for the US residents with High school diplome is significantly higher than those with GED.  

Step-by-step explanation:

Data given and notation

$\bar X_{1}=35621$ represent the mean for sample 1  

$\bar X_{2}=33498$ represent the mean for sample 2  

$\sigma_{1}=4310$ represent the population standard deviation for 1  

$\sigma_{2}=4310$ represent the population standard deviation for 2

$n_{1}=100$ sample size for the group 2  

$n_{2}=120$ sample size for the group 2  

$\alpha=0.01$ Significance level provided

z would represent the statistic (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the mean for US residents (sample 1) is higher than the mean for sample 2, the system of hypothesis would be :  

Null hypothesis:$\mu_{1}-\mu_{2}\leq0$  

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

We have the population standard deviation's, so for this case is better apply a z test to compare means, and the statistic is given by:  

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

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.  

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

$z=\frac{(35621-33498)-0}{\sqrt{\frac{4310^2}{100}+\frac{4310^2}{120}}}}=3.637$  

P value  

Since is a right tailed test the p value would be:  

$p_v =P(z>3.637)=0.000138$  

Comparing the p value with the significance level $\alpha=0.01$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and the mean for the US residents with High school diplome is significantly higher than those with GED.