Question

From a random sample of 87 us adults with no more than a high school education that mean weekly income is 678 with a sample standard deviation of 197 from another random sample of 73 us adults with no more than a bachelor's degree the mean weekly income is 1837 with a standard deviation of $328 construct pain 95% confidence interval for the mean difference in the weekly income levels between us adults with no more than a high-school diploma and those with no more than a bachelor's degree there are a hundred 13 degrees of freedom in the appropriate probability distribution

Answer 1

Answer:

So on this case the 95% confidence interval would be given by $-1235.756 \leq \mu_1 -\mu_2 \leq -1072.245$. So we can conclude that we have a significant difference, with the mean of population two higher than the mean for the population 1, because the interval just contains negative values.  

Step-by-step explanation:

Notation and previous concepts

$n_1 =87$ represent the sample of us adults with no more than a high school education

$n_2 =73$ represent the sample of us adults with no more than a bachelor's degree

$\bar x_1 =678$ represent the mean sample of us adults with no more than a high school education

$\bar x_2 =1837$ represent the mean sample of us adults with no more than a bachelor's degree

$s_1 =197$ represent the sample deviation of us adults with no more than a high school education

$s_2 =328$ represent the sample deviation of us adults with no more than a bachelor's degree

$\alpha=0.05$ represent the significance level

Confidence =95% or 0.95

The confidence interval for the difference of means is given by the following formula:  

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{(\frac{s^2_1}{n_s}+\frac{s^2_2}{n_s})}$ (1)  

The point of estimate for $\mu_1 -\mu_2$ is just given by:  

$\bar X_1 -\bar X_2 =678-1837=-1159$  

The appropiate degrees of freedom are $df=n_1+ n_2 -2=158$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,158)".And we see that $t_{\alpha/2}=1.98$  

The standard error is given by the following formula:  

$SE=\sqrt{(\frac{s^2_1}{n_s}+\frac{s^2_2}{n_s})}$  

And replacing we have:  

$SE=\sqrt{(\frac{197^2}{87}+\frac{328^2}{73})}=43.816$  

Confidence interval  

Now we have everything in order to replace into formula (1):  

$-1159-1.98\sqrt{(\frac{197^2}{87}+\frac{328^2}{73})}=-1235.756$  

$-1159+1.98\sqrt{(\frac{197^2}{87}+\frac{328^2}{73})}=-1072.245$  

So on this case the 95% confidence interval would be given by $-1235.756 \leq \mu_1 -\mu_2 \leq -1072.245$. So we can conclude that we have a significant difference with the mean of population two higher than the mean for the population 1, because the interval contains just negative values.