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:So on this case the 95% confidence interval would be given by [tex]-1235.756 \leq \mu_1 -\mu_2 \leq -1072.245[/tex]. 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[tex]n_1 =87[/tex] represent the sample of us adults with no more than a high school education[tex]n_2 =73[/tex] represent the sample of us adults with no more than a bachelor's degree [tex]\bar x_1 =678[/tex] represent the mean sample of us adults with no more than a high school education[tex]\bar x_2 =1837[/tex] represent the mean sample of us adults with no more than a bachelor's degree [tex]s_1 =197[/tex] represent the sample deviation of us adults with no more than a high school education[tex]s_2 =328[/tex] represent the sample deviation of us adults with no more than a bachelor's degree [tex]\alpha=0.05[/tex] represent the significance levelConfidence =95% or 0.95The confidence interval for the difference of means is given by the following formula:  [tex](\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{(\frac{s^2_1}{n_s}+\frac{s^2_2}{n_s})}[/tex] (1)  The point of estimate for [tex]\mu_1 -\mu_2[/tex] is just given by:  [tex]\bar X_1 -\bar X_2 =678-1837=-1159[/tex]  The appropiate degrees of freedom are [tex]df=n_1+ n_2 -2=158[/tex]Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], 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 [tex]t_{\alpha/2}=1.98[/tex]  The standard error is given by the following formula:  [tex]SE=\sqrt{(\frac{s^2_1}{n_s}+\frac{s^2_2}{n_s})}[/tex]  And replacing we have:  [tex]SE=\sqrt{(\frac{197^2}{87}+\frac{328^2}{73})}=43.816[/tex]  Confidence interval  Now we have everything in order to replace into formula (1):  [tex]-1159-1.98\sqrt{(\frac{197^2}{87}+\frac{328^2}{73})}=-1235.756[/tex]  [tex]-1159+1.98\sqrt{(\frac{197^2}{87}+\frac{328^2}{73})}=-1072.245[/tex]  So on this case the 95% confidence interval would be given by [tex]-1235.756 \leq \mu_1 -\mu_2 \leq -1072.245[/tex]. 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.