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# 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

Accepted Solution

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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 levelConfidence =95% or 0.95The 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.