Respuesta :
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 level
Confidence =95% or 0.95
The 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.
