Respuesta :
Answer:
[tex] (42700- 36275) -2.374 \sqrt{\frac{2030^2}{41} +\frac{1360^2}{41}}= 5519.071[/tex]
[tex] (42700- 36275) +2.374 \sqrt{\frac{2030^2}{41} +\frac{1360^2}{41}}=7330.929[/tex]
Step-by-step explanation:
For this case we have the following info given:
[tex]n_1 = 41 , \bar X_1 =42700 , s_1 = 2030[/tex]
[tex]n_2 = 41 , \bar X_2 =36375 , s_2 = 1360[/tex]
And for this case we want a 98% confidence interval. The significance would be:
[tex] \alpha= 1-0.98=0.02[/tex]
The degrees of freedom are:
[tex] df = n_1 +n_2 -2= 41+41 -2= 80[/tex]
And the critical value for this case is:
[tex] t_{\alpha/2}= 2.374[/tex]
And the confidence interval would be given by:
[tex](\bar X_1 -\bar X_2) \pm t_{\alpha/2} \sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}[/tex]
And replacing we got:
[tex] (42700- 36275) -2.374 \sqrt{\frac{2030^2}{41} +\frac{1360^2}{41}}= 5519.071[/tex]
[tex] (42700- 36275) +2.374 \sqrt{\frac{2030^2}{41} +\frac{1360^2}{41}}=7330.929[/tex]
