Respuesta :
Answer:
[tex] (14.5-13.9) -1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = -1.63[/tex]
[tex] (14.5-13.9) +1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = 2.83[/tex]
And the confidence interval for this case [tex] -1.63 \leq \mu_1 -\mu_2 \leq 2.83[/tex]
Step-by-step explanation:
We know the following info from the problem
[tex]\bar X_1 = 14.5[/tex] sample mean for the group 1
[tex]s_1 = 3.98[/tex] the standard deviation for the group 1
[tex] n_1= 20[/tex] the sample size for group 1
[tex]\bar X_2 = 13.9[/tex] sample mean for the group 2
[tex]s_1 = 4.03[/tex] the standard deviation for the group 2
[tex] n_2= 17[/tex] the sample size for group 2
We have all the conditions satisifed since we have random samples.
We want to construct a confidence interval for the true difference of means and the correct formula for this case is:
[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]
The degrees of freedom are given :
[tex] df = n_1 +n_2- 2 = 20+17-2=35[/tex]
The confidence level is 0.9 or 90% and the significance level is [tex]\alpha=1-0.9=0.1[/tex] and [tex]\alpha/2 = 0.05[/tex] and the critical value for this case is:
[tex] t_{\alpha/2} = 1.69[/tex]
And replacing the info given we got:
[tex] (14.5-13.9) -1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = -1.63[/tex]
[tex] (14.5-13.9) +1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = 2.83[/tex]
And the confidence interval for this case [tex] -1.63 \leq \mu_1 -\mu_2 \leq 2.83[/tex]
