Respuesta :
Answer:
[tex] (21.9-19.8) -1.966\sqrt{\frac{3.4^2}{200} +\frac{3.5^2}{200}}= 1.422[/tex]
[tex] (21.9-19.8) -1.966 \sqrt{\frac{3.4^2}{200} +\frac{3.5^2}{200}}= 2.778[/tex]
And we are 95% confident that the true difference means are between [tex] 1.422 \leq \mu_1 -\mu_2 \leq 2.778[/tex]
Step-by-step explanation:
We know the following info:
[tex]\bar X_1 = 21.9[/tex] sample mean for group 1
[tex]\bar X_2 = 19.8[/tex] sample mean for group 2
[tex]s_1 = 3.4[/tex] sample standard deviation for group 1
[tex]s_2 = 3.5[/tex] sample standard deviation for group 2
[tex]n_1 = 200[/tex] sample size group 1
[tex]n_2 = 200[/tex] sample size group 2
We want to find a confidence interval for the difference of means and the correct formula to do this 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]
Now we just need to find the critical value. The confidence level is 0.95 then the significance is [tex]1-0.95 =0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. The degrees of freedom are given by:
[tex]df= n_1 +n_2 -2= 200+200-2= 398[/tex]
The critical value for this case would be :[tex] t_{\alpha/2}=1.966[/tex]
And replacing into the confidence interval formula we got:
[tex] (21.9-19.8) -1.966\sqrt{\frac{3.4^2}{200} +\frac{3.5^2}{200}}= 1.422[/tex]
[tex] (21.9-19.8) -1.966 \sqrt{\frac{3.4^2}{200} +\frac{3.5^2}{200}}= 2.778[/tex]
And we are 95% confident that the true difference means are between [tex] 1.422 \leq \mu_1 -\mu_2 \leq 2.778[/tex]
