Respuesta :
Answer:
[tex] (9-8) -2.02 \sqrt{\frac{2^2}{25} +\frac{1^2}{20}}= 0.0743[/tex]
[tex] (9-8) +2.02 \sqrt{\frac{2^2}{25} +\frac{1^2}{20}}= 1.926[/tex]
And we are 9% confidence that the true mean for the difference of the population means is given by:
[tex] 0.0743 \leq \mu_1 -\mu_2 \leq 1.926[/tex]
Step-by-step explanation:
For this problem we have the following data given:
[tex]\bar X_1 = 9[/tex] represent the sample mean for one of the departments
[tex]\bar X_2 = 8[/tex] represent the sample mean for the other department
[tex]n_1 = 25[/tex] represent the sample size for the first group
[tex]n_2 = 20[/tex] represent the sample size for the second group
[tex]s_1 = 2[/tex] represent the deviation for the first group
[tex]s_2 =1[/tex] represent the deviation for the second group
Confidence interval
The confidence interval for the difference in the true means is 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]
The confidence given is 95% or 9.5, then the significance level is [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. The degrees of freedom are given by:
[tex] df=n_1 +n_2 -2= 20+25-2= 43[/tex]
And the critical value for this case is [tex] t_{\alpha/2}=2.02[/tex]
And replacing we got:
[tex] (9-8) -2.02 \sqrt{\frac{2^2}{25} +\frac{1^2}{20}}= 0.0743[/tex]
[tex] (9-8) +2.02 \sqrt{\frac{2^2}{25} +\frac{1^2}{20}}= 1.926[/tex]
And we are 9% confidence that the true mean for the difference of the population means is given by:
[tex] 0.0743 \leq \mu_1 -\mu_2 \leq 1.926[/tex]
