Respuesta :
Answer:
[tex] (52-49) -4.300= -1.300[/tex]
[tex] (52-49) +4.300= 7.300[/tex]
And the 95% confidence would be :
[tex]-1.300 \leq \mu_1 -\mu_2 \leq 7.300[/tex]
Step-by-step explanation:
We have the following info given from the problem
[tex]\bar X_1 = 52[/tex] sample mean for this year
[tex]s_1= 13[/tex] sample deviation for this year
[tex]n_1 = 75[/tex] random sample selected for this year
[tex]\bar X_2 = 49[/tex] sample mean for last year
[tex]s_2= 11[/tex] sample deviation for last year
[tex]n_1 = 53[/tex] random sample selected for last year
And we want to construct a 95% confidence interval estimate of the difference μ1−μ2, where μ1 is the mean of this year's sales and μ2 is the mean of last year's sales
For this case the formula that we need to use 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 by:
[tex] df= n_1 +n_2 -2 = 75+53-2= 126[/tex]
The confidence level is 0.95 and the significance would be [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex] so then the critical value for this case is :
[tex]t_{\alpha/2}= 1.979[/tex]
The margin of error would be:
[tex] ME = 1.979 \sqrt{\frac{13^2}{75} +\frac{11^2}{49}}= 4.300[/tex]
And the confidence interval would be given by:
[tex] (52-49) -4.300= -1.300[/tex]
[tex] (52-49) +4.300= 7.300[/tex]
And the 95% confidence would be :
[tex]-1.300 \leq \mu_1 -\mu_2 \leq 7.300[/tex]
