Answer:
Step-by-step explanation:
Hello!
You have two populations of interest and want to compare them. If you define the study variables as:
Xā: average hourly wages of an employee of the Downtown store.
nā= 25
X[bar]ā= $9
Sā= $2
Xā: average hourly wages of an employee of the North Mall store.
nā= 20
X[bar]ā= $8
Sā= $1
Both samples taken are independent, assuming that both populations are normal and that their population variances are equal I'll use the Student's-t statistic with a pooled sample variance to calculate the Confidence interval:
95% CI for Ī¼ā - Ī¼ā
(X[bar]ā-X[bar]ā) Ā± [tex]t_{n_1+n_2-2; 1-\alpha /2} * (Sa*\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } )[/tex]
[tex]Sa^2= \frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2}[/tex]
[tex]Sa^2= \frac{24*4+19*1}{25+20-2}= 2.67[/tex]
Sa= 1.64
[tex]t_{n_1-n_2-2;1-\alpha /2} = t_{43; 0.975} = 2.017[/tex]
(9-8)Ā±2.017*[tex](1.64*\sqrt{\frac{1}{25} +\frac{1}{20} } )[/tex]
[0.007636;1.9923]
I hope it helps!
