In order to estimate the difference between the average hourly wages of employees of two branches of a department store, two independent random samples were selected and the following statistics were calculated. Downtown Store - North Mall Store Sample size: 25 - 20 Sample mean: $9 - $8 Sample standard deviation: $2 - $1A 95% interval estimate for the difference between the two population means is
a. 1.09 to 4.08
b. 0.07 to 1.9
c. 1.08 to 2.92
d. 1.9 to 2.08

Respuesta :

Answer:

Step-by-step explanation:

Hello!

You have two populations of interest and want to compare them. If you define the study variables as:

X₁: average hourly wages of an employee of the Downtown store.

n₁= 25

X[bar]₁= $9

S₁= $2

X₂: average hourly wages of an employee of the North Mall store.

n₂= 20

X[bar]₂= $8

S₂= $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 μ₁ - μ₂

(X[bar]₁-X[bar]₂) ± [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)±2.017*[tex](1.64*\sqrt{\frac{1}{25} +\frac{1}{20} } )[/tex]

[0.007636;1.9923]

I hope it helps!