Question

In order to estimate the difference between the average hourly wages of employees of two branches of a department store, the following data have been gathered. Downtown StoreNorth Mall Store Sample size2520 Sample mean$12$6 Sample standard deviation$3$1 The point estimate for the difference between the two population means is 6. Find a 95% interval estimate for the difference between the two population means. 0.654 to 2.615 3.958 to 8.042 4.692 to 7.308 5.699 to 8.301

Answer 1

4.692 to 7.308

Explanation:

Given :

n1 = 25 ; n2 = 20 ;

Mean x1 = $12 ; mean x2 = $6

Standard deviation, s1 = 3 ; s2 = 1

Confidence interval :

(x1 - x2) ± t*√(s1²/n1) + (s2²/n2)

x1 - x2 = (12 - 6) = 6

√(s1²/n1) + (s2²/n2) = √(3²/25) + (1²/20) = 0.640

Degree of freedom :

((s1²/n1) + (s2²/n2))² / (s1²/n1)²/n1-1 + (s2²/n2)²/n2-1

Numerator = 0.1681

Denominator = 0.0054 + 0.0001315 = 0.0055315

Degree of freedom = 0.1681 / 0.0055315 = 30.38 = 30

Tcritical at 0.05, df = 30 = 2.042272

Hence ;

6 ± 2.042272(0.640)

6 ± 1.30705408

Lower boundary = 6 - 1.30705408 = 4.69294592

Upper boundary = 6 + 1.30705408 = 7.30705408

(4.693 ; 7.307)