Question

A national consumer agency selected independent random samples of 45 owners of newer cars (less than five years old) and 40 owners of older cars (more than five years old) to estimate the difference in mean dollar cost of yearly routine maintenance, such as oil changes, tire rotations, filters, and wiper blades. The agency found the mean dollar cost per year for newer cars was $195 with a standard deviation of $46. For older cars, the mean was $286 with a standard deviation of $58. Which of the following represents the 95 percent confidence interval to estimate the difference (newer minus older) in the mean dollar cost of routine maintenance between newer and older cars?
A. (195 - 286) + 1.992 underroot46/45 + 58/40
B. (286 - 195) + 1.992 underroot46^2/45+ 58^2/40
C. (195 - 286)+ 1.992 underrot 140² 158²/45+40
D. (286 - 195) + 1.992 46 1582 45 +40
E. (195-286) +1.992 462 + 58 45 40

Answer 1

Answer:

(195 - 286) ± 1.992 * sqrt((46^2/45) + (58^2/40))

Step-by-step explanation:

Given that:

NEWER CARS:

Sample size = n1 = 45

Standard deviation s1 = 46

Mean = m1 = 195

OLDER CARS:

Sample size = n2 = 40

Standard DEVIATION s2 = 58

Mean = m2 = 286

Confidence interval at 95% ; α = 1 - 0.95 = 0.05 ; 0.05 / 2 = 0.025

Confidence interval is calculated thus : (newer--older)

(m1 - m2) ± Tcritical * standard error

Mean difference = m1 - m2; (195 - 286)

Tcritical = Tn1+n2-2, α/2 = T(45+40)-2 = T83, 0.025 = 1.99 (T value calculator)

Standard error (E) = sqrt((s1²/n1) + (s2²/n2))

E = sqrt((46^2/45) + (58^2/40))

Hence, confidence interval:

(195 - 286) ± 1.992 * sqrt((46^2/45) + (58^2/40))