A taxi company is trying to decide whether to purchase brand A or brand B tires for its fleet of taxis. To estimate the difference in the two brands, an experiment is conducted using 8 of each brand, assigned at random to the left and right rear wheels of 8 taxis. The tires are run until they wear out and the distances, in kilometers, are recorded in the accompanying data set. Find a 95% confidence interval for mu1minusmu2. Assume that the differences of the distances are approximately normally distributed.

A taxi company is trying to decide whether to purchase brand A or brand B tires for its fleet of taxis. To estimate the difference in the two brands, an experiment is conducted using 8 of each brand, assigned at random to the left and right rear wheels of 8 taxis. The tires are run until they wear out and the distances, in kilometers, are recorded in the accompanying data set. Find a 95% confidence interval for Hx - Hy. Assume that the differences of the distances are approximately normally distributed Data Set Let, Hx be the population mean for brand A and let Hy be the population mean for brand B. The confidence interval is -42- (Round to one decimal place as needed.) Taxi Brand A 37,700 46,800 36,500 40,600 43.600 31,300 37,600 43.800 Brand B 39,800 47,500 37.100 39,200 42,900 36,000 38,700 45,000

Solution:

For brand A,

Mean,x1 = (37700 + 46800 + 36500 + 40600 + 43600 + 31300 + 37600 + 43800)/8 = 39737.5

Standard deviation = √(summation(x - mean)²/n

Summation(x - mean)² = (37700 - 39737.5)^2 + (46800 - 39737.5)^2 + (36500 - 39737.5)^2 + (40600 - 39737.5)^2 + (43600 - 39737.5)^2 + (31300 - 39737.5)^2 + (37600 - 39737.5)^2 + (43800 - 39737.5)^2 = 172438750

Standard deviation = √(172438750/8

s1 = 4642.72

For brand B,

Mean,x2 = (39800 + 47500 + 37100 + 39200 + 42900 + 36000 + 38700 + 45000)/8 = 40775

Standard deviation = √(summation(x - mean)²/n

Summation(x - mean)² = (39800 - 40775)^2 + (47500 - 40775)^2 + (37100 - 40775)^2 + (39200 - 40775)^2 + (42900 - 40775)^2 + (36000 - 40775)^2 + (38700 - 40775)^2 + (45000 - 40775)^2 = 111635000

Standard deviation = √(111635000/8

s2 = 3735.56

The formula for determining the confidence interval for the difference of two population means is expressed as

Confidence interval = (x1 - x2) ± z√(s²/n1 + s2²/n2)

For a 95% confidence interval, we would determine the z score from the t distribution table because the number of samples are small

Degree of freedom =

(n1 - 1) + (n2 - 1) = (8 - 1) + (8 - 1) = 14

z = 2.145

x1 - x2 = 39737.5 - 40775 = -1037.5

z√(s1²/n1 + s2²/n2) = 2.145√(4642.72²/8 + 3735.56²/8)

= 4519.1

95% Confidence interval = - 1037.5 ± 4519.1