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.

Step-by-step explanation:

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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

the answer is (4,3)

since the rule for r0, 90º is (-y,x) and it was already applied to form the image, all you have to do is reverse the answer using this formula.

the results from a pre-test for students for the year 2000 and the year 2010 are illustrated in the box plot. what do these results tell us about how students performed on the 29 question pre-test for the two years?

if we compare only the lowest and highest scores between the two years, we might conclude that the students in 2010 did better than the students in 2010. this conclusion seems to follow since the lowest score of 8 in 2010 is greater in value than the lowest score of 6 in 2000. also, the highest score of 28 in 2010 is greater in value than the highest score of 27 in 2000.

but the box portion of the illustration gives us more detailed information. the middle bar in each box shows us that the median score of 20 in 2000 is greater in value than the median score of 17 in 2010. further, we note that the box and whiskers divide the illustration into four pieces. each of these four pieces represents the same portion of students. so, the upper half of the students in 2000 scored in the same score range as the upper one-fourth of the students in 2010, see the illustration at a score of 20.

by considering the upper one-fourth, upper half, and upper three-fourths instead of just the lowest and highest scores, we would conclude that the students as a whole did much better in 2000 than in 2010. we would conclude that as a whole the students in 2010 are less prepared than the students in 2000.

in this section, we discuss box-and-whisker plots and the five key values used in constructing a box-and-whisker plot. the key values are called a five-number summary, which consists of the minimum, first quartile, median, third quartile, and maximum.

step-by-step explanation: