Question

he manager of a fleet of automobiles is testing two brands of radial tires and assigns one tire of each brand at random to the two rear wheels of eight cars and runs the cars until the tires wear out. The data (in kilometers) follow. Find a 99% confidence interval on the difference in the mean life.

Car	Brand 1	Brand 2
1	36,925	34,318
2	45,300	42,280
3	36,239	35,524
4	32,100	31,950
5	37,210	38,015
6	48,360	47,800
7	38,200	37,810
8	33,500	33,215

Answer 1

Mean of difference and standard deviation of difference was calculated using MS Excel.

Functions that were used to calculate mean and std. dev. are shown below:

Following is the result:

Confidence Interval Calculation

Step 1: Find α/2
Level of Confidence = 99%
α = 100% - (Level of Confidence) = 1%
α/2 = 0.5% = 0.005

Step 2: Find tα/2
Calculate tα/2 by using t-distribution with degrees of freedom (DF) as n - 1 = 8 - 1 = 7 and α/2 = 0.005 as right-tailed area and left-tailed

tα/2 = 3.4992

Step 3: Calculate Confidence Interval

Lower Bound = d̄ - tα/2•(sd/√n) = 865.25 - (3.4992)(1290.4187/√8) = -731.1967
Upper Bound = d̄ + tα/2•(sd/√n) = 865.25 + (3.4992)(1290.4187/√8) = 2461.6967

A 99% confidence interval on the difference in the mean life is (-731.20,2461.70) - Answer rounded to 2 decimal places