A tire company guarantees that a particular brand of tire has a mean useful lifetime miles or more. A consumer testing agency sampled n = 10 tires on a test wheel that simulated conditions. The sample mean lifetimes (in thousands of miles) were as follows: 42 36 46 43 41 35 43 45 40 39 Construct a 98% CI for the mean lifetime of this particular brand of tire.

Answer:Step-by-step explanation:Number of sampled tyres = 10To determine the mean, we will divide the total number of miles by the total number of tyres.Mean, u = (42+36+46+43+41+35+43+45+40+39)/10 = 410/10 = 41Standard deviation = √[summation(u - ub)^2]/nub = deviation from the mean, uSummation(u - ub)^2] = (42-41)^2 + (36-41)^2 + (46-41)^2 +(43-41)^2 + (41-41)^2 + (35-41)^2 + (43-41)^2 + (45 -41)^2 + (40-41)^2 + (39-41)^2= 1 + 25 + 25 + 4 + 0 + 36 + 4 + 16 + 1 + 4 = 116Standard deviation = √116/10 = √11.6= 3.41We want to determine a 98% confidence interval for the mean mean lifetime of a particular brand of tire.For a confidence level of 98%, the corresponding z value is 2.33. This is determined from the normal distribution table.We will apply the formulaConfidence interval= mean +/- z ×standard deviation/√nIt becomes 41 +/- 2.33 × 3.41/√10= 41 +/- 2.33 × 10.78= 41 +/- 25.11The lower end of the confidence interval is 41 - 25.11 =15.89The upper end of the confidence interval is 41 + 25.11 = 66.11Therefore, with 98% confidence interval, the mean lifetime of a particular brand of tire is between 15.89 miles and 66.11 miles