A tire manufacturer has been producing tires with an average life expectancy of 26,000 miles. Now the company is advertising that its new tires' life expectancy has increased. In order to test the legitimacy of the advertising campaign, an independent testing agency tested a sample of 6 of their tires and has provided the following data.

Answer:a) Mean = 27 (in thousand miles) = 27,000 milesStandard deviation = 1.5275 (in thousand miles) = 1527.5 milesb) 99% Confidence interval = (24.49, 29.51) in thousand milesThis range of values constitute the values that the life expectancy for all of the company's tires can take on. Since values that are lower than the old life expectancy are in this range, then the company isn't using the right advertisement. c) p-value = 0.085248 > significance level.Hence, we fail to reject the null hypothesis & say that there isn't enough evidence to conclude that the life expectancy for the tires is more than the old life expectancy of 26,000 miles.More proof that the company isn't using the right advertisement!Step-by-step explanation:Life Expectancy (In Thousands of Miles)282725282925a) Mean = (Σx)/N = (28+27+25+28+29+25)/6= 27 (in thousand miles)= 27,000 miles.Standard deviation = σ = √[Σ(x - xbar)²/N]Σ(x - xbar)² = (28-27)² + (27-27)² + (25-27)² + (28-27)² + (29-27)² + (25-27)² = 14N = 6σ = √(14/6) = 1.5275 (in thousand miles) = 1527.5 milesb) Confidence Interval for the population mean is basically an interval of range of values where the true population mean can be found with a certain level of confidence.Mathematically,Confidence Interval = (Sample mean) ± (Margin of error)Sample Mean = 27 (in thousand miles)Margin of Error is the width of the confidence interval about the mean.It is given mathematically as,Margin of Error = (Critical value) × (standard Error of the mean)Critical value will be obtained using the t-distribution. This is because there is no information provided for the population mean and standard deviation.To find the critical value from the t-tables, we first find the degree of freedom and the significance level.Degree of freedom = df = n - 1 = 6 - 1 = 5.Significance level for 99% confidence interval(100% - 99%)/2 = 0.5% = 0.005t (0.005, 5) = 4.032 (from the t-tables)Standard error of the mean = σₓ = (σ/√n)σ = standard deviation of the sample = 1.5275n = sample size = 6σₓ = (1.5275/√6) = 0.623599% Confidence Interval = (Sample mean) ± [(Critical value) × (standard Error of the mean)]CI = 27 ± (4.032 × 0.6235)CI = 27 ± 2.51395299% CI = (24.486048, 29.513952)99% Confidence interval = (24.49, 29.51) in thousand milesThis range of values constitute the values that the life expectancy for all of the company's tires can take on. Since values that are lower than the old life expectancy are in this range, then the company isn't using the right advertisement. c) For hypothesis testing, the first thing to define is the null and alternative hypothesis.The null hypothesis plays the devil's advocate and usually takes the form of the opposite of the theory to be tested. It usually contains the signs =, ≤ and ≥ depending on the directions of the test.While, the alternative hypothesis usually confirms the the theory being tested by the experimental setup. It usually contains the signs ≠, < and > depending on the directions of the test.For this question, we want to prove that the life expectancy for the new tires is more than 26,000 miles.Hence, the null hypothesis is that there isn't significant evidence to say that the life expectancy for the new tires is more than 26,000 miles.The alternative hypothesis will now be that there is significant evidence to say that the life expectancy for the new tires is more than 26,000 miles.Mathematically,The null hypothesis is represented asH₀: μ ≤ 26,000 milesThe alternative hypothesis is represented asHₐ: μ > 26,000 miles To do this test, we will use the t-distribution because no information on the population standard deviation is knownSo, we compute the t-test statistict = (x - μ₀)/σₓx = sample mean = 27μ₀ = 26σₓ = standard error = 0.6235t = (27 - 26) ÷ 0.6235t = 1.60checking the tables for the p-value of this t-statisticDegree of freedom = df = n - 1 = 6 - 1 = 5Significance level = 0.01This is because the hypothesis test uses a one-tailed condition because we're testing only in one direction.p-value (for t = 1.60, at 0.01 significance level, df = 5, with a one tailed condition) = 0.085248The interpretation of p-values is thatWhen the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.So, for this question, significance level = 0.01p-value = 0.0852480.085248 > 0.01Hence,p-value > significance levelThis means that we fail to reject the null hypothesis & say that there isn't enough evidence to conclude that the life expectancy for the tires is more than the old life expectancy of 26,000 miles.More proof that the company isn't using the right advertisement. Hope this Helps!!!