A trucking firm suspects that the mean life of a certain tire. it uses is less than 33,000 miles. To check the claim, the firm randomly selects and tests 18 of these tires in gets a mean lifetime of 32, 450 miles with a standard deviation of 1200 miles. At α = 0.05, test the trucking firms claim.a. State Hypothesis and Identify Claim.b. Identify level of significance.c. Choose correct probability distribution, locate critical values.identify rejection region.d. Calculate test statistic.e. Make decisionf. Write conclusion.SHOW ALL YOUR WORK

Answer:We accept the alternate hypothesis. We conclude that the mean lifetime of tires is  is less than 33,000 miles. Step-by-step explanation:We are given the following in the question:  Population mean, μ = 33,000 milesSample mean, [tex]\bar{x}[/tex] =  32, 450 milesSample size, n = 18Alpha, α = 0.05 Sample standard deviation, s = 1200 milesa) First, we design the null and the alternate hypothesis [tex]H_{0}: \mu = 33000\text{ miles}\\H_A: \mu < 33000\text{ miles}[/tex] b) Level of significance:[tex]\alpha = 0.05[/tex]c) We use One-tailed t test to perform this hypothesis. d) Formula: [tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex] Putting all the values, we have [tex]t_{stat} = \displaystyle\frac{32450 - 33000}{\frac{1200}{\sqrt{18}} } = -1.9445[/tex] Now, [tex]t_{critical} \text{ at 0.05 level of significance, 17 degree of freedom } = -1.7396[/tex] Rejection area: [tex]t < -1.7396[/tex]Since,                  [tex]t_{stat} < t_{critical}[/tex] e) We fail to accept the null hypothesis and reject it as the calculated value of t lies in the rejection area. f) We accept the alternate hypothesis. We conclude that the mean lifetime of tires is  is less than 33,000 miles.