A bearing used in an automotive application is required to have a nominal inside diameter of 1.5 inches. A random sample of 25 bearings is selected and the average inside diameter of these bearings is 1.4975 inches. Bearing diameter is known to be normally distributed with standard deviation inch. (a) Test the hypotheses versus using The true mean hole diameter significantly different from 1.5 in. at alpha equals 0.01. (b) What is the P-value for the test in part (a)

Answer:Step-by-step explanation:In the question, there is an incomplete sentence. The complete sentence is:"Bearing diameter is known to be normally distributed with standard deviation of 0.01 inch"We would set up the hypothesis test. For the null hypothesis, H0 : µ = 1.5For the alternative hypothesis, H1 : µ ≠ 1.5It is a two tailed testSince the population standard deviation is given, z score would be determined from the normal distribution table. The formula is z = (x - µ)/(σ/√n)Wherex = sample mean µ = population mean σ = population standard deviationn = number of samplesFrom the information given,µ = 1.5x = 1.4975σ = 0.01n = 25z = (1.4975 - 1.5)/(0.01/√25) = - 1.25Since it is a 2 tailed test, we would find the p value by doubling the area to the left of the z score to include the area to right. Area to the left from the normal distribution table is 0.11P value = 0.11 × 2 = 0.22Assuming a significance level of 5%Since 0.05 < 0.22, we would accept the null hypothesisTherefore, at 5% significance level, we can conclude that the true mean hole diameter is 1.5 inches