# A Car retailer thinks that a 40,000 mile claim for the tire life by the manufacturer is too high.A Car retailer thinks that a 40,000 mile claim for the tire life by the manufacturer is too high....

A Car retailer thinks that a 40,000 mile claim for the tire life by the manufacturer is too high.

A Car retailer thinks that a 40,000 mile claim for the tire life by the manufacturer is too high. She carefully records the mileage obtained from a sample of 64 such tires. The mean turns out to be 38,500 miles. The standard deviation of the life of all tires of this type has previously been calculated by the manufacturer to be 7,600 miles. Assuming that the mileage is normally distributed, determine the largest significance leave at which we should accept the manufacturer's mileage claim, that is, at which we would not conclude the mileage is significantly less than 40,000 miles.

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The tyre life is assumed to be normally distributed with mean M = 40000 miles and standard deviation 7600 miles.

The mean of sample size n also follows the normal distribution with the same population mean M and a standard deviation equal to population standard deviation/n^(1/2). Therefore the mean of the sample of size 64 should follow a normal distribution with mean M and standard deviation (7600/sqrt 64) miles = 7600/8 miles = 950 miles.

Therefore the distribution of the sample mean m N(M , 950).

Therefore P{ |(m-M)|/950 < C } > p.

Or P{(M-38500) /950 > - C } < p

Therefore for M = 40000, P(M-38000)/950 > 1.5789) = 94.3%.

So 1-p = 5.7%.

So it is not possible to reject the difference of 40000-38500 = 1500 miles if the largest level of significant limit at 5.7%.