Question

A consumer organization wants to estimate the actual treadwear index of a brand name of tires that claims “graded 150” on the sidewall of the tire. A random sample of n=15 indicate a sample mean treadwear index of 141.8 and a sample standard deviation of 19.3.

a. Assuming that the population of treadwear indexes is normally distributed, construct a 90% confidence interval estimate of the population mean treadwear index for tires produced by this manufacturer under this brand name. (Round two decimal places

b. Do you think that the consumer organization should accuse the manufacturer of producing tires that do not need the performance information on the sidewall of the tire?

c. Explain why and observe treadwear index of 206 for particular tire is not an usual, even though it is outside of the confidence interval developed in (a)

Answer 1

A random sample of n=15 indicate a sample mean treadwear index of xbar =141.8 and a sample standard deviation of s= 19.3.

90% confidence interval of population mean

Xbar +/- [ tcritical * ( s / sqrt ( n ) ) ]

141.8 +/- [ 1.7613 * ( 19.3 / sqrt ( 15 ) ) ]

133.02 < mu < 150.58

No, there is no grounds to allegate wrong-doing because the 90% confidence interval contains 150, which means that the sample mean is not statistically significantly different from the claimed tread wear index.

Comparing means by z and t distributions is an endeavor to compute probabilities; these methods cannot be legitimately used to determine whether a single observation is within or outside of expected limits.