Question

A consumer organization wants to estimate the actual tread wear index of a brand name of tires that claims "graded 150" on the sidewall of the tire. A random sample of n=21 indicates a sample mean tread wear index of 136.7 and a sample standard deviation of 29.3. Complete parts​ (a) through​ (c).

a.  Assuming that the population of tread wear indexes is normally​ distributed, construct a 99% confidence interval estimate of the population mean tread wear index for tires produced by this manufacturer under this brand name.

b. Do you think that the consumer organization should accuse the manufacturer of producing tires that do not meet the perfomance information on the sidewall of the​ tire? Explain.

c. Explain why an observed tread wear index of 102 for a particular tire is not​ unusual, even though it is outside the confidence interval developed in​ (a).

Answer 1

a)

99% Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.01 /2, 21- 1 ) = 2.845
136.7 ± 2.845 * 29.3/√(21)
Lower Limit = 136.7 - 2.845 * 29.3/√(21)
Lower Limit = 118.51
Upper Limit = 136.7 + 2.845 * 29.3/√(21)
Upper Limit = 154.89
99% Confidence interval is ( 118.51 , 154.89 )

b)

Some values in confidence interval are less than 150.

The consumer organization should accuse the manufacturer of producing tires that do not

meet the perfomance information on the sidewall of the​ tire.

c)

z = ( x - mean) / SD

= ( 102 - 136.7) / 29.3

= -1.18

Since 102 is within 2 standard deviations of the mean,

an observed tread wear index of 102 for a particular tire is not​ unusual, even though it is outside the

confidence interval developed in​ (a).