A consumer organization wants to estimate the actual tread wear index of a brand

Question

A consumer organization wants to estimate the actual tread wear index of a brand name of tires that claims "graded 250" on the sidewall of the tire. A random sample of n=20
indicates a sample mean tread wear index of 238.7 and a sample standard deviation of 22.1

a)Assuming that the population of tread wear indexes is normally distributed, construct a 95 % 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.
A.
Yes, because a grade of 250 is in the interval.
B.
Yes comma because a grade of 250 is not in the interval.
C.
No, because a grade of 250 is not in the interval.
D.
No comma because a grade of 250 is in the interval.

c)Explain why an observed tread wear index of 254 for a particular tire is not unusual, even though it is outside the confidence interval developed in (a).
A.
It is not unusual because it is just outside the confidence interval.
B.
It is not unusual because it is only 0.69 standard deviations above the sample mean.
C.
It is not unusual because it is actually in the confidence interval.
D.
It is not unusual because it is only 0.22 standard deviations above the confidence interval.

Explanation / Answer

here degree of freedom =20-1 =19

for 95% CI and 19 degree of freedom t value =2.093

std error =std deviation/(n)1/2 =4.9417

hence confidence interval =mean +/- t*std error =228.357 ; 249.043

b) there a significant difference in memory between the two conditions

c)D.
It is not unusual because it is only 0.22 standard deviations above the confidence interval.

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