A tire manufacturer has been producing tires with an average life expectancy of
ID: 2956644 • Letter: A
Question
A tire manufacturer has been producing tires with an average life expectancy of 26,000 miles. Now the company is advertising that its new tires' life expectancy has increased . In order to test the legitmacy of the advertising campaign, an independent testing agency tested a sample of 6 of their tires and had provided the following data.Life Expectancy
(In Thousands of Miles)
28
27
25
28
29
25
a. Determine the mean and the standard deviation. b. At 95% confidence using the critical value approach, test to determine whether or not the tire company is using legitimate advertising. Assume the population is normally distributed. c. Repeat the test using the p-value approach.
A tire manufacturer has been producing tires with an average life expectancy of 26,000 miles. Now the company is advertising that its new tires' life expectancy has increased . In order to test the legitmacy of the advertising campaign, an independent testing agency tested a sample of 6 of their tires and had provided the following data.
Life Expectancy
(In Thousands of Miles)
28
27
25
28
29
25
a. Determine the mean and the standard deviation. b. At 95% confidence using the critical value approach, test to determine whether or not the tire company is using legitimate advertising. Assume the population is normally distributed. c. Repeat the test using the p-value approach.
Explanation / Answer
(a) mean=(28+27+25+28+29+25)/6=27
standard deviation=[(28-27)^2 +...+ (25-27)^2]/(6-1) = 1.67
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(b) The test hypothesis is
Ho:<=26
Ha:>26
The test statistic is
t=(xbar - )/(s/n)
=(27-26)/(1.67/sqrt(6))
= 1.47
Given a=0.01, the critical value is |t(0.01, df=n-1=5)|=3.36 (check student t table)
Since t=1.47 < 3.36, we do not reject Ho.
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(c) The p-value is P(t>1.47)= 0.1008 (check student t table)
Since p-value is larger than a=0.01, we do not reject Ho.
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