Tire Inc claims that its new brand of tire has the average l

Tire Inc. claims that its new brand of tire has the average life expectancy of at least 40,000 miles. To check validity of this claim, a consumer protection group tested 100 tires and based on this sample data determined that the average life of was 39,600 miles. Assume that the population standard deviation for the life of the tires is 2500 miles. Is Tire Inc.\'s claim valid at significance level of 5%? Which of the following is the appropriate test of hypothesis that would allow testing the validity of the claim made by Tire Inc.?

1.

A. HO: > 40,000 H1: 40,000

B. HO: 40,000 H1: > 40,000

C. HO: < 40,000 H1: 40,000

D. HO: 40,000 H1: < 40,000

E. None of the above answers is correct

2.

What is the test statistic for this problem?

Select one:

A. 1.53

B. -1.6

C. 1.89

D. 2.32

3.

What statement should be made about the null and alternative hypothesis based on sample data and significance level?

Select one:

A. fail to reject H_O

B. reject H_O

C. fail to reject H_a

D. reject H_a

E. None of the above answers is correct

4

What statement can made about the Tire Inc.\'s claim based on the results of test of hypothesis?

Select one:

A. the claim is valid

B. the claim is not valid

C. additional information is needed to determine whether the claim is valid

D. None of the above answers is correct

5.

provided that a specific conclusion can be drawn about Tire Inc.\'s claim based on values determined for sample test statistic and critical value(s) of test statistics, what is the probability that the specific conclusion drawn is wrong?

Select one:

A. 100%

B. 0%

C. 5%

D. There is not enough information to calculate this probability

E. None of the above answers is correct

Solution

1.

Formulating the null and alternative hypotheses,              
              
Ho:   u   >=   40000  
Ha:    u   <   40000

Some symbols did not register in your choices, so please look for this answer.

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2.

Getting the test statistic, as              
              
X = sample mean =    39600          
uo = hypothesized mean =    40000          
n = sample size =    100          
s = standard deviation =    2500          
              
Thus, z = (X - uo) * sqrt(n) / s =    -1.6   [ANSWER, OPTION B]

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3.  
              
As we can see, this is a    left   tailed test.      
              
Thus, getting the critical z, as alpha =    0.05   ,      
alpha =    0.05          
zcrit =    -   1.644853627      

Comparing z > zcrit, we   FAIL TO REJECT THE NULL HYPOTHESIS.   [OPTION A]

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4.

OPTION A. the claim is valid. [ANSWER]

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5.

It is the level of significance,

OPTION C: 5% [ANSWER]