SOLUTION: The manufacturer of Winston Tire Company claims its new tires last for an average 40,000 miles. An independent testing agency road-tested 20 tires to substantiate the claim made b
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Question 180477: The manufacturer of Winston Tire Company claims its new tires last for an average 40,000 miles. An independent testing agency road-tested 20 tires to substantiate the claim made by the Winston Tire Company. The sample mean was 39,000 miles with a sample standard deviation of 5,000 miles. Using the .05 level of significance and the five-step hypothesis testing procedure, determine if there is reason to reject the claim made by Winston Tire Company and conclude that tires last for less than 40,000 miles.
My question: Am I on the right track and/or if not, what are the steps and formulas to use to solve for this?
I came up with:
H0: (Mu) = 40,000
H1: (Mu) not equal to 40,000
Z = 39,000 – 40,000/5,000/ sqrt 20 = - 4
I would say:
Reject the null in favor of the alternative, since – 4 falls in the rejection region.
Thank you
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The manufacturer of Winston Tire Company claims its new tires last for an average 40,000 miles. An independent testing agency road-tested 20 tires to substantiate the claim made by the Winston Tire Company. The sample mean was 39,000 miles with a sample standard deviation of 5,000 miles. Using the .05 level of significance and the five-step hypothesis testing procedure, determine if there is reason to reject the claim made by Winston Tire Company and conclude that tires last for less than 40,000 miles.
My question: Am I on the right track and/or if not, what are the steps and formulas to use to solve for this?
I came up with:
H0: (Mu) = 40,000
H1: (Mu) < 40,000
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Note: I changed to a left-tail test because of the words "..conclude that
tires last for LESS than 40,000 miles".
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Critical Value for a left-tail test with alpha=5% is -1.645
Rule: Reject Ho if test statistic is < -1.645
Z = (39,000 – 40,000)/[5,000/ sqrt 20] = -0.8944.
p-value: P(z < -0.8944) = 0.186 = 18.6%
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Note: I changed your test statistic result.
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I would say:
Reject the null in favor of the alternative, since – 4 falls in the rejection region.
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Coclusion should be: Since the test statistic is not in the rejection interval,
Fail to reject Ho.
You could also say: Since p-value is greater than 5% Fail to reject Ho.
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Interpretatation of the p-value: 18.6% of test result could have provided
stronger evidence for rejecting Ho.
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Cheers,
Stan H.
P.S. Please let me know if this answer is any help to you. I see quite
a few Stat questions but get little feedback from the questioners.
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