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2) A trucking firm suspects that the mean lifetime of a certain tire it uses is less

than 40,000 miles.To check the claim, the firm randomly selects and tests 54 of these tires and gets a mean lifetime of 39,460 miles with a population standard deviation of 1200 miles. At alpha = 0.05, test the trucking firm’s claim.

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Answer:

We conclude that the mean lifetime of a certain tire it uses is less than 40,000 miles which means that the trucking firm’s claim was correct.

Explanation:

We are given that a trucking firm suspects that the mean lifetime of a certain tire it uses is less than 40,000 miles.

To check the claim, the firm randomly selects and tests 54 of these tires and gets a mean lifetime of 39,460 miles with a population standard deviation of 1200 miles.

Let
\mu = mean lifetime of a certain tire.

SO, Null Hypothesis,
H_0 :
\mu \geq 40,000 miles {means that the mean lifetime of a certain tire it uses is more than or equal to 40,000 miles}

Alternate Hypothesis,
H_A :
\mu < 40,000 miles {means that the mean lifetime of a certain tire it uses is less than 40,000 miles}

The test statistics that will be used here is One-sample z test statistics as we know about the population standard deviation;

T.S. =
(\bar X -\mu)/((\sigma)/(√(n) ) ) ~ N(0,1)

where,
\bar X = sample mean lifetime of 54 tires = 39,460 miles


\sigma = population standard deviation = 1200 miles

n = sample of tires = 54

So, test statistics =
(39,460-40,000)/((1200)/(√(54) ) )

= -3.307

Hence, the value of test statistics is -3.307.

Now at 0.05 significance level, the z table gives critical value of -1.6449 at for left-tailed test. Since our test statistics is less than the critical value of t as -3.307 < -1.6449, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the mean lifetime of a certain tire it uses is less than 40,000 miles which means that the trucking firm’s claim was correct.

User Mike Stonis
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