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A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8000. A survey of owners of that tire design is conducted. Of the 29 tires in the survey, the average lifespan was 46,800 miles with a standard deviation of 9800 miles. Do the data support the claim at the 5% level? State the distribution to use for the test. (Round your standard deviation to two decimal places.)

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Answer: We reject the null hypothesis, and we use Normal distribution for the test.

Explanation:

Since we have given that

We claim that

Null hypothesis :
H_0:\mu\geq 50000

Alternate hypothesis :
H_1:\mu<50000

There is 5% level of significance.


\bar{X}=46800\\\\\sigma=9800\\\\n=29

So, the test statistic would be


z=\frac{\bar{X}-\mu}{(\sigma)/(√(n))}\\\\z=(46800-50000)/((9800)/(√(29)))\\\\z=-1.75

Since alternate hypothesis is left tailed test.

So, p-value = P(z≤-2.31)=0.0401

And the P-value =0.0401 is less than the given level of significance i.e. 5% 0.05.

So, we reject the null hypothesis, and we use Normal distribution for the test.

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