Question 1194298: A trucking firm suspects that the average lifetime of 25,000 miles claimed for certain tires is too high.
To test the claim, the firm puts a random sample of 40 of these tires on its trucks and later finds that their mean lifetime is 4, 421 miles and the standard deviation is 1,349 miles. What can it conclude at the 0.01 level of significance?
Formula 𝑧 = 𝑥̅ − 𝜇𝑜 𝜎/√𝑛
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Hypotheses:
H0: mu = 25,000
H1: mu < 25,000
The claim is in the alternate hypothesis. The inequality sign here tells us we have a left-tailed test.
n = 40 = sample size
xbar = 4421 = sample mean
s = 1349 = sample standard deviation
Compute the z score
z = (xbar - mu)/( s/sqrt(n) )
z = (4421 - 25000)/( 1349/sqrt(40) )
z = -96.4811148533804
We get an extremely small z value
Usually the z scores are between -3 and 3, so we can use a Z table to find the p-value.
I have a feeling there's a typo in one of the numbers given.
The 4421 seems a bit too small.
Did you mean to say something like 44210 perhaps?
Please carefully double-check the entire problem and get back to me, or make a new post.
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