Question 1176547: A microwave oven repairer says that the mean repair cost for damaged microwave ovens is less than $100. You find that a random sample of five microwave ovens has a mean repair cost of $75 and a standard deviation of $12.50. At α = 0.01 , do you have enough evidence to support the repairer's claim? What is the p-value?
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
mu = population mean
mu = mean repair cost of every microwave made by the company
We're doing a hypothesis test on the claim that mu < 100.
So we have these two hypotheses
H0: mu = 100
H1: mu < 100
where H0 and H1 are the null and alternative hypothesis respectively. We're doing a left-tailed test because of what the alternative hypothesis inequality sign is saying.
Some textbooks will write the hypotheses as
H0: mu >= 100
H1: mu < 100
but I'll go with the first format since the null hypothesis has one, and only one, fixed center. Otherwise, we'll generate infinitely many normal distributions to use with the null. So the first format is explicitly more clear that we have one fixed mu value to set up our distribution around.
The claim is made in the alternative hypothesis. So if we reject the null, then we accept the claim. If we fail to reject the null, then we don't have enough evidence to overturn the null, and therefore must reject the claim.
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We have a sample of n = 5 microwaves.
In this sample, the mean repair cost is $75
So xbar = 75
The sample standard deviation is s = 12.50
The test statistic is
t = (xbar - mu)/(s/sqrt(n))
t = (75 - 100)/(12.50/sqrt(5))
t = -4.47213595499958
t = -4.47
Note the use of the T distribution test statistic here. This is because n = 5 is not larger than 30, and we also don't know sigma.
Now use a calculator to compute the p-value.
If you are using a TI84 or similar, then the steps are as follows
1) Hit the 2ND key
2) Hit the VARS key
3) Select tcdf(
4) Type in "-99,-4.47,4)" without quotes
5) Hit enter
You should have tcdf(-99,-4.47,4) typed in your calculator and the result should be roughly 0.0055, which is the approximate p-value.
The -99,-4.47 portion means we want to find the area from -99 to -4.47; you can replace the -99 with any large negative number.
The "4" at the very end represents the degrees of freedom (df), since df = n-1 = 5-1 = 4.
An alternative calculator you can use is something like this
https://www.statology.org/t-score-p-value-calculator/
You'll type in -4.47 as the t score and 4 as the degrees of freedom. Select "one-tailed" because we're doing a left-tailed test.
This p-value 0.0055 is less than alpha = 0.01, so we reject the null. You can think of the phrase "if the p value is low, then the null must go" to help remember what the decision rule should be.
Or here's another way to think about it. This train of thought isn't 100% accurate but it's good enough in my opinion. The p-value represents the probability that the null is correct (again, not entirely accurate but its roughly true). The higher the p-value, the more chances the null is the case. We use the alpha value as the threshold we set. So if the p-value is smaller than alpha, then we toss the null and go with the alternative.
Because we rejected the null, this means we go with the alternative hypothesis that mu < 100. So there appears to be enough evidence to support the claim that the mean repair cost is under $100.
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Answers:
There is enough evidence to support the repairer's claim that the mean repair cost is less than $100
p-value = 0.0055
p-value is less than alpha, so we reject the null
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