Question 324572: An ice cream store estimates that the mean consumption of their waffle cones by their customers is more than 3 per year. In a sample of 80 customers, you find that the mean consumption of waffle cones is 4 per year with a standard deviation of 5.0. Let the level of significance be α = 0.06.
Decide whether to reject the null hypothesis or fail to reject the null hypothesis. Also interpret the decision in terms of the original claim. Choose two answers below.
answers
A.Fail to reject the null hypothesis
B.Reject the null hypothesis
C.At the α = 0.06 level, there is enough evidence to support the ice cream store’s claim.
D. At the α = 0.06 level, there is not enough evidence to support the ice cream store’s claim.
Found 2 solutions by stanbon, jrfrunner:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! An ice cream store estimates that the mean consumption of their waffle cones by their customers is more than 3 per year. In a sample of 80 customers, you find that the mean consumption of waffle cones is 4 per year with a standard deviation of 5.0. Let the level of significance be α = 0.06.
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Ho: u = 3
Ha: u > 3 (store estimate)
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sample mean = 4
t(4) = (4-3)/5 = 0.2
p-value for right tail test with df = 79 is tcdf(0.2,100,79) = 0.42
Note: The p-value is greater than 6% so fail to reject Ho.
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Decide whether to reject the null hypothesis or fail to reject the null hypothesis. Also interpret the decision in terms of the original claim.
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Choose two answers below.
A and D
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Cheers,
stan H.
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answers
A.Fail to reject the null hypothesis
B.Reject the null hypothesis
C.At the α = 0.06 level, there is enough evidence to support the ice cream store’s claim.
D. At the α = 0.06 level, there is not enough evidence to support the ice cream store’s claim.
Answer by jrfrunner(365)  (Show Source):
You can put this solution on YOUR website! Ho: Mu=3
Ha: Mu not equal 3 (the claim)
given n=80, xbar=4, s=5 alpha =0.06
Since sigma is not known, the test statistic used would be a t, but since the sample size is large, either Z or t statistic could be used.
Furthermore, since analysis is on the means "and" sample size is large, the normality assumption is acceptable.
test statistic: Z=(Xbar - Mu)/SE where SE = s/sqrt(n)
Z = (4-3)/(5/sqrt(80))=1.7888
This is a two tail test, so the critical values for alpha=.06 are -1.88, 1.88
The region between these two critical values is the "cannot reject" region
Since the test statistic falls in this region we conclude "cannot reject Ho, no evidence that Mu is not equal to 3" in other words fail to reject Ho.
in your list of choices: A & D
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