Question 128715: Plese help me. I am stuck with this question. Thank you kindly
A sample of 25 concession stand purchases at the October 22 matinee of Bride of Chucky showed a mean purchase of $5.29 with a standard deviation of $3.02. For the October 26 evening showing of the same movie, for a sample of 25 purchases the mean was $5.12 with a standard deviation of $2.14. The means appear to be very close, but not the variances. At α = .05, is there a difference in variances? Show all steps clearly, including an illustration of the decision rule
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A sample of 25 concession stand purchases at the October 22 matinee of Bride of Chucky showed a mean purchase of $5.29 with a standard deviation of $3.02. For the October 26 evening showing of the same movie, for a sample of 25 purchases the mean was $5.12 with a standard deviation of $2.14. The means appear to be very close, but not the variances. At α = .05, is there a difference in variances? Show all steps clearly, including an illustration of the decision rule
---------------
Ho: variance(matinee)-variance(evening) = 0
Ha: variance (matinee)-variance (evening) is not equal to 0
-------------------
Matinee DATA: n1=22 ; variance = 3.02^2 = 9.12 ; df=21-----denominator
Evening DATA; n1=25 ; variance = 2.14^2 = 4.58 ; df=24-----numerator
Two-tailed F-test with alpha = 5%
----------
Test statistic: F= 4.58/9.12=0.5021
p-value = 0.1056
------------
Conclusion: Fail to reject Ho.
-----------------------
Comment: I do not have an F-chart that shows critical
values for a two-tailed test with alpha=5%.
-------------------------
Cheers,
Stan H.
|
|
|
