Question 135744: I'm in an online class so there is no textbook, please help, thanks.
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?
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Ho: s^2(matinee)-s^2(evening) = 0
Ho: difference is not 0
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Test statistic: F= s1^2/s2^2 = (2.14)^2/3.02^2 = 0.5021...
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Critical values for two-tail test with alpha-5%:
left tail(F=0.44)
right tail (F=2.27)
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Conclusion: Since test stat is not in either rejection interval,
Fail to reject Ho; variances are statistically the same, based
on this test.
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Cheers,
Stan H.
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