SOLUTION: I need help some help in solving this problem, please? A sample of 25 concession stand purchases at the October 22 matinee of Bride of Chucky showed a mean purchase of $5.29 with

Question 126188This question is from textbook Applied Statistics in Business and Economics
: I need help some help in solving this problem, please?
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 the .05 level of significance, is there a difference in variances? Show all steps clearly, including an illustration of the decision rule. This question is from textbook Applied Statistics in Business and Economics

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
I need help some help in solving this problem, please?
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 the .05 level of significance, is there a difference in variances? Show all steps clearly, including an illustration of the decision rule.
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Ho: s^2(matinee)=s^2(evening)
Ha: s^2(matinee) is not equal to s^2(evening)
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Two-tailed test with alpha = 0.05
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Test Statistic: F = (2.14)^2/(3.02)^2 = 0.502127
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Critical Value:
numerator df's = 24 ; denominator df's = 24
F = 2.27
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Conclusion:
Since the Test Statistic is less than the Critical Value, Fail to
reject Ho.
The variances for matinee and evening are statistically equivalent.
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Cheers,
Stan H.
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