document.write( "Question 176058: 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. (Data are from a project by statistics students Kim Dyer, Amy Pease, and Lyndsey Smith.)\r
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Algebra.Com's Answer #134550 by stanbon(75887)\"\" \"About 
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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
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\n" ); document.write( "Ho: variance matinee = variance evening
\n" ); document.write( "Ha: they are not equal
\n" ); document.write( "Run 2-Sample F-Test to get:
\n" ); document.write( "Test Stat: F = 1.9915...
\n" ); document.write( "p-value: 0.09811...
\n" ); document.write( "df numerator = 24 = df denominator
\n" ); document.write( "Critical Value: F(right) = 2.23 ; F(left)= 1/2.33 = 0.448..
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\n" ); document.write( "Conclusion: Since p-value is greater than 5%, Fail to reject Ho.
\n" ); document.write( "The variances are not statistically different.
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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