10.56 A sample of 25 concession stand purchases at the October 22 matinee of Bri

Question

10.56 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 a = .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.)

Explanation / Answer

Given n1=25, xbar1=5.29, s1=3.02
n2=25, xbar2=5.12, s2=2.14

The test hypothesis is
Ho:1^2=2^2
Ha:1^2 not equal to 2^2

The test statistic is

F=s1^2/s2^2

=3.02^2/2.14^2

=1.99

Given a=0.05, the critical value is F(0.025, df1=n1-1=24, df2=n2-1=24)=0.44 (check F table)

F(0.975, df1=n1-1=24, df2=n2-1=24)= 2.27 (check F table)

Since 0.44<F=1.99<2.27, we do not reject Ho.

So we can not conclude that there is a difference in variances

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