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.)
To determine if there is a difference in variances between the two sample sets, we can perform a hypothesis test using the F-test. The F-test compares the ratio of the variances from two populations.
Let's define our null and alternative hypotheses:
Null Hypothesis (H0): The variances of the two populations are equal.
Alternative Hypothesis (H1): The variances of the two populations are not equal.
We can use the F-test statistic to test this hypothesis. The formula for the F-test statistic is:
F = (s1^2) / (s2^2)
Where s1^2 and s2^2 are the variances of the two samples, respectively.
Given the following information:
Sample 1 (October 22 matinee):
n1 = 25
x̄1 (mean) = $5.29
s1 (standard deviation) = $3.02
Sample 2 (October 26 evening):
n2 = 25
x̄2 (mean) = $5.12
s2 (standard deviation) = $2.14
We can calculate the F-test statistic:
F = (s1^2) / (s2^2)
F = ($3.02^2) / ($2.14^2)
F ≈ 1.602
To determine if there is a significant difference in variances, we need to compare this test statistic to the F-distribution with degrees of freedom (df1, df2).
The degrees of freedom for the numerator (df1) is equal to the sample size of the first sample minus 1:
df1 = n1 - 1 = 25 - 1 = 24
The degrees of freedom for the denominator (df2) is equal to the sample size of the second sample minus 1:
df2 = n2 - 1 = 25 - 1 = 24
At α = .05 significance level, we need to find the critical value from the F-distribution table or use statistical software. Let's assume it is 2.23.
The decision rule is:
If F > 2.23 or F < 1/2.23, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
Since F (1.602) does not fall outside the critical region (F > 2.23 or F < 1/2.23), we fail to reject the null hypothesis. Therefore, we do not have enough evidence to conclude that there is a significant difference in variances between the two samples.
In summary, at α = .05 level of significance, there is no difference in variances between the two concession stand purchase samples at the October 22 matinee and October 26 evening showings of the movie "Bride of Chucky."