The film industry rates movies as G, PG, PG-13, and R. A total of 70 movies were randomly selected, and reviewers noted the amount of time on violent scenes spent in each movie (in minutes). The researcher tested
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0
:
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1
=
𝜇
2
=
𝜇
3
=
𝜇
4
, where
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the true mean time spent in violent scenes for each of the four movie ratings, and got a test statistic of 3.0. The value from this test statistic's distribution at which point the upper tail area under the curve is 0.05 is 3.07. What can we say about the degrees of freedom and the p-value for this test?

Question

The film industry rates movies as G, PG, PG-13, and R. A total of 70 movies were randomly selected, and reviewers noted the amount of time on violent scenes spent in each movie (in minutes). The researcher tested 
𝐻
0
:
𝜇
1
=
𝜇
2
=
𝜇
3
=
𝜇
4
, where 
𝜇
=
 the true mean time spent in violent scenes for each of the four movie ratings, and got a test statistic of 3.0. The value from this test statistic's distribution at which point the upper tail area under the curve is 0.05 is 3.07. What can we say about the degrees of freedom and the p-value for this test?

BlackTom AI · Accepted Answer

First, restating the scenario: we have a one-way ANOVA comparing the mean time spent in violent scenes across four movie ratings (G, PG, PG-13, R), with n = 70 total observations. The null hypothesis H0 is that all four group means are equal: μ1 = μ2 = μ3 = μ4. The reported test statistic is 3.0, and the critical value for the upper tail area of 0.05 is 3.07. We need to determine the degrees of freedom (df) and the p-value interpretation.

Option by option analysis:

Option A: df1 = 3, df2 = 67, p-value > 0.05
- In a one-way ANOVA with k groups, the numerator degrees of freedom df1 is k - 1. Here k = 4, so df1 should be 3, which matches the intended structure. However, the denominator degrees of freedom df2 is N - k, where N is the total sample size. With N = 70 and k = 4, df2 = 70 - 4 = 66, not 67. Therefore df2 = 67 is incorrect, making this option inaccurate even before assessing the p-value.
- Regarding p-value, even if df2 were correct, the observed F-statistic corresponds to a value 3.0, which is slightly below the critical value 3.07 for alpha = 0.05. This would yield a p-value greater than 0.05. While the p-value statement could be correct, the df specification is already incorrect, so this option is flawed.

Option B: df1 = 4, df2 = 65, p-value > 0.05
- Here df1 is given as 4, but for four groups the correct df1 is k - 1 = 3. A df1 of 4 would imply five groups, which contradicts the problem setup. This makes this option inconsistent with the described design. The p-value claim > 0.05 could be true, but the df values are not.

Option C: df1 = 3, df2 = 66, p-value < 0.05
- This option has the correct df1 and df2 values for a 4-group one-way ANOVA with N = 70, but the p-value is claimed to be less than 0.05. Since the test statistic is 3.0 and the 0.05 upper-tail critical value is 3.07, the observed statistic does not exceed the critical value, so the p-value should be greater than 0.05, not less. Therefore this option misstates the p-value.

Option D: df1 = 3, df2 = 66, p-value > 0.05
- This option has the correct df1 and df2: df1 = 4 - 1 = 3 and df2 = 70 - 4 = 66. The test statistic is 3.0, which is below the 0.05 upper-tail critical value of 3.07, indicating that the p-value is greater than 0.05. Thus, this option accurately reflects both the correct degrees of freedom and the correct p-value interpretation given the data.

Summary of reasoning: The correct ANOVA design yields df1 = 3 and df2 = 66, since there are four groups and 70 observations. The statistic 3.0 falls short of the 3.07 cutoff for p = 0.05, so the p-value is greater than 0.05. Among the options, only Option D correctly matches the df values and the p-value direction; the other options either misstate df values or misstate the p-value in light of the provided critical value.

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