Question

Question 11 (4 points) If your one factor ANOVA table shows that 'group' has a p-value = 0.0079, it would be appropriate to then conduct a Levene's F test retain the null hypothesis and report that none of the means are significantly different conduct a Hartley's F-max test conduct pairwise t-tests with an alpha of 0.05 conduct a Tukey's Honest Significant Difference test conduct a Bartlett's K-squared test

          Question 11 (4 points)
If your one factor ANOVA table shows that 'group' has a p-value = 0.0079, it would
be appropriate to then

conduct a Levene's F test

retain the null hypothesis and report that none of the means are significantly
different

conduct a Hartley's F-max test

conduct pairwise t-tests with an alpha of 0.05

conduct a Tukey's Honest Significant Difference test

conduct a Bartlett's K-squared test

Question 11 (4 points)
If your one factor ANOVA table shows that 'group' has a p-value = 0.0079, it would
be appropriate to then

conduct a Levene's F test

retain the null hypothesis and report that none of the means are significantly
different

conduct a Hartley's F-max test

conduct pairwise t-tests with an alpha of 0.05

conduct a Tukey's Honest Significant Difference test

conduct a Bartlett's K-squared test

Added by Katherine M.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Jacob Fry

Step 1

The p-value of 0.0079 is less than the significance level (alpha) of 0.05, so we reject the null hypothesis of the one-factor ANOVA test. This means that there is a significant difference among the group means. Show more…

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Question 11 (4 points) If your one-factor ANOVA table shows that 'group' has a p-value of 0.0079, it would be appropriate to then conduct a Levene's F test, retain the null hypothesis, and report that none of the means are significantly different. Alternatively, you could conduct pairwise t-tests with an alpha of 0.05 or a Tukey's Honest Significant Difference test. Conducting a Hartley's F-max test or a Bartlett's K-squared test would not be appropriate in this scenario.