a. Our null hypothesis is that there will be no differences in average bagworm numbers between any of the genera. b. Our alternate hypothesis is that the average number of bagworms of at least one genus will be different compared to the average number of bagworms of at least one other genus. 9. Answer the following questions about your results (please always round to thenearest 100th place)a. What is your p-value for your ANOVA test (found in the ANOVA Summary table)?0.36b. What does this p-value tell you regarding your null hypothesis (should youaccept or reject the null and what does this mean)?c.When you ran this test, you should have seen that there were no Tukey HSDresults in the box under "Tukey HSD Test". Why were there no post-hoc tests runafter the ANOVA was run? Data SummaryNΣΧMean11641125.6875EX² 25797Variance 1015.9625Std.Dev.Std.Err.31.87427.9685219114460.21052187488325.953291.246720.9334327622.2383321865639.03572388782791.7442Samples52.837df37.0606ANOVA Summary Independent Samples k=4SourceSSTreatment11326.5373[between groups]ErrorSs/BlTotal338948.775597Ss/Bl= Subjects or Blocks depending on the design.Applicable only to correlated-samples ANOVA.94Tukey HSD TestThis test will be performed only ifK>2 and the analysis of varianceyields a significant F-ratio.4733948.4286253871494.952438.664614.6139MS3775.51243485.3435F1.08Total98408041.63275088103494.317359.11285.9713P0.361483M1 = mean of Sample 1M2 = mean of Sample 2and so forth.Graph MakerHSD = the absolute [unsigned]difference between any twosample means required forsignificance at the designatedlevel. HSD[.05] for the .05 level;HSD[.01] for the .01 level.

As the above question does not discretely specify the degrees of freedom for the numerator…

a. What is your p-value for your ANOVA test (found in the ANOVA Summary table)? 0.36 b. What does this p-value tell you regarding your null hypothesis (should you accept or reject the null and what does this mean)? c. When you ran this test, you should have seen that there were no Tukey HSD results in the box under "Tukey HSD Test". Why were there no post-hoc tests run after the ANOVA was run?

a. What is your p-value for your ANOVA test (found in the ANOVA Summary table)? 0.36 b. What does this p-value tell you regarding your null hypothesis (should you accept or reject the null and what does this mean)? c. When you ran this test, you should have seen that there were no Tukey HSD results in the box under "Tukey HSD Test". Why were there no post-hoc tests run after the ANOVA was run?

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter11: Genome Alterations: Mutation And Epigenetics

Section: Chapter Questions

Problem 8QP: Bruce Ames and his colleagues have pointed out that although detailed toxicological analysis has...

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