Life Span of Ma but not for males An Longevity as a resalt of increased repeoduc

Question

Life Span of Ma but not for males An Longevity as a resalt of increased repeoduction has been stadied for e fruit flies was manipulated. There were ive groups of groupl- no companion, group 2- one newly pregnant companion. group 3-one vergin companion, groupi- eight newly pregnant companions group 5- eight vigin companions All males weretreated The longevity in days for each fire fly is recorded. experiement was conducted in which was identically with respect to the anesthetizations and the provision of fresh food medium Refer to the SISS Fire Fly Longevity output provided to answer the fol Leu) Reponse variable hen LegA US kar a. Identify: Treatment: Variable. . b, write the ANOVA Hypothsis Ill: o.hu., ‘wts1hc2 (as.A+ L Rott K>Baan c.Is there a significant diffrence in longevity between the groupE No Report test statistic and p-valuea d. If select yes in c, check the pair of groups that differ -64) 35) 4.5) f. Rank the groups by longevity ( 1-higest) h. Now lets check the ANOV A conditions i Is Normality assumpion violated ? Yes No Explain why i. Is Homogeniety of variances violated? Yes No Explain Why i. Based on h. Select one: A) All assumptions arwe met hence ANOVA is valid. B) ANOVA is invalid. C) ANOVA is robust against the violated assumption. Pts

Explanation / Answer

f)

As, per the box plot, the ranks of the groups based on longevity are

1. Group 3

2. Group 4

3. Group 2

4. Group 1

5. Group 5

h)

i) Yes, the normality assumption violated. The p-value of Shapiro-Wilk test for group 3 is 0.016 which is less than the significance level of 0.05 and we reject the null hypothesis and conclude that the data of group 3 does not follow normal distribution.

ii) No, the homogeneity of variances assumption is not violated. The p-value of Levene Statistic is 0.388 which is greater than the significance level of 0.05 and we fail to reject the null hypothesis and conclude that the data follow homogeneity of variances assumption.

i)

If the populations involved did not follow a normal distribution, an ANOVA test could not be used to examine the equality of the sample means. Instead, one would have to use a non-parametric test (or distribution-free test), which is a more general form of hypothesis testing that does not rely on distributional assumptions. But, ANOVA including repeated measures is robust to violations to normality of errors assumptions. As, there are 25 mesures taken for each group, the anova test in the problem includes repeated measures.

So, the correct option is C) ANOVA is robust against the violated assumptions.

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