As the \"baby boom\" generation ages, interest in finding treatments that extend
ID: 3155238 • Letter: A
Question
As the "baby boom" generation ages, interest in finding treatments that extend life span has surged. Experimental research mainly uses nonhuman animals. In a recent experiment, tested the influence of the anticonvulsant medication trimethadione on the life span of the nematode worm, Caenorhabditis elegans. The study compared the effects of three trimedthadione treatments (provided at the larval stage, at the adult stage, and at both stages and a water treatment (the control). The resulting life spans are shown in the following histograms for 50 worms in each treatment. Assume that each worm was treated independently.
a. ANOVA would be the preferred method to compare the means of each group. What problem or problems do you see in applying this method to the data shown in the preceding histograms?
b. The data were analyzed with a Kruskal-Wallis test. What are the null and alternative hypotheses?
c. The Kruskal-Wallis test yielded the following rank sums.
Life stage of trimethadione treatment.
None(water)
Larval Stage
Adult Stage
Both Stages
4201
3672
6003.5
6223.5
H-=29.27. Can we conclude from the result that the means of the treatment groups and unequal? Explain.
None(water)
Larval Stage
Adult Stage
Both Stages
4201
3672
6003.5
6223.5
Explanation / Answer
As the "baby boom" generation ages, interest in finding treatments that extend life span has surged. Experimental research mainly uses nonhuman animals. In a recent experiment, tested the influence of the anticonvulsant medication trimethadione on the life span of the nematode worm, Caenorhabditis elegans. The study compared the effects of three trimedthadione treatments (provided at the larval stage, at the adult stage, and at both stages and a water treatment (the control). The resulting life spans are shown in the following histograms for 50 worms in each treatment. Assume that each worm was treated independently. a. ANOVA would be the preferred method to compare the means of each group. What problem or problems do you see in applying this method to the data shown in the preceding histograms? When the data not follows normal distribution , one of the assumption for the ANOVA is violated. Non parametric method is appropriate. b. The data were analyzed with a Kruskal-Wallis test. What are the null and alternative hypotheses? Null hypothesis: Null hypothesis : The samples are from identical populations. Alternative hypothesis: Alternative hypothesis : The samples come from different populations. c. The Kruskal-Wallis test yielded the following rank sums. Life stage of trimethadione treatment. None(water) Larval Stage Adult Stage Both Stages 4201 3672 6003.5 6223.5 H-=29.27. Can we conclude from the result that the means of the treatment groups are unequal? Explain. The Kruskal-Wallis test statistic H is approximately a chi-square distribution, with k-1 =3-1 =2 degrees of freedom Critical value of chi square at 0.05 level =5.991 Calculated H=29.27 > 5.991. The null hypothesis is rejected. We conclude from the result that the means of the treatment groups are unequal.
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