Data for the problem is given below: crabType bodyTemperature Female: 1.9, 1.6,

Question

Data for the problem is given below:

crabType bodyTemperature
Female: 1.9, 1.6, 1.4, 1.1, 1.6, 1.8, 1.9, 1.7, 1.5, 1.8, 1.7, 1.7, 1.8, 1.7, 1.8, 2, 1.8, 1.7, 1.6, 1.6, 1.5

intact male: 1.9, 1.2, 1, 0.9, 1.4, 1, 1.3, 1.4, 1.1, 1, 1.4, 1.2, 1.4, 1.4, 1.5, 1.5, 1.1, 1.4, 1.3, 1.3, 1.3

male minor removed: 1.2, 1, 0.9, 0.8, 1.2, 0.9, 1.1, 1.1, 1.3, 1.3, 1.3, 1.1, 1.4, 1.5, 1.4, 1.4, 1.2, 1.4, 1.3, 1.2, 1.4

male major removed: 1.2, 0.9, 1.4, 1.2, 1.2, 1.6, 1.9, 1.4, 1.4, 1.4, 1.6, 1.4, 1.7, 1.3, 1.5, 1.2, 1.3, 1.6, 1.5, 1.5, 1.5

Fiddler crabs are so called because males have a greatly enlarged "major" claw, which is used to attract females and to defend a burrow. Darnell and Munguia (2011) recently suggested that this appendage might also act as a heat sink, keeping males cooler while out of the burrow on hot days. To test this, they placed four groups of crabs into separate plastic cups and supplied a source of radiant heat (60-watt light bulb) from above. The four groups were intact male crabs; male crabs with the major claw removed; male crabs with the other (minor) claw removed (control); and intact female fiddler crabs. They measured body temperature of crabs every 10 minutes for 1.5 hours These measurements were used to calculate a rate of heat gain for every individual crab in degree log minute. Rates of heat gain for all crabs are provided in the datafile "fiddler crabs s C/ R minute. Rates of heat gain for all crabs are provided in the datafile "fiddler crabs.csv. a) (2 pts) State the null and alternative hypotheses. (3 pts) Graph the data. (5 pts) Perform the appropriate analysis. (8 pts) Test the assumptions of the statistical model and state your conclusions. (For this problem, it should not be necessary to perform any remedial measures.) (5 pts) Partition the Treatment Sums of Squares using orthogonal contrasts. You may choose whichever set of orthogonal contrasts you think is most appropriate. (2 pts) Perform any other desired comparisons of means using the Tukey-Kramer procedure. (5 pts) State your overall conclusions. b) c) d) e) f) g)

Explanation / Answer

Here there are four groups namely female, intact male, male minor removed and male major removed.

Here we have to test four groups.

And we test the four groups by using ANOVA.

The test of hypothesis is,

H0 : mu1 = mu2 = mu3 = mu4

H1 : At least one of the mean is differ than 0.

Assume alpha = level of significance = 5% = 0.05

ANOVA we can done by using EXCEL.

steps :

Enter all the data in EXCEL sheet --> Data --> Data analysis --> Anova : Single Factor --> ok -->Input Range : select all range --> Grouped by : columns --> Alpha : 0.05 --> output range : select one empty cell --> ok

Test statitstic F = 20.31219

P-value = 7E-10

P-value < alpha

Reject H0 at 5% level of significance.

Conclusion :  At least one of the mean is differ than 0.

We can dome Tukey's test for testing which groups are differ than 0.

Anova: Single Factor SUMMARY Groups Count Sum Average Variance Column 1 21 35.2 1.67619 0.038905 Column 2 21 27 1.285714 0.051286 Column 3 21 25.4 1.209524 0.036905 Column 4 21 29.7 1.414286 0.046286 ANOVA Source of Variation SS df MS F P-value F crit Between Groups 2.64131 3 0.880437 20.31219 7E-10 2.718785 Within Groups 3.467619 80 0.043345 Total 6.108929 83

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