Investigators gave caffeine to fruit flies to see if it affected their rest. The

Question

Investigators gave caffeine to fruit flies to see if it affected their rest. The four treatments were a control, a low caffeine dose of 1 mg/ml of blood, a medium dose of 3 mg/ml of blood, and a highor caffeine dose of 5 mg/ml of blood. Twelve fruit flies were assigned at random to the four treatments, three to each treatment, and the minutes of rest measured over a 24-hour period were recorded. The data follow. Assume the data are four independent SRSs, one from each of the four populations of caffeine levels, and that the distribution of the yields is Normal. A partial ANOVA table produced by Minitab follows, along with the means and standard deviation of the yields for the four groups One-way ANOVA. Rest Versus Caffeine The value of the ANOVA F statistic for testing equality of the population means of the average rest time for the four caffeine levels is: A. 4.73 B. 7.41. C. 2.78. D. 4.82.

Explanation / Answer

There are 4 treatments here. So the degrees of freedom for Caffeine will be 4 - 1 = 3.

The total number of observations is 12. Hence, the total degrees of freedom is 12 - 1 = 11.

Therefore, the error degree of freedom is, df(Total) - df(Caffeine) = 11 - 3 = 8.

Also, MS = SS / DF

and F statistic = MS( Caffeine) / MS( Error )

Therefore, using the above formulae we get the following ANOVA table,

P = 2 * min ( P (F<7.41), P(F>7.41))

Hence the correct option is (B) 7.41

Source DF SS MS F P Caffeine 3 11976 3992 7.41 0.02142505 Error 8 4310 538.75 Total 11 16286

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