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 higher caffeine does 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 Source DF SS MS F P Caffeine 11976 Error 538.75 Total Level N Mean StDev Control 3 427.00 20.07 Low 3 441.00 22.61 Medium 3 431.00 19.08 High 3 361.00 29.61 The numerator degrees of freedom for the ANOVA F-test are 11 8 3 2

Explanation / Answer

The numerator degrees of freedom = K - 1    , where K   = number of treatment

D.f = 4 - 1 = 3

The numerator degrees of freedom for ANOVA F test are : 3

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