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 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 P-value of this test is: A. greater than 0.1. B. less than 0.05 C. between 0.05 and 0.1. D. It is not possible to determine the P-value from the information provided.

Explanation / Answer

From the ANOVA table

df of Total = 12-1 = 11

df of Caffenie = 4 -1 = 3

Sum of Square of Caffeine = 11976 / 3 = 3992

df of Error = 11 - 3 = 8

Sum of Square of Error = 538.75 * 8 = 4310

F - Value = 3992 / 538.75 = 7.409745

P-value = 0.010714

Here P-value < alpha 0.05, so we reject H0

Correct Answer: Option (B)

Source df Sum of square mean Square F P-value Caffeine 3 11976 3992 7.409745 0.010714 Error 8 4310 538.75 Total 11 16286

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