Investigators gave caffeine to fruit flies to see if it affected their rest. The
ID: 3216861 • Letter: I
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.
Treatment Minutes of rest
Control. 450 413 418
Low dose 466 422 435
Medium dose 421 453 419
High dose. 364 330 389
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
1. The numerator degrees of freedom for the ANOVA F- test are:
a. 2
b. 3
c. 8
d. 11
2. The denominator degrees of freedom for the ANOVA F-test are:
a. 2
b. 3
c. 8
d. 11
3. 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. 2.78
b. 4.73
c. 4.82
d. 7.41
4. The P-value of this test is:
a. greater than 0.1
b. between 0.05 and 0.1
c. less than 0.05
d. it is not possible to determine the P-value from the info. provided
Explanation / Answer
1. The numerator degrees of freedom for the ANOVA F- test are:
Answer : K-1 = 4-1 = 3
2. The denominator degrees of freedom for the ANOVA F-test are:
Answer : n - K = 12 - 4 = 8
3. 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:
SSTr = 11976 (Given)
MSTr = SSTr / DF = 11976 / 3 = 3992
MSE = 538.75
F = MSTr / MSE = 7.41
Answer : 7.41
4. The P-value of this test is:
Answer : less than 0.05
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