Investigators gave caffeine to fruit flies to see if it affected their rest. The four treatments were controlled, 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.
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 information provided.

Respuesta :

Answer:

The P-value of this test is less than 0.05 (option C).

Step-by-step explanation:

Source df Sum of square mean Square     F- value         P-value

Caffeine 3 11976                  3992               7.409745        0.010714

Error 8 4310                  538.75  

Total 11 16286

Note: df = degree of freedom

The df of total = 12-1 = 11

The df of Caffeine = 4 -1 = 3

The sum of square of Caffeine = 11976 / 3 = 3992

The df of error = 11 - 3 = 8

The sum ofsSquare of Error = 538.75 * 8 = 4310

F - value = 3992 / 538.75 = 7.409745

P-value = 0.010714

P-value < alpha 0.05

Therefore, the P-value of this test is less than 0.05 (option C).