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).