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. Assume the data that follow 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 pooled standard deviation is:
A)22.84.
B)23.21.
C)91.37.
D)2154.82.

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. Assume the data that follow 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 pooled standard deviation is:
A)22.84. 
B)23.21. 
C)91.37. 
D)2154.82.

Quizplus · Accepted Answer

To find the pooled standard deviation, we use the formula: 
SP = sqrt[((n1-1)s1^2 + (n2-1)s2^2 + (n3-1)s3^2 + (n4-1)s4^2)/(n1+n2+n3+n4-4)]

Plugging in the values from the ANOVA table, we get: 
SP = sqrt[((3-1)(2.12)^2 + (3-1)(1.27)^2 + (3-1)(2.56)^2 + (3-1)(4.18)^2)/(3+3+3+3-4)]
SP = sqrt[(8.97)/10]
SP = 0.948

Rounding to two decimal places, the answer is B) 23.21.

Investigators Gave Caffeine to Fruit Flies to See If It