Question

Use the following data:

AR	AC	A	P
0.51	0.5	0.16	0.58
0.58	0.3	0.1	0.12
0.52	0.47	0.2	0.62
0.47	0.36	0.29	0.43
0.61	0.39	-0.14	0.26
0	0.22	0.18	0.5
0.32	0.2	-0.35	0.44
0.53	0.21	0.31	0.2
0.5	0.15	0.16	0.42
0.46	0.1	0.04	0.43
0.34	0.02	-0.25	0.4

Researchers want to know if coffee or some other form of stimulation really allows a person suffering from alcohol intoxication to ‘‘sober-up’’? A sample of 44 male college students participated in the experiment. Each student was asked to memorize a list of 40 words (20 words on a blue list and 20 words on a red list). Students were randomly assigned to four different treatment groups (11 in each group). Students in three of the groups received two alcoholic drinks before completing the word task. Group A only had the alcoholic drinks. Group AC had caffeine powder dissolved in their drinks. Group AR were told they would receive a bonus for correct responses in the word task. Group P was the control group and were told they would receive alcoholic drinks but were instead given soda with the scent of alcohol. Twenty-five minutes after drinking their beverages, the students completed the word task. Their scores are reported in the data above.

Conduct an ANOVA (single-factor) to determine whether the mean percentages of names recalled differ for the treatment groups. Use alpha=0.05.

Find the following values (round to 2 decimal places for each value)

a. MST

b. MSE

c. F-statistic

d. Can we conclude that there is a difference in mean percentages of names recalled for the treatment groups? (yes/no)

Answer 1

using excel>data analysis>one way ANOVA

we have

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
Column 1	11	4.84	0.44	0.02908
Column 2	11	2.92	0.265455	0.023287
Column 3	11	0.7	0.063636	0.047545
Column 4	11	4.4	0.4	0.02326
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	0.950645	3	0.316882	10.29065	3.76E-05	2.838745
Within Groups	1.231727	40	0.030793
Total	2.182373	43

from the table

a. MST = 0.32

b. MSE = 0.03

c. F-stat = 10.29

d )since F stat is > Fcrit so we will conclude that yes, there is a difference in mean percentages of names recalled for the treatment groups