Question

Memory drug Placebo No treatment 70 37 43 50 10 17 23 83 90 97 57 63 30 Mean (Y) Variance (4 (Y Y) Grand Mean () Grand Variance ((Yy Y)2) 83.4 50.0 16.6 112.3 109.0 112.3 50 892.14 3. (10 points) Students were given different drug treatments before revising for their exams. Some were given a memory drug, some a placebo drug and some no treatment. The exam score are shown below for the three different groups (ive students in each group). Carry out ANOVA to test the hypothesis that θ1-02 = 03 with level α 0.05. (The to be very accurate but you need to write down the steps. For the F statistic, write down the degree of freedom.) computation doesn't have

Please type the answer by computer

Answer 1

test statistic

F=Between(MS)/Within(MS)=50.16 with (2,12) df.

the critical F(0.05,2,12)=3.89 is less than calculated F=50.16, so we fail to accept (or reject) null hypothesis H0 and conclude that atleast one of the $\theta$ is different from others

						within SS		between SS
	Group	nj	mean(xj-)	s2	nj*xj-	(n-1)s2	(xj--x-)	nj(xj--x-)2
	1	5	83.4	112.3	417	449.2	33.4	5577.8
	2	5	50	109	250	436	0	0
	3	5	16.6	112.3	83	449.2	-33.4	5577.8
	sum	15	150	333.6	750	1334.4	0	11155.6
	grand mean(x-)	50
			ANOVA
	SOURCE	DF	SS	MS	F	CRITICAL F(0.05)	p-value
	BETWEEN	2	11155.60	5577.8	50.16	3.89	1.49E-06
	WITHIN(ERROR)	12	1334.40	111.2
	TOTAL	14	12490.00

this can be done using ms-excel data analysis tool kit

Anova: Single Factor
SUMMARY
Groups	Count	Sum	Average	Variance
Column 1	5	417	83.4	112.3
Column 2	5	250	50	109
Column 3	5	83	16.6	112.3
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	11155.6	2	5577.8	50.16007	1.49E-06	3.885294
Within Groups	1334.4	12	111.2
Total	12490	14