Question

Three real estate agents were each asked to assess the values of five houses in a neighbourhood. The results, in thousands of Kwacha, are given in the table below

House Agent

A . B C

1 . 210. 218 . 226

2. 192 . 190. 198

3. 183. 187. 185

4 . 227. 223. 237

5. 242 . 240. 237

For the two way analysis of variance model with one observation per cell, we write the observation from the jth group and ith block as

Yij= µ + βi + τ j + Ԑ i j

CLEARLY AND NEATLY,

(a) Explain each term in the above model in the context of the given information.

(b) Consider the observation on agent B and house 1 (Y21 = 218)

i. Estimate µ

ii. Estimate and interpret β2

iii. Estimate and interpret τ2

iv. Estimate Ԑ21

(c) Name the blocking variable and treatment in this experiment.

(d) What is the purpose of the blocks in this experiment?

(e) Prepare a two way analysis of variance table

(f) Test at the 5% level the null hypothesis that population mean valuations are the same for the three real estate agents.

(g) Test at the 5% level the null hypothesis that population mean valuations are the same for the five houses.

Answer 1

(a) Let, Yij = Observation corresponding ith house and jth Agent; j = 1, 2, 3, 4, 5; i = A,B,C H= general mean Bi = effect due to ith Agent Tj = effect due to ith House Eij = error corresponding (ii)th observation. b. i. À = (210+218 +226 + 192 + 190 + 198 + 183 +187 + 185 +227 +223+ 237 +242 +240 +237)/15 = 213 ü. BB = B2 = (218 +190+187+223 +240)/5 - = (218+ 190+187+223+ 240)/5 - 213 = -1.4 ii. 12 = (192 + 190 + 198)/3 - À = -19.6667 iv. €21 = Y21 - - B2-71 = 218 - 213 +1.4 - ((210+218 +226)/3-213) = 1.4 (c) Blocking variable = Agent Treatment = House (d) To answer this question : Is there significant effect due Agent?

		Agent
House	A	B	C		Anova: Two-Factor Without Replication
1	210	218	226
2	192	190	198		SUMMARY	Count	Sum	Average	Variance
3	183	187	185		House 1	3	654	218	64
4	227	223	237		House 2	3	580	193.3333	17.33333333
5	242	240	237		House 3	3	555	185	4
					House 4	3	687	229	52
					House 5	3	719	239.6667	6.333333333
					Agent 1	5	1054	210.8	590.7
					Agent 2	5	1058	211.6	512.3
					Agent 3	5	1083	216.6	566.3
					ANOVA
					Source of Variation	SS	df	MS	F	P-value	F crit
					House	6488.666667	4	1622.167	68.8330976	3.1062E-06	3.837853
					Agent	98.8	2	49.4	2.096181047	0.18535681	4.45897
					Error	188.5333333	8	23.56667
					Total	6776	14

(f) Since p-value=0.18535681>0.05 hence we fail to reject null hypothesis that population mean valuations are the same for the three real estate agents.

(g) Since p-value=0.000003<0.05 hence we reject the null hypothesis that that population mean valuations are the same for the five houses.