The city of Tucson, Arizona, employs people to assess the value of homes for the

ID: 3154352 • Letter: T

Question

The city of Tucson, Arizona, employs people to assess the value of homes for the purpose of establishing real estate tax. The city manager sends each assessor to the same five homes and then compares the results. The information is given below, in thousands of dollars. Can we conclude that there is a difference in the assessors? Use the 0.05 significance level.

Is there a difference in the treatment means? (Round your answer to 2 decimal places.)

Is there a difference in the block means? (Round your answer to 2 decimal places.)

a difference in the block means.

Explanation / Answer

This a a two way anova model without replication with treatment as the assessor and the blocks as the homes. We will do that in R. Here is the code to run two way anova in R.

setwd('D:/rex') # directory where you have saved the csv file.

asmnt<-read.csv("assess.csv") # reading the dataset.
asmnt

This takes the data R. with the following output.

Output:

Homes Assessor Values
1 A Zawodny 56
2 A Norman 54
3 A Cingle 49
4 A Holiday 43
5 B Zawodny 55
6 B Norman 57
7 B Cingle 52
8 B Holiday 54
9 C Zawodny 44
10 C Norman 54
11 C Cingle 46
12 C Holiday 57
13 D Zawodny 77
14 D Norman 67
15 D Cingle 63
16 D Holiday 61
17 E Zawodny 82
18 E Norman 89
19 E Cingle 94
20 E Holiday 81

Now following is the code to write two-way anova:

asmnt$Assessor=as.factor(asmnt$Assessor)
asmnt$Homes=as.factor(asmnt$Homes)

anova(lm(Values ~ Homes + Assessor , asmnt))

Output:

Analysis of Variance Table

Response: Values
Df Sum Sq Mean Sq F value Pr(>F)
Homes 4 3806.0 951.50 26.9802 6.425e-06 ***
Assessor 3 72.5 24.18 0.6857 0.5778
Residuals 12 423.2 35.27
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

From the ANOVA table the computed F value for treatment is F(assessor) = 0.6857

As the p-value for Assessor is 0.5778 which is greater than 0.05, so at 5% signifcance level we will fail to reject the null hypothesis to conclude that there is no significant difference in treatment means.

Computed F value for blocks = F(homes) = 26.9802

As the p-value for for homes is close to 0.0000 which is much less than 0.05, at 5% significance level we will reject null hypothesis to conclude that there is a significant difference in means of the blocks.

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The city of Tucson, Arizona, employs people to assess the value of homes for the

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