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 ten compares the results. The information is given below, in thousands of dollars. Can we conclude that there is a difference in the assessors, at a significance level of 0.05?

Home	Smith	Norman	Thomas	Holiday
A	53	55	49	45
B	50	51	52	53
C	48	52	47	53
D	70	68	65	64
E	84	89	92	86

Part A:

Assessor A.

What is the null hypothesis statement for this problem?

B. What is the alternative hypothesis statement for this problem?

C. What is alpha for this analysis?

D. What is the most appropriate test for this problem? (choose one of the following)

a. F-Test Two-Sample for Variance

b. Anova: Single Factor

c. Anova: Two-Factor with Replication

d. Anova: Two-Factor without Replication

E. What is the value of the test statistic for the most appropriate analysis?

F. What is the value of the critical value for the most appropriate analysis?

G. Is it reasonable to conclude that there is a difference in the mean home value based on the assessor? (choose one of the following)

a. Yes

b. No

h. What is the p-value for this analysis?

Part B:

Home

A. What is the null hypothesis statement for this problem?

B. What is the alternative hypothesis statement for this problem?

C. What is alpha for this analysis?

D. What is the most appropriate test for this problem? (choose one of the following)

a. F-Test Two-Sample for Variance

b. Anova: Single Factor

c. Anova: Two-Factor with Replication

d. Anova: Two-Factor without Replication

E. What is the value of the test statistic for the most appropriate analysis?

F. What is the value of the critical value for the most appropriate analysis?

G. Is it reasonable to conclude that there is a difference in the value based on the home? (choose one of the following)

a. Yes

b. No

h. What is the p-value for this analysis?

you must show your work and you must use the appropriate formulas in Excel to answer the questions. SHOW ALL FORMULAS USED IN DETAILS.

Answer 1

Anova: Two-Factor Without Replication

Anova: Two-Factor Without Replication						row data, j			Factor A
	Smith	Norman	Thomas	Holiday	total		count, n j	mean , x̅ j	std. dev., sj	sample variances, sj^2	( x̅ j - x̅̅)²	nj( x̅j - x̅̅)²
A	53	55	49	45	202	A	4	50.5	4.4347	19.667	116.640	466.560
B	50	51	52	53	206	B	4	51.5	1.2910	1.667	96.040	384.160
C	48	52	47	53	200	C	4	50	2.9439	8.667	127.690	510.760
D	70	68	65	64	267	D	4	66.75	2.7538	7.583	29.703	118.810
E	84	89	92	86	351	E	4	87.75	3.5000	12.250	699.603	2798.410
6						6
7						7
total	305	315	305	301	1226						SSA = total=	4278.700
											SSA=Σnj( x̅j - x̅̅)² =	4278.700
column data , i (Factor B)	Smith	Norman	Thomas	Holiday						SST =	Σ(Xij - X̅̅ ) =	4428.2
count, ni =	5	5	5	5
mean , x̅ i =	61.000	63.000	61.000	60.200						SSE =	SST-SSA-SSB =	128.100
std. dev., si =	15.524	16.047	18.695	15.928
sample variances, si^2 =	241.000	257.500	349.500	253.700
grand mean , x̅̅ =	ΣXij/Σn =				61.3
square of deviation of sample mean from grand mean,( x̅ - x̅̅)²	0.09	2.89	0.09	1.21
					TOTAL
SS(between)= SSB = Σni( x̅ i - x̅̅)² =	0.45	14.45	0.45	6.05	21.4

so, SSA=   4278.700
SSB=   21.4
SSE=   128.100
df factor A(row) = r-1 =   4
df factor B(column) = c-1 =    3
here, N =    20
df error = (r-1)(c-1) =   12
  
mean square factor A , MSA = SSA/df=    1069.675
  
mean square(factor B) =MSB= SSB/df =    7.133333333
  
mean square error = MSE =SSE/df =    10.675
  
F-statistics  
Factor A = MSA/MSE =    100.2037471
factor B = MSB/MSE=   0.668227947

ANOVA
Source of Variation	SS	df	MS	F-stat	p-value	F-critical	Result
Rows	4278.7000	4	1069.6750	100.204	0.0000	3.2591667	significant
Columns	21.4000	3	7.1333	0.668	0.5876	3.4902948	not significant
Error	128.1000	12	10.6750
Total	4428.2000	19
					α =	0.05

using above result, we can answer following question:

Part A:

Assessor A.

What is the null hypothesis statement for this problem?

there is no difference in the mean home value based on the assessor

B. What is the alternative hypothesis statement for this problem?

there is a difference in the mean home value based on the assessor

C. What is alpha for this analysis?

0.05

D. What is the most appropriate test for this problem? (choose one of the following)

d. Anova: Two-Factor without Replication

E. What is the value of the test statistic for the most appropriate analysis?

F-stat=0.6682

F. What is the value of the critical value for the most appropriate analysis?

F-critical=3.4903

G. Is it reasonable to conclude that there is a difference in the mean home value based on the assessor? (choose one of the following)

since F-stat<F-critical, do not reject Ho,

so, answer is

b. No

h. What is the p-value for this analysis?

0.5876

Part B:

Home

A. What is the null hypothesis statement for this problem?

there is no difference in the value based on the home

B. What is the alternative hypothesis statement for this problem?

there is a difference in the value based on the home

C. What is alpha for this analysis?

0.05

D. What is the most appropriate test for this problem? (choose one of the following)

d. Anova: Two-Factor without Replication

E. What is the value of the test statistic for the most appropriate analysis?

F-stat=100.2037471

F. What is the value of the critical value for the most appropriate analysis?

F-critical=3.2592

G. Is it reasonable to conclude that there is a difference in the value based on the home? (choose one of the following)

since, F-stat>f-critical, so, reject Ho

hence answer is

a. Yes

h. What is the p-value for this analysis?

p-value=0.000

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please revert for doubts