House Type Appraiser 1 2 3 Cape Cod 425 415 430 Colonial 530 550 540 Ranch 390 4
ID: 3371236 • Letter: H
Question
House Type
Appraiser
1
2
3
Cape Cod
425
415
430
Colonial
530
550
540
Ranch
390
400
380
House Type
Appraiser
1
2
3
Cape Cod
425
415
430
Colonial
530
550
540
Ranch
390
400
380
2eNov Refer to Chapter 13 problem 41 on p. 457. Refer to Houses data on Connect or Webcam pus. At the 5% significance level, can you conclude that the a differ by appraiser? (20 points) 2. verage values State the hypothesis: What is the test statistic? What is the critical value rejection rule? What is the p-value? What is the p-value rejection rule? What is the decision to the test? Interpret your decision: Note. Attach Excel output that shows the test results. Hint: thesame output is used for #2Explanation / Answer
Solution:
Here, we have to use two way analysis of variance or two way ANOVA F test for checking the given hypothesis. The null and alternative hypotheses for this test are given as below:
Null hypothesis: H0: The average values do not differ by appraiser.
Alternative hypothesis: Ha: The average values differ by appraiser.
We are given 5% level of significance. ? = 0.05
The test statistic for this test is given as below:
F = MS treatment / MS error
The test statistic value from excel output is given as below:
F for row = 0.325
F for column = 167.5
The critical value for this test is given as below:
Critical value F = 6.944272
(Critical value is calculated by using excel or F-table. Excel command: =Finv(0.05,2,4). ? = 0.05, df1 = 2, df2 = 4)
Rejection rule:
We reject the null hypothesis if the test statistic value is more than critical value, otherwise we do not reject the null hypothesis.
We reject the null hypothesis if the p-value is less than the alpha value 0.05, otherwise we do not reject the null hypothesis.
Excel data and output for this ANOVA is given as below:
House Type
Appraiser
Cape Cod
Colonial
Ranch
1
425
530
390
2
415
550
400
3
430
540
380
Anova: Two-Factor Without Replication
SUMMARY
Count
Sum
Average
Variance
1
3
1345
448.3333
5308.333
2
3
1365
455
6825
3
3
1350
450
6700
Cape Cod
3
1270
423.3333
58.33333
Colonial
3
1620
540
100
Ranch
3
1170
390
100
ANOVA
Source of Variation
SS
df
MS
F
P-value
F crit
Rows
72.22222
2
36.11111
0.325
0.73997
6.944272
Columns
37222.22
2
18611.11
167.5
0.000139
6.944272
Error
444.4444
4
111.1111
Total
37738.89
8
P-value for rows is given as 0.73997 > ? = 0.05, so we do not reject the null hypothesis that average values do not differ by appraiser.
There is sufficient evidence to conclude that average values do not differ by appraiser.
P-value for column is given as 0.000139 < ? = 0.05, so we reject the null hypothesis that average values do not differ by house type.
There is sufficient evidence to conclude that average values are differ by house type.
House Type
Appraiser
Cape Cod
Colonial
Ranch
1
425
530
390
2
415
550
400
3
430
540
380
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