Four suppliers were asked to quote prices for seven different building materials
ID: 3173376 • Letter: F
Question
Four suppliers were asked to quote prices for seven different building materials. The average quote of supplier A was 1315.8. The average quote of suppliers B, C, and D were 1238.4, 1225.8, and 1200.0 respectively. The following is calculated ANOVA table with some entries missing.
(a) Complete the table using the information provided above
(b) Is there a significant difference between the quotes of different suppliers?
(c) Construct a 90% confidence interval for the difference between the mean quotes of suppliers A and D. Would you say there is evidence of a difference?
(d) Construct a 95% simultaneous confidence interval for the difference between suppliers A and D and between suppliers B and C.
Source df sum of squares mean square F-value Materials 17800 Suppliers Error Total 358700Explanation / Answer
Given,
Source
df
sum of squares
mean square
F-value
Materials
17800
Suppliers
Error
Total
358700
Quotes of suppliers A,B,C,D are 1315.8,1238.4,1225.8 and 1200
Averaage of qutes= (1315.8 + 1238.4 + 1225.8 + 1200)/4=1245
sum of squares(SS) =7*× [(1315.8 1245)2 + (1238.4 1245)2 + (1225.8 1245)2 + (1200 1245)2 ] sum of squares(SS) for suppliers= 52148.88
Mean within squares(MS)=52418.88/4-1
Mean within squares(MS)=17382.96
Degrees of freedom are 7-1=6 and 4-1=3
error=6*3=18
SS for materials=Df*mean square
SS for materials =6*17800=106800
SS for error=358700- SS for materials- SS for suppliers
SS for error =358700-106800-52148.88
SS for error =199751.12
Mean square (MS) for error=199751.12/18=11097.28
F value for materials=MS(materials)/MS(error)
=17800/11097.28
F value for materials =1.604
F value for suppliers=MS(suppliers)/MS(error)
17382.96/11097.28
F value for suppliers =1.567
A.a) now the complete table looks like
Source
df
sum of squares
mean square
F-value
Materials
6
106800
17800
1.604
Suppliers
3
52148.88
17382.96
1.567
Error
18
199751.12
11097.28
Total
27
358700
A.b) Null hypothesis-H0: There is no difference between quotes given by suppliers
Alternate hypothesis-H1: There is a significant difference between quotes given by suppliers
we already have F-value for suppliers from the anova table 1.567
at 5% significance level the critical value from table is 3.16
As F-value is less htan critical value(1.567<3.16), we do not reject the null hypothesis
Hence, There is no difference between quotes given by suppliers
A.c) for confidence intervals in 2-way analysis, we use t-statistic
at 10% significance level and df=18
we have critical value of 1.734 from t-distribution table
average quote(A)=1315.8
average quote(D)=1200
confidence interval is given as,
1315.8 1200 ± 1.734*sqrt(11097.28(1/7+1/7)
solving, we get (18.16,213.44)
both lower and upper limits are positive
this indicates definitely there is a difference between quotes of suppliers A and D
A.d) ) for simultaneous confidence intervals in 2-way analysis, we use F-statistic
At 5% significance level
F-value critical=3.16
the confidence intervals are
A and D
CI=1315.8 1200 ± sqrt(3.16 × 11097.28 × 3 × 2/7)
solving we get
CI=(-57.57,289.17)
B and C
CI= 1238.4 1225.8 ± sqrt( 3.16 × 11097.28 × 3 × 2/7)
solving we get
CI=(-160.77,185.97)
Source
df
sum of squares
mean square
F-value
Materials
17800
Suppliers
Error
Total
358700
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