Respuesta :
Answer:
A.) At α = 0.05, is there a significant difference in the mean miles-per-gallon characteristics of the three brands of gasoline.
B.)The advantage of attempting to remove the block effect is
The completely randomized designs does not prove that H0 is incorrect only that it cannot be rejected.
Step-by-step explanation:
Using Two-way ANOVA method
Given problem
Observation I II III Row total (xr)
A 18 21 20 59
B 24 26 27 77
C 30 29 34 93
D 22 25 24 71
E 20 23 24 63
Col total (xc) 114 124 129 367
∑x²=9233→(A)
∑x²c/r
=1/5(114²+124²+129²)
=1/5(12996+15376+16641)
=1/5(45013)
=9002.6→(B)
∑x²r/c
=1/3(59²+77²+93²+71²+67²)
=1/3(3481+5929+8649+5041+4489)
=1/3(27589)
=9196.3333→(C)
(∑x)²/n
=(367)²/15
=134689/15
=8979.2667→(D)
Sum of squares total
SST=∑x²-(∑x)²/n
=(A)-(D)
=9233-8979.2667
=253.7333
Sum of squares between rows
SSR=∑x²r/c-(∑x)²/n
=(C)-(D)
=9196.3333-8979.2667
=217.0667
Sum of squares between columns
SSC=∑x²c/r-(∑x)²/n
=(B)-(D)
=9002.6-8979.2667
=23.3333
Sum of squares Error (residual)
SSE=SST-SSR-SSC
=253.7333-217.0667-23.3333
=13.3333
ANOVA table
Source Sums Degrees Mean Squares
of Variation of Squares of freedom
SS DF MS F p-value
B/ w SSR=217.0667 4 MSR=54.2667 32.56 0.0001
rows
B/w SSC=23.3333 c-1=2 MSC=11.6667 7 0.01
columns
Error (residual)SSE=13.3333 (r-1)(c-1)=8 MSE=1.6667
Total SST=253.7333 rc-1=14
Conclusion:
1. F for between Rows
The critical region for F(4,8) at 0.05 level of significance=3.8379
The calculated F for Rows=32.56>3.8379
Therefore H0 is rejected
2. F for between Columns
The critical region for F(2,8) at 0.05 level of significance=4.459
We see that the calculated F for Colums=7>4.459
therefore H0 is rejected,and concluded that there is significant differentiating between columns
Part B:
To analyze the data for completely randomized designs click on anova two factor without replication in the data analysis dialog box of the excel spreadsheet.
The following table is obtained
Source DF Sum Mean F Statistic
(df1,df2) of Square (SS) Square (MS) P-value
Factor A 1 1496.5444 1496.5444 769.6514 0.001297
Rows
Factor B - 2 19.4444 9.7222 5 0.1667
Columns
Interaction
AB 2 3.8889 1.9444 0.1013 0.9045
Error 12 230.4 19.2
Total 17 1750.2778 102.9575
Factor - A- Rows
Since p-value < α, H0 is rejected.
Factor - B- Columns
Since p-value > α, H0 can not be rejected.
The averages of all groups assume to be equal.
Interaction AB
Since p-value > α, H0 can not be rejected.
The advantage of attempting to remove the block effect is
The completely randomized designs does not prove that H0 is incorrect only that it cannot be rejected.
