Question

A research firm tests the miles-per-gallon characteristics of three brands of gasoline. Because of different gasoline performance characteristics in different brands of automobile, five brands of automobiles are selected and treated as blocks in the experiment; that is, each brand of automobile is tested with each type of gasoline. The results of the experiment (in miles per gallon) follow.
Gasoline Brands
Automobiles I II III
A 18 21 20
B 24 26 27
C 30 29 34
D 22 25 24
E 20 23 24
A.) At ? = .05, is there a significant difference in the mean miles-per-gallon characteristics of the three brands of gasoline?
B.) Analyze the experimental data using the ANOVA procedure for completely randomized designs. Compare your findings with those obtained in part (A). what is the advantage of attempting to remove the block effect?

Answer 1

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.