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.)<u><em>The advantage of attempting to remove the block effect is</em></u>
The completely randomized designs does not prove that H0 is incorrect only that it cannot be rejected.
Step-by-step explanation:
<em><u /></em>
<em><u>Using Two-way ANOVA method</u></em>
Given problem
<em><u>Observation              I          II       III          Row total (xr)</u></em>
A                              18	21	20            59
B                            24	26	27             77
C                            30	29	34             93
D                            22	25	24            71
<u>E                            20	23	24           63                      </u>
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
<u>ANOVA table</u>
Source                 Sums         Degrees      Mean Squares
of Variation       of Squares   of freedom
<u>                               SS                 DF              MS       F	p-value</u>
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	
<u>Error (residual)SSE=13.3333	(r-1)(c-1)=8	MSE=1.6667                  </u>
<u>Total	SST=253.7333	rc-1=14                                                        </u>
Conclusion:
<u> 1. F for between Rows</u>
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
<u>2. F for between Columns</u>
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
<u><em>Part B:</em></u>
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 
<u>                 (df1,df2)    of Square (SS)	Square (MS)                    P-value</u>
 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
<u> Error     12   230.4            19.2                                           </u>
<u>Total	17	1750.2778	102.9575                                                         </u>
<u />
<u>Factor - A- Rows</u>
 Since p-value < α, H0 is rejected.
<u>Factor - B- Columns</u>
Since p-value > α, H0 can not be rejected.
The averages of all groups assume to be equal.
<u>Interaction AB</u>
 Since p-value > α, H0 can not be rejected.
<u><em>The advantage of attempting to remove the block effect is</em></u>
The completely randomized designs does not prove that H0 is incorrect only that it cannot be rejected.