An ANOVA test is conducted to compare three different income tax software packages to determine whether there is any difference in the average time it takes to prepare income tax returns using the three different software packages. Ten different individuals' income tax returns are done by each of the three software packages and the time is recorded for each. Because individual tax returns are different the analysis has decided to block on individual returns. The computer results are shown below. The level of significance α = 0.05 for all tests. Complete the ANOVA table as needed to answer the following questions.
SUMMARY |
Count |
Sum |
Average |
Variance |
1 |
3 |
9 |
3 |
1 |
2 |
3 |
30 |
10 |
1 |
3 |
3 |
12 |
4 |
0 |
4 |
3 |
6.5 |
2.17 |
0.58 |
5 |
3 |
25 |
8.33 |
2.33 |
6 |
3 |
7 |
2.33 |
1.08 |
7 |
3 |
10 |
3.33 |
0.33 |
8 |
3 |
18 |
6 |
1 |
9 |
3 |
33.5 |
11.17 |
0.58 |
10 |
3 |
4.5 |
1.5 |
0.25 |
Software A |
10 |
47.5 |
4.75 |
12.96 |
Software B |
10 |
47.5 |
4.75 |
10.79 |
Software C |
10 |
60.5 |
6.05 |
13.47 |
ANOVA
Source of Variation |
SS |
df |
MS |
F |
F crit |
Returns (Blocking) |
329.91 |
9 |
36.66 |
||
Software |
|||||
Error (within) |
5.07 |
18 |
|||
Total |
346.25 |
29 |
What is the calculated value of the F test statistic (select the closest answer) for conducting the hypothesis test to determine whether there is a difference in the software packages?
a.F = 2.46
b.F = 3.55
c.F = 130.93
d.F = 20.11
At the α = 0.05 level of significance what conclusions would you make as to whether the type of software used has an effect on the time required to complete individual tax returns?
a.Because F calculated = 130.93 > F critical = 3.55 do not reject the null hypothesis and conclude that type of software used does not make a difference in average tax preparation time.
b.Because F calculated = 20.11 > F critical = 3.55 reject the null hypothesis and conclude that type of software used does make a difference in average tax preparation time.
c.Because F calculated = 2.46 is < F critical = 3.55, do not reject the null hypothesis and conclude that type of software used does not make a difference in average tax preparation time.
d.No decision can be made as to whether the type of software used makes a difference or not in average tax preparation time without conducting a post-test comparison of all possible contrasts.
Suppose in the tax preparation ANOVA problem above, a post-test comparison is conducted to determine where differences exist between all possible contrasts (all pairs) of software packages. For this question focus only on the contrast between Software A and Software C. If Fisher’s Least Significant Difference is computed to be 0.497 then what conclusion would you make concerning whether there is a significant different between Software A and Software C?
a.No conclusion can be reached unless a pooled t-test is conducted.
b.Because the absolute difference in sample means between Software A and Software C are greater than Fisher’s Least Significant Difference we conclude the average tax preparation time for the two software packages are equal.
c.Because the absolute difference in sample means between Software A and Software C are greater than Fisher’s Least Significant Difference we conclude the average tax preparation time for the two software packages are different.
d.The confidence interval for the difference between the two software packages would need to include the value of 0 (zero) for us to conclude that the average tax preparation time for the two software packages are different.
Source of Variation | SS | df | MS | F | F crit |
Returns (Blocking) | 329.91 | 9 | 36.66 | 130.153846 | 2.45628115 |
Software | 11.27 | 2 | 5.635 | 20.0059172 | 3.55455715 |
Error (within) | 5.07 | 18 | 0.28166667 | ||
Total | 346.25 | 29 |
Source of Variation | SS | df | MS | F | F crit |
Returns (Blocking) | 329.91 | 9 | 36.66 | =D2/D4 | =F.INV.RT(0.05,C2,18) |
Software | =B6-B2-B4 | 2 | =B3/C3 | =D3/D4 | =F.INV.RT(0.05,C3,18) |
Error (within) | 5.07 | 18 | =B4/C4 | ||
Total | 346.25 | 29 |
for software
option D) F = 20.11 is correct
critical value = 3.55
option B) is correct
F = 20.11 > 3.55
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