THE RELATIONSHIP BETWEEN THE ANALYSIS OF VARIANCE
THE RELATIONSHIP BETWEEN THE ANALYSIS OF VARIANCE
AND THE t TEST
When a study involves just two independent groups and we are testing the null hypothesis that μ1 = μ2, we can use either the t test for independent groups or the analysis of variance. In such situations, it turns out that t2 = F. To illustrate this point, let’s analyze a suitable situation with both the t test and the analysis of variance.
Example: Dating – Sorority Versus Dormitory Women
A sociologist wants to determine whether sorority or dormitory women date more often. He randomly samples 12 women who live in sororities and 12 women who live in dormitories and determines the number of dates they each have during the ensuing month. The results are shown in Table 15.4.
table 15.4 Dating experiment results
|Sorority Women |Dormitory Women |
|X1 |X12 |X2 |X22 |
| 8 | 64 | 9 | 81 |
|5 |25 |7 |49 |
|6 |36 |3 |9 |
|4 |16 |4 |16 |
|12 |144 |4 |16 |
|7 |49 |8 |64 |
|9 |81 |7 |49 |
|10 |100 |5 |25 |
|5 |25 |8 |64 |
|3 |9 |6 |36 |
|7 |49 |3 |9 |
|5 |25 |5 |25 443 |
|81 |623 |69 | |
| n1 = 12 n2 = 12 |
|[pic]= 6.750 [pic]= 5.750 |
Based on the data in Table 15.4, what is his conclusion? Use α = 0.052 tail.
SOLUTION
The null hypothesis is that μ1 = μ2. This is an independent groups design, so we can use either the t test for independent groups or the analysis of variance to evaluate this hypothesis. We shall first analyze the data with the t test, as follows:
| [pic] | [pic] |
|[pic] | |
|[pic] | |
|[pic] |
With α = 0.052 tail and df = N – 2 = 22, from Table D in Appendix D,
tcrit = ±2.074
Since ItobtI < 2.074, we retain H0. We cannot conclude that the sorority or dormitory women date more often.
This problem can also be solved with the analysis of variance. The solution is shown here.
STEP 1: Calculate SSB:
| [pic] |
STEP 2: Calculate SSW:
| [pic] |
STEP 3: Calculate SST:
| [pic] |
As a check,
| [pic] |
STEP 4: Calculate df:
[pic]
STEP 5: Calculate sB2:
| [pic] |
STEP 6: Calculate sW2:
| [pic] |
STEP 7: Calculate Fobt:
| [pic] |
With α = 0.05, dfnumerator = 1, and dfdenominator = 22, from Table F,
Fcrit = 4.30
Since Fobt < 4.30, we retain H0. We cannot conclude that sorority or dormitory women date more often.
Note that we have reached the same conclusion with both the t test and the F test. A comparison of the two obtained statistics and the associated critical values are shown here:
| tobt2 = Fobt | tcrit2 = Fcrit |
|(1.04)2 = 1.08 |(2.074)2 = 4.30 |
|1.08 = 1.08 |4.30 = 4.30 |
Thus, in the independent groups design, when there are two groups and we are testing the null hypothesis that μ1 = μ2, tobt2 = Fobt and tcrit2 = Fcrit. Therefore, both tests give the same result.
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