The t-test is widely used in research to verify whether the observed difference between two averages obtained in the samples is considered large to be significant.
Assuming two establishments discuss which one has the most satisfied customers. To measure the degree of satisfaction with customers, it is decided to conduct a satisfaction survey by applying a questionnaire with 5-point interval questions.
Client A obtained an overall average of 2.85 and Client B obtained an overall average of 3.45. Supposedly, it turns out that customer B has more satisfied customers than A.
The t-test aims precisely to verify if such difference is significant and to explain if the differences between the means occur due to the sampling error or not.
When working with small samples, there is a tendency for sample averages to actually differ, even if they originate from the same population. In this case the t test aims to verify if the degree of difference between the two sets may be due to factors other than the sampling error.
Conditions for application
- Only for interval questions;
- When the population variance is unknown;
- cannot be any size.
Enforcement Procedure
1. Determine H_{0}, with no differences between the means;
2. Determine H_{1}, for the existence of difference between the means;
3. Establish a level of significance;
4. Calculate t, where the degrees of freedom, φ = n_{1} + n_{2} - 2
_{ }
where SQ is the sum of squares and x1 and x2 are the means of each group.
The above formula may differ in some statistic books that deal with unequal samples, however, it includes samples of equal or not equal sizes.
Compare the tabulated t with calculated t and reject the null hypothesis in favor of the alternative, if t calculated higher than the tabulated t.
Example
THE | B | |||
1 | 3 | 2 | 3 | |
2 | 4 | 4 | 3 | |
3 | 3 | 5 | 4 | |
2 | 2 | 3 | 5 | |
4 | 1 | 3 | 3 | |
3 | 3 | 4 | 2 | |
3 | 4 | 2 | 1 | |
3 | 2 | 5 | 3 | |
4 | 4 | 4 | 4 | |
4 | 2 | 5 | 2 | |
T = 57 | T = 69 | |||
n = 20 | n = 20 | |||
X = 2,85 | X = 3,45 | |||
SQ = 18.55 | SQ = 24.95 |
Since t is 2.02 with 38 degrees of freedom and t is calculated as 1.77, the null hypothesis is rejected in favor of the true hypothesis.
Therefore, it is concluded that both groups of customers are satisfied and that the differences between the means are probably due to sampling error.
Next: Analysis of Variance