T-Test for Correlated Groups by STR Grp. 2
Download as DOC, PDF5 likes23,616 views
The t-test for two correlated groups is used to determine if there is a significant difference between the means of two groups that are correlated, such as samples tested before and after an intervention. In an example, students' sensory-motor coordination was tested with two types of music. A t-test found no significant difference between students' performance on the two music types, as the calculated t value of -3.0 was smaller than the critical value of ±2.145 for 14 degrees of freedom and a significance level of 0.05. Therefore, the null hypothesis that there is no difference between the music types was retained.
T-Test for Correlated Groups by STR Grp. 2
- 1. T-Test for Two Correlated Groups WHAT IS IT? It is a test that checks whether there is any significant difference between the population means of two groups. WHEN IS IT USED? It is used when two correlated groups of data (interval or ratio) are being compared. Two groups are correlated if: - The same samples are observed before and after the introduction of the independent variable - The samples in both the experimental and control groups are matched on some variable known to or correlated to the dependent variable Correlated groups are used to remove the effects of chance factors statistically, and to help us determine the effects of the independent variable on the dependent variable. EXAMPLE Suppose we were interested in determining whether two types of music, A and B, differ with respect to their effects on sensory-motor coordination. We test some subjects in the presence of type- A music and other subjects in the presence of type-B music. With the design for correlated samples we test all subjects in both conditions and focus on the difference between the two measures for each subject. To obviate the potential effects of practice and test sequence in this case, we would also want to arrange that half the subjects are tested first in the type-A condition, then later in the type-B condition, and vice versa for the other half. Results: MUSIC TYPE STUDENT A B 1 10.2 13.2 2 8.4 7.4 3 17.8 16.6 4 25.2 27.0 5 23.8 27.5 6 25.7 26.6 7 16.2 18.0 8 21.5 21.2 9 21.1 23.4 10 16.9 21.1
- 2. 11 24.6 23.8 12 20.4 20.2 13 25.8 29.1 14 17.1 17.7 15 14.4 19.2 Steps: 1. Set-up and test appropriate statistic 2. Formulate and interpret your conclusion. Appropriate test is t-test because of the two correlated samples (same samples before and after independent variable is introduced). 3. State null hypothesis and alternative hypothesis NULL: There is no significant difference between the performance o f students based on the two music types. ALTERNATIVE: There is a significant difference between the performance o f students based on the two music types. 4. State the level of significance (probability that test statistic falls on rejection region) α = 0.05 5. Solve for deviation (D) and D2 MUSIC TYPE STUDENT A B D (Xa-Xb) D2 1 10.2 13.2 -3.0 9 2 8.4 7.4 +1.0 1 3 17.8 16.6 +1.2 1.44 4 25.2 27.0 -1.8 3.24 5 23.8 27.5 -3.7 13.69 6 25.7 26.6 -0.9 0.81 7 16.2 18.0 -1.8 3.24 8 21.5 21.2 +0.3 0.09 9 21.1 23.4 -2.3 5.29 10 16.9 21.1 -4.2 17.64 11 24.6 23.8 +0.8 0.64
- 3. 12 20.4 20.2 +0.2 0.64 13 25.8 29.1 -3.3 10.89 14 17.1 17.7 -0.6 0.36 15 14.4 19.2 -4.8 23.04 __________________________________ Σ D -22.9 Σ D2 90.41 6. Compute t a. Calculate the sum of squares of the difference score Σ d2 = Σ D2 – ((Σ D)2 /N) = (90.41)-((-22.9)2 /N) = 55.45 b. Calculate standard error of the mean difference Standard error of mean difference = sqrt (Σ d2 /N(N-1)) = sqrt (55.45/(15)(14)) = 0.5139 _ c. Calculate D _ D = Σ D / N = -22.9/15 = -1.53 d. Calculate t _ t = D/ standard error of mean difference = -1.53/ 0.5139
- 4. = -2.97 = -3.0 6. Find critical value of t, df=14, α = 0.05 t critical = t 0.05 = +2.145 or -2.145 7. Conclusion Since calculated t is smaller than t critical, there is no significant difference between the performance of students between the two music types.