Correlated T-Test
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This document describes the steps to conduct a paired t-test to determine if a special shoe increases runners' speeds compared to ordinary shoes. Seven runners' speeds were measured with both shoes. The null hypothesis is the special shoe provides no speed change. The alternative is it increases speed. Calculations find the t-statistic, which is compared to the critical value to determine if the null can be rejected in favor of the alternative hypothesis.
Correlated T-Test
- 1. Mg-10 and Mg- Perez, Sugay, Torres, Villafuerte, Alcos, Manaid
- 2. The t-Test is used when you want to compare the means of two groups. Two distributions would be shown: the control group and the treated group. The test answers whether the two groups are statistically different.
- 3. In this type of t-Test, the two groups are tested before and after the independent variable is introduced to the treated group. This is done to statistically remove the chance of coincidence.
- 4. For example, an experiment is done to determine whether a certain shoe would make people run faster than ordinary shoes. Statistics are shown as follows: Runner Speed w/ordinary Speed w/ special shoes shoes A 12.1 mi/h 12.5 mi/h B 14.4 mi/h 15.0 mi/h C 11.6 mi/h 11.4 mi/h D 11.2 mi/h 11.6 mi/h E 12.4 mi/h 11.8 mi/h F 11.6 mi/h 12.0 mi/h G 15.0 mi/h 14.9 mi/h
- 5. Runner Speed Speed w/ w/ordinary special shoes shoes A 12.1 mi/h 12.5 mi/h B 14.4 mi/h 15.0 mi/h C 11.6 mi/h 11.4 mi/h D 11.2 mi/h 11.6 mi/h E 12.4 mi/h 11.8 mi/h F 11.6 mi/h 12.0 mi/h G 15.0 mi/h 14.9 mi/h We then follow the procedures in testing the statistics: State the null hypothesis to be tested: “ Ho = uA = uB ” This states that the special shoe provides no significant change in the speeds of the people tested
- 6. Runner Speed Speed w/ w/ordinary special shoes shoes A 12.1 mi/h 12.5 mi/h B 14.4 mi/h 15.0 mi/h C 11.6 mi/h 11.4 mi/h D 11.2 mi/h 11.6 mi/h E 12.4 mi/h 11.8 mi/h F 11.6 mi/h 12.0 mi/h G 15.0 mi/h 14.9 mi/h State the alternative hypothesis: The special shoe provides an increase in speed for the runners “ HA : uA = uB ”
- 7. State the level of significance, α = 0.05 Set-up the following table: Runner Speed Speed w/ Speed before – Speed after w/ordinary special shoes shoes D D^2 A 12.1 mi/h 12.5 mi/h -0.4 0.16 ∑D = -1.9 ∑D2 = 1.05 B 14.4 mi/h 15.0 mi/h -0.6 0.36 C 11.6 mi/h 11.4 mi/h 0.2 0.04 D 11.2 mi/h 11.6 mi/h -0.4 0.16 E 12.4 mi/h 11.8 mi/h -0.4 0.16 F 11.6 mi/h 12.0 mi/h -0.4 0.16 G 15.0 mi/h 14.9 mi/h 0.1 0.01
- 8. Compute for t using the following steps: Calculate sum of squares of the difference score: (∑D)2 ∑d2 = ∑d2 - = 1.05 – (-1.9)2 = 0.534 N 7 -Calculate the standard error of the mean difference sqrt(∑d2 / N(N-1)) = sqrt(0.534/7(6)) = 0.11 -Calculate D = ∑D = (-1.9) = -0.27 N 7
- 9. Calculate t = D = -0.27 = -2.45 SD 0.11 Find the critical or tabular value of t, df = 6, α = 0.05 t critical = t0.05 = +/- Formulate your conclusion: Since the calculated t is larger than the critical t, we shall accept HA, which states that the special shoe is effective in increasing the speed of runners wearing them.