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ADDITIONAL PP 14.4 We are now going to practice doing independent groups, t test problems. First, let’s do a bare bones problem. An investigator conducts an experiment to determine the effect of the independent variable on the dependent variable. A two group experiment is run. Twenty subjects are randomly sampled from the population and randomly assigned 10 to each condition. The conditions for the two groups are as similar as possible, except that Group 1 gets the independent variable and Group 2 does not. The dependent variable is recorded for each subject in each group. There is no basis for pairing of scores. The results are shown in the following table. Group 1 10 12 18 16 10 8 15 20 14 17 a. b. c. Group 2 6 8 12 5 9 10 14 6 3 10 What is the alternative hypothesis? Assume a nondirectional hypothesis is appropriate. What is the null hypothesis? Using = 0.052 tail, what do you conclude? SOLUTION is on next page. SOLUTION Group 1 Group 2 2 X2 X2 2 6 36 8 64 12 144 5 25 9 81 10 100 14 196 6 36 3 9 10 100 83 791 n2 = 10 X 2 83 8.3 X2 10 n2 X1 X1 10 100 12 144 18 324 16 256 10 100 8 64 15 225 20 400 14 196 17 289 140 2098 n1 = 10 X 1 140 14.0 X1 10 n1 X 1 X 2 5.7 a. b. c. Alternative hypothesis: The alternative hypothesis asserts that the independent variable has an effect on the dependent variable. The sample mean difference of 5.7 is due to random sampling from populations where 1 2. Null hypothesis: The null hypothesis states that the independent variable has no effect on the dependent variable. The sample mean difference of 5.7 is due to random sampling from populations where 1 = 2. Conclusion, using = 0.052 tail. STEP 1: Calculate the appropriate statistic. From the table above, n1 = 10, n2 = 10, 2 2 X 1 14.0, X 2 8.3, X1 = 140, X1 = 2098, X2 = 83, and X2 = 791 X 2 SS 1 X 2 1 1 n1 X 2 SS 2 X 2 2 2 140 2098 2 83 791 138.0 102.1 10 2 n1 10 Substituting these values in the equation for tobt with n1 = n2, we have t obt STEP X 1 X 2 14.0 8.3 3.49 138.0 102.1 SS 1 SS 2 nn 1 109 2: Evaluate the statistic. From Table D, with = 0.052 tail and df = N – 2 = 20 – 2 = 18, tcrit = 2.101 Because ItobtI > 2.365, we conclude by rejecting H0. The independent variable affects the dependent variable; it appears to produce an increase in the value of the dependent variable.