Question

The scenario is as follows You are an IO psychology researcher, and you have developed a new type of employee training program. You conduct an experiment in which you administer the training program to one group (treatment group) and do nothing to the other group (control group). You then give each group a task to complete. The training program was developed to improve performance on the experimental task; as a researcher, then, you expect the treatment group to perform better than the control group. You want to be confident in your findings, so- in your NHST (t-test)-you use a.01 alpha level (aka significant level). Do the following: Compute the mean, median, mode, range, and standard deviation for each sample. (Hint: you will need some of these values for the independent samples t-test) Answer the following questions: a. What is the null hypothesis? b. What is the alternative hypothesis? c Is this a 1-or 2-tail hypothesis test? d. What are the degrees of freedom (an? 1. 2. e What is the t-critical value? Compute an independent samples t-test on the data. After step 3, answer the following questions: a. What is the resulting t-value from the t-test? b. Is there significance (ie, do you reject the null, or do you fail to reject the null)? c What do you, as a researcher, conclude from this experiment? (please use a couple 3. 4. sentences for this portion)

Answer 1

(2)(a) Null hypothesis: Ho: 4 = 42 That is, treatment group performed same as control group. Alternative hypothesis: H:44 > 2 That is, treatment group performed better than the control group. This is 1-tail hypothesis test. (d) Degrees of freedom = 37.

The value of t-critical value is teri = 2.4314. EXCEL (Data Analysis) output:

t-Test: Two-Sample Assuming Unequal Variances Mean Variance Observations Hypothesized Mean Difference TREATMENT CONTROL 17.9047619 9.19047619 33.69047619 19.46190476 21 t Stat P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail t Critical two-tail 37 5.477470921 1.60153E-06 2.431447397 3.20306E-06 2.715408718 (4)(a) The value of test statistic is t=5.4775. Since teac (= 5.4775) >terit (= 2.4314), then reject the null hypothesis.

That is, treatment group performed better than the control group.