Question

1. Assume you are planning to compare the Indirect Responsibility (IR) group with the Direct Responsibility (DR1 and DR2) groups. Without using SPSS (5 pts)

   1. Provide the linear contrast to perform this comparison.

   2. Compute the contrast.

   3. Test whether the contrast is significant or not.

DATA:

A study by Staub, 1970, was concerned with the effects of instructions to young children and their subsequent attempts to help another child (apparently) in distress. Twenty-four first-grade students were randomly assigned to one of three groups. The first group was labeled as indirect responsibility (IR). Students in the IR group were informed that another child was alone in an adjoining room and had been warned not to climb up on a chair. The second group was labeled direct responsibility one (DR1). Students in the DR1 group were told the same story as in the IR condition, but was also told that they were left in charge and to take care of anything that happened. The students were given a simple task, and the researcher left the room. The students then heard a loud crash in the adjoining room followed by a minute of sobbing and crying. Students in the third group, direct responsibility two (DR2), had the same instructions as the DR1 group, but the sounds of distress also included calls for help. Ratings from 1 (no help) to 5 (went to the adjoining room) were given to each student by an observer sitting behind a one-way mirror. The ratings are given below. Perform a one-way ANOVA in SPSS with α = .05 and answer the following questions: (35 pts)

IR	DR1	DR2
3	5	4
4	4	4
2	5	3
1	4	3
1	5	4
2	5	2
1	4	5
1	3	3

Answer 1

Null and Alternative Hypothesis:

H₀: µ_IR = µ_{DR 1} = µ_{DR 2}

H₁: Not all Means are equal

Alpha = 0.05

N = 24

n = 8

Degress of Freedom:

df_Between = a – 1 = 3-1 =2

df_Within = N-a = 24-3 = 21

df_Total = N-1 = 24-1 = 23

Critical Values:

Time (df_Between, df_Within): (2,21) = 3.47

Decision Rule:

If F is greater than 3.47, reject the null hypothesis

Test Statistics:

SS_Betwen = ∑(∑a_i)²/n - T²/N = 25.75

SS_Within = ∑(Y)² - ∑(∑a_i)²/n = 18.75

SS_Total = SS_{Betwen ­­}+ SS_Within = 44.5

MS = SS/df

F = MS_effect / MS_error

Hence,

F = 12.88/0.89 = 14.42

Result:

Our F = 14.42, we reject the null hypothesis

Conclusion:

Not all means are equal.