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
nn  1
109
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.