When only two treatments are involved, ANOVA and the Student’s t test (Chapter 1
ID: 3242523 • Letter: W
Question
When only two treatments are involved, ANOVA and the Student’s t test (Chapter 11) result in the same conclusions. Also, formula120.mml. As an example, 14 randomly selected Introduction to History students were divided into two groups, one consisting of 6 students and the other of 8. One group took the course in the normal lecture format. The other group of 8 students took the course via the Internet. At the end of the course, each group was given a 50-item test. The following is a list of the number correct for each of the two groups.
Traditional Lecture Internet
a-1. Complete the ANOVA table. (Round SS, MS and F values to 2 decimal places.)
Source SS df MS F
Factors Error Total
a-2. Use a ALPHA = .01 level of significance. (Round your answer to 2 decimal places.)
The test statistic is F ________
b. Using the t test from Chapter 11, compute t. (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)
t _______
There is ________ (A DIFFERENCE OR NO DIFFERENCE) in the mean test scores.
13 35 11 21 25 33 27 23 14 28 12 26 22 22Explanation / Answer
Answer:
When only two treatments are involved, ANOVA and the Student’s t test (Chapter 11) result in the same conclusions. Also, formula120.mml. As an example, 14 randomly selected Introduction to History students were divided into two groups, one consisting of 6 students and the other of 8. One group took the course in the normal lecture format. The other group of 8 students took the course via the Internet. At the end of the course, each group was given a 50-item test. The following is a list of the number correct for each of the two groups.
a-1. Complete the ANOVA table. (Round SS, MS and F values to 2 decimal places.)
One factor ANOVA
Mean
n
Std. Dev
17.0
6
7.07
Traditional
26.3
8
5.34
Internet
22.3
14
7.56
Total
ANOVA table
Source
SS
df
MS
F
p-value
Treatment
293.36
1
293.36
7.83
.0161
Error
449.50
12
37.46
Total
742.86
13
a-2. Use a ALPHA = .01 level of significance. (Round your answer to 2 decimal places.)
The test statistic is F = 7.83
b. Using the t test from Chapter 11, compute t. (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)
t = -2.80
There is (A DIFFERENCE ) in the mean test scores.
Pooled-Variance t Test for the Difference Between Two Means
(assumes equal population variances)
Data
Hypothesized Difference
0
Level of Significance
0.05
Population 1 Sample
Sample Size
6
Sample Mean
17
Sample Standard Deviation
7.0711
Population 2 Sample
Sample Size
8
Sample Mean
26.25
Sample Standard Deviation
5.3385
Intermediate Calculations
Population 1 Sample Degrees of Freedom
5
Population 2 Sample Degrees of Freedom
7
Total Degrees of Freedom
12
Pooled Variance
37.4583
Standard Error
3.3054
Difference in Sample Means
-9.2500
t Test Statistic
-2.7985
Two-Tail Test
Lower Critical Value
-2.1788
Upper Critical Value
2.1788
p-Value
0.0161
Reject the null hypothesis
One factor ANOVA
Mean
n
Std. Dev
17.0
6
7.07
Traditional
26.3
8
5.34
Internet
22.3
14
7.56
Total
ANOVA table
Source
SS
df
MS
F
p-value
Treatment
293.36
1
293.36
7.83
.0161
Error
449.50
12
37.46
Total
742.86
13
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