I was leaning toward 1-way ANOVA being a better option for testing instead of a
ID: 3204760 • Letter: I
Question
I was leaning toward 1-way ANOVA being a better option for testing instead of a t-test. But I don't know why.
A sample of 12 subjects were tested on ability in a spatial rotation task, both before and after the training period. The resulting data listed from subject 1 to subject 12, were: Before 25 11 39 26 21 28 20 24 47 25 26 30 After 28 14 46 33 25 37 18 32 49 30 31 34 The researchers reported that the pooled-variance t statistic equaled 1.54, not statistically significant at a 0.05. Do you agree with the choice of the test statistic? lf not, why not? Assuming that you disagree with the choice of the test statistic in question #1 indicate a more appropriate hypothesis test for these data. Is the conclusion the same or different? Please explainExplanation / Answer
Here, we can apply the paired difference t-test which can be calculated as follows:-
H0: D = 0
H1: D ¹ 0
Subject
Before
After
Di
Di-D
(Di-D)^2
1
25
28
3
-1.58
2.4964
2
11
14
3
-1.58
2.4964
3
39
46
7
2.42
5.8564
4
26
33
7
2.42
5.8564
5
21
25
4
-0.58
0.3364
6
28
37
9
4.42
19.5364
7
20
18
-2
-6.58
43.2964
8
24
32
8
3.42
11.6964
9
47
49
2
-2.58
6.6564
10
25
30
5
0.42
0.1764
11
26
31
5
0.42
0.1764
12
30
34
4
-0.58
0.3364
Sum
55
Sum
98.9168
D=55/12=4.58
SD=sqrt(E (Di-D)^2)/n-1
=sqrt(98.9168/11)
=sqrt(8.99)
=2.9987 and approx. 3
t-STAT=D-µD/SD/sqrt(n)
=4.58-0/3/sqrt(12)
=4.58/3/3.464
=4.58*3.464/3
=5.29
tCRIT=2.201
Since 5.29>2.201, it lies in the rejection region and hence reject null hypothesis.
Subject
Before
After
Di
Di-D
(Di-D)^2
1
25
28
3
-1.58
2.4964
2
11
14
3
-1.58
2.4964
3
39
46
7
2.42
5.8564
4
26
33
7
2.42
5.8564
5
21
25
4
-0.58
0.3364
6
28
37
9
4.42
19.5364
7
20
18
-2
-6.58
43.2964
8
24
32
8
3.42
11.6964
9
47
49
2
-2.58
6.6564
10
25
30
5
0.42
0.1764
11
26
31
5
0.42
0.1764
12
30
34
4
-0.58
0.3364
Sum
55
Sum
98.9168
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