MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN: 9781119256830
Author: Amos Gilat
Publisher: John Wiley & Sons Inc
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Given this already solved problem, how could you determine whether homogeneity of variance is satisfied for this analysis, and be clear on how you decided this. 

In Exercise 7.25, we saw data from Everitt that showed that girls receiving cognitive behavior therapy gained weight over
the course of that therapy. However, it is possible that they just gained weight because they got older. One way to control
for this is to look at the amount of weight gained by the cognitive therapy group (n = 29) in contrast with the amount
gained by girls in a Control group (n = 26), who received no therapy. The data on weight gain for the two groups is shown
below.
Control
Cognitive Therapy
-0.5
3.3
1.7
-9.1
-9.3
11.3
0.7
2.1
-5.4
0.0
-0.1
-1.4
12.3
-1.0
-0.7
1.4
-2.0
- 10.6
-3.5
-0.3
-3.7
- 10.2
- 12.2
-4.6
14.9
-6.7
3.5
-0.8
11.6
2.8
17.1
2.4
-7.1
0.3
-7.6
12.6
6.2
-0.2
1.8
1.6
1.9
3.7
11.7
3.9
-9.2
15.9
6.1
0.1
8.3
- 10.2
1.1
15.4
-4.0
-0.7
20.9
Мean
-0.45
3.01
St Dev.
7.99
7.31
Variance
63.82
53.41
Run the appropriate test to compare the group means. What would
you
conclude?
expand button
Transcribed Image Text:In Exercise 7.25, we saw data from Everitt that showed that girls receiving cognitive behavior therapy gained weight over the course of that therapy. However, it is possible that they just gained weight because they got older. One way to control for this is to look at the amount of weight gained by the cognitive therapy group (n = 29) in contrast with the amount gained by girls in a Control group (n = 26), who received no therapy. The data on weight gain for the two groups is shown below. Control Cognitive Therapy -0.5 3.3 1.7 -9.1 -9.3 11.3 0.7 2.1 -5.4 0.0 -0.1 -1.4 12.3 -1.0 -0.7 1.4 -2.0 - 10.6 -3.5 -0.3 -3.7 - 10.2 - 12.2 -4.6 14.9 -6.7 3.5 -0.8 11.6 2.8 17.1 2.4 -7.1 0.3 -7.6 12.6 6.2 -0.2 1.8 1.6 1.9 3.7 11.7 3.9 -9.2 15.9 6.1 0.1 8.3 - 10.2 1.1 15.4 -4.0 -0.7 20.9 Мean -0.45 3.01 St Dev. 7.99 7.31 Variance 63.82 53.41 Run the appropriate test to compare the group means. What would you conclude?
The t statistic is,
(+),
where, X= mean of first group
mean of second group
y
and combined variance,
(ns; + n3s;)
S* =
Step 2
We have,
Ho : Hs = Hy: mean of both group are not significantly different
> H: 'H
where H, = mean of girls receiving no therapy, i.e control
Hy = mean of girls receiving cognitive therapy
Step 3
From given,
I = -0. 45 y = 3.01
sI = 7.99 2 = 7.31
NI = 26
n2 = 29
Now the combined variance,
(26 * 7.99 + 29 * 7.3
26429-2
32094395
53
= 60. 5564
Step 4
The test statistic,
60.5564
= -1. 6462
As it is a left tailed test, and n, + 12 - 2 = 26 + 29 – 2 = 53 > 30,
hence t-Z-
The calculated t=-1.6462
The tabulated Z for left-tailed at 0.05 level of significance=-1.645
As
i.e - 1.646 < -1. 645
The
is rejected, and the mean weight of girls getting cognitive therapy is greater than the
Ho
mean weight of girls from control group.
expand button
Transcribed Image Text:The t statistic is, (+), where, X= mean of first group mean of second group y and combined variance, (ns; + n3s;) S* = Step 2 We have, Ho : Hs = Hy: mean of both group are not significantly different > H: 'H where H, = mean of girls receiving no therapy, i.e control Hy = mean of girls receiving cognitive therapy Step 3 From given, I = -0. 45 y = 3.01 sI = 7.99 2 = 7.31 NI = 26 n2 = 29 Now the combined variance, (26 * 7.99 + 29 * 7.3 26429-2 32094395 53 = 60. 5564 Step 4 The test statistic, 60.5564 = -1. 6462 As it is a left tailed test, and n, + 12 - 2 = 26 + 29 – 2 = 53 > 30, hence t-Z- The calculated t=-1.6462 The tabulated Z for left-tailed at 0.05 level of significance=-1.645 As i.e - 1.646 < -1. 645 The is rejected, and the mean weight of girls getting cognitive therapy is greater than the Ho mean weight of girls from control group.
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