A study was conducted to examine the effect of heat and duration of stretching on the extensibility of hamstring muscles and their electromyographic responses to passive stretch in children with hypertonia and severe mental retardation. Twenty subjects were randomly assigned to the following four treatments with five subjects per treatment: (A) 10-second stretching, (B) 30-second stretching, (C) hot pack followed by 10-second stretching and (D) hot pack followed by 30-second stretching. The outcome measure is the increase in hamstring extensibility (Y) and the data are given in Table 4. Table 4
A study was conducted to examine the effect of heat and duration of stretching on the extensibility of hamstring muscles and their electromyographic responses to passive stretch in children with hypertonia and severe mental retardation. Twenty subjects were randomly assigned to the following four treatments with five subjects per treatment: (A) 10-second stretching, (B) 30-second stretching, (C) hot pack followed by 10-second stretching and (D) hot pack followed by 30-second stretching. The outcome measure is the increase in hamstring extensibility (Y) and the data are given in Table 4. Table 4
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
Question

Transcribed Image Text:850
85b
15c
B5d
85e
μ = 0.678
M₂ = 0.414
Me = 1.316
Me = 1.382
Source
Between group
Residual
Total
SS
3.4092
307122
701214
SSE =
MSR = 3.4082/3 - 101364
MSE
3-7/22/16= 0.2320
Y = 0.9475
22 Y₁² = 0.07² +124² +
+ 1.11² + 1.79² = 25.0765
SST EE(Y₁-F²
EE Y² -n Y² - 25.0765 - 20 (0.9475)² = 7.1214
SSR = En (√₁₂-F)² = 5 (0.678-0-7474)² + 5 (0.414-0.9476)² +5 (10316-0.9474) + 5 (1-382-0-94+1)
= 3.4092
=
SST - SSR = 7.1214 - 3.4092 = 3.7122
Ho. The mean
H₁ He is not true.
·r-1.3.
・ny. -r. = 16
M²₁ -1=19
96% confidence internal of un
MA ± to.rs, N-r [. 5²/MA.
•√3520
0.678 ± 2.120
- (0.22131-1347).
ms.
1.1364
0.2320
F = 4.8983 > Fo. 05, 3,16 = 3.24.
It means that at least two of the means
each group is not significant different MAM = μc = μ
0.678 0.414 I 0.6458.
= (-0.3818.7.0.9098).
Therefore, we are
F
-M³R = 1.1364
MSE
= 4.8983
Increase T
-0.3818 and 0.3088.
95% confidence interval for
MA-MB.
-0.22/3+1=1347 -0.414 ± 2.120 (0.2320 ( 33 + 5)
0.2320
reject He at 5% level of significant.
sigaficant different.
are
95% confident that the true difference between the mean
hamstring extensibility for treatment A and B. is between

Transcribed Image Text:A study was conducted to examine the effect of heat and duration of stretching on the
extensibility of hamstring muscles and their electromyographic responses to passive stretch in
children with hypertonia and severe mental retardation. Twenty subjects were randomly
assigned to the following four treatments with five subjects per treatment: (A) 10-second
stretching, (B) 30-second stretching, (C) hot pack followed by 10-second stretching and (D) hot
pack followed by 30-second stretching. The outcome measure is the increase in hamstring
extensibility (Y) and the data are given in Table 4.
Table 4
Treatment
A
B
C
D
0.07
-0.19
1.11
0.57
1.24
0.5
1.28
2.19
Y
0.62
0.64
1.08
1.25
0.34
0.48
2.03
1.11
Consider a one-way ANOVA to study the effects of the four treatments. Denote μA to μD as the
means of increase in hamstring extensibility under treatments A to D respectively.
the 95% confidence interval for µ^ — µB.
1.12
0.64
1.08
1.79
(f) Estimate and comment on the contrast of the treatment means : L =
With a 90% confidence interval.
stensibili
MA + MB
2
eq
μc + UD
2
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