Question

In: Statistics and Probability

2) The purpose of a study by Tem et al (A-15) was to investigate the wheelchair...

2) The purpose of a study by Tem et al (A-15) was to investigate the wheelchair maneuvering in individuals with lower-level spinal cord injury (SCI) and health controls. Subject used a modified wheelchair to incoporate a rigid seat surface to facilitate the specified experimental measurements.Interface pressure measurement was recorded by using a high -resolution pressure-sensitive mat with a spatial resolution of 4 sensors per square centimeter taped on the rigid seat support. During static sitting conditions, average pressures were recorded under the ischial tuberosities . The data for measurements of the left ischial tuberosity (in mm Hg) for the SCI and control groups are shown below

Control 131 115 124 131 122. 117. 88. 114 150 169

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SCI 60 150. 130 180 163 130 121 119 130 148

a) Find the mean, median, variance, and standard deviation for the controls.

b)Find the mean, median variance, and standard deviation for SCI group.

c)Construct a box-and whisker plot for the controls.

d)construct a box-and whisker plot for the SCI.

e)Do you believe there is a difference in pressure readings for controls and SCI subjects in this study?

Solutions

Expert Solution

Solution-

problem solved using statistical analysis software Minitab.

a)

Descriptive Statistics: Control

Variable Mean StDev Variance Median
Control 125.10 22.25 495.21 120.50

b)

Descriptive Statistics: SCI

Variable Mean StDev Variance Median
SCI   133.1 32.2 1035.4 130.0

c)

d)

e)

if we observe both boxplots then we can say both data are different.

for checking there is a difference in pressure readings for controls and SCI subjects in this study two sample t test is performed.

Two-Sample T-Test and CI: Control, SCI

Two-sample T for Control vs SCI

N Mean StDev SE Mean
Control 10 125.1 22.3 7.0
SCI 10 133.1 32.2 10


Difference = μ (Control) - μ (SCI)
Estimate for difference: -8.0
95% CI for difference: (-34.2, 18.2)
T-Test of difference = 0 (vs ≠): T-Value = -0.65 P-Value = 0.527 DF = 16

by above test result we can not say both are different.

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Minitab paths-


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