Data Analysis and Interpretation in Healthcare: Assignment

This document presents a comprehensive solution to a data analysis homework assignment, employing various statistical techniques to analyze healthcare data. The analysis begins with an examination of BMI differences between OA and control participants using t-tests to determine if there is a statistically significant difference. Subsequent analyses explore heart rate changes before and after a 400-meter walk, again utilizing t-tests to assess significance. The assignment then investigates differences in 400m walk times across different weight categories and visits using ANOVA. Furthermore, the document includes an analysis of the correlation between KOOS pain and function scores using Pearson correlation, and a regression analysis is performed to determine if age predicts the time taken to complete the 400-meter walk. The document provides detailed results, interpretations, and conclusions for each analysis, offering a complete understanding of the data and statistical methods applied.

Data analysis 1
Student Name:
Student number:
Lecturer:

Data analysis 2
QUESTION 1
To examine if there is a difference in BMI (kg/m2) between the OA and Control participants, we
determine first whether the data is normally distributed or not. If the data ids found to be
normally distributed, then a parametric test is employed to determine whether there is a
significant difference in BMI between OA and Control participants. There are various methods
used to determine normality of data. In this case, kurtosis and skew method has been employed.
Skewness measures amount of departure from the line of symmetry and the direction. The
direction can be positive or negative or positive. A skew value of 0 (zero) means perfect
normality.
Test for normality
summary statistics
Mean 28.8879661
Standard Error 0.671148243
Median 28.7
Mode 34
Standard Deviation 5.155187471
Sample Variance 26.57595786
Kurtosis -0.570124636
Skewness 0.219884827
Count 59
Table 1
It can be observed from the table above that the value of skewness is .2 which means that the
data is relatively normally distributed. Therefore a parametric test can be employed to establish
whether there is a significant difference in BMI between OA and Control participants. Since the
sample size of the data is greater than 30 and we are only testing difference between two
variables only, then a paired sample t-test which is a parametric test is appropriate. T-test
normally test the hypothesis as illustrated below,

Data analysis 3
Hypothesis
Null hypothesis: There is no difference in mean BMI between OA and Control participants
Alternative hypothesis: There is a significant difference in mean BMI between OA and Control
participants
The t-test results are as in the table below;
t-Test: Paired Two Sample for Means
control OA
Mean 28.25862
1
29.61
Variance 24.51179
8
29.3301
9
Observations 29 29
Pearson Correlation 0.112891
8
Hypothesized Mean
Difference
0
df 28
t Stat -1.052729
P(T<=t) one-tail 0.150732
9
t Critical one-tail 1.701130
9
P(T<=t) two-tail 0.301465
7
t Critical two-tail 2.048407
1
Table 2
From the t-test results in the table above, it can be seen that the p-value computed (.3) is greater
than the level of significance which is .05. This means that we fail to reject the null hypothesis.
The conclusion therefore is that there is no difference in mean BMI between OA and Control
participants.

Data analysis 4
Question 2
To examine if there is a difference in heart rate before and after walking, we determine first
whether the data is normally distributed or not. If the data ids found to be normally distributed,
then a parametric test is employed to determine whether there is a significant difference in in
heart rate before and after walking. Normality has been checked in this test using skewness.
Skewness measures amount of departure from the line of symmetry and the direction. The
direction can be positive or negative or positive. A skew value of 0 (zero) means perfect
normality.
Test for normality
test for normality at rest test for normality for heart rate after 400m walk
descriptive statistics of heart rate at rest descriptive statistics for heart rate after 400m
walk
Mean 77.54237288 Mean 99.79661017
Standard Error 1.678119588 Standard Error 2.216744584
Median 75 Median 101
Mode 70 Mode 107
Standard Deviation 12.88988113 Standard Deviation 17.02713824
Sample Variance 166.1490357 Sample Variance 289.9234366
Kurtosis -0.20900471 Kurtosis 0.738911335
Skewness 0.556200949 Skewness -0.368718759
Range 53 Range 87
Minimum 57 Minimum 50
Maximum 110 Maximum 137
Sum 4575 Sum 5888
Count 59 Count 59
Table 3
As can be observed from table 3 above, the two variables can be said to be normally distributed
since they have skewness values of close to zero. This means that a parametric test can be

Data analysis 5
employed to establish whether there is a significant difference between heart rate at rest and
heart rate after walking for 400 metres.
Since the sample size of the data is greater than 30 and we are only testing difference between
two variables only, then a paired sample t-test which is a parametric test is appropriate. T-test
normally test the hypothesis as illustrated below,
Hypothesis
Null hypothesis: There is no significant difference between heart rate at rest and heart rate after
walking for 400 metres.
Alternative hypothesis: There is a significant difference between heart rate at rest and heart
rate after walking for 400 metres.
The t-test results are as in the table below;
t-Test: Paired Two Sample for Means
at rest after 400 walk
Mean 77.5423728
8
99.79661017
Variance 166.149035
7
289.9234366
Observations 59 59
Pearson Correlation 0.64844425
9
Hypothesized Mean
Difference
0
df 58
t Stat -
13.0553922
8
P(T<=t) one-tail 3.26086E-19
t Critical one-tail 1.67155276
2
P(T<=t) two-tail 6.52172E-19

Data analysis 6
t Critical two-tail 2.00171748
4
Table 4
From the t-test results in the table above, it can be seen that the p-value computed (.00) is less
than the level of significance which is .05. This means that we fail to accept the null hypothesis
and accept the alternative. The conclusion therefore is that there is a significant difference
between heart rate at rest and heart rate after walking for 400 metres.
QUESTION 3
Test to determine if there is a difference in time to complete 400m Walk Test (s) between
the three weight categories (obese, overweight & heavyweight) in OA participants
Since the test involves comparison of means of three variables, an analysis of variance test is
employed. Since ANOVA is a parametric test which is very sensitive to normality, we must first
ensure that the data is normally distributed. From the descriptive statistics in the earlier question,
it has been established the data is normally distributed.
ANOVA usually tests the null hypothesis that the means of the three variables are the same. The
hypothesis is as below,
Hypothesis
Null hypothesis: There is no difference in mean time to complete 400m Walk Test (s) between
the three weight categories (obese, overweight & heavyweight) in OA participants
Alternative hypothesis: At least one or more mean time is difference.
Anova: Single Factor

Data analysis 7
SUMMARY
Groups Count Sum Average Variance
Overweight 9
2953.6
4
328.182
2
2117.70
3
Heavyweight 7 2366.9
338.128
6
1403.17
6
Obese 14
4088.0
5
292.003
6
938.563
2
ANOVA
Source of Variation SS df MS F P-value F crit
Between Groups 12655 2
6327.49
8
4.54827
8
0.01984
4
3.35413
1
Within Groups 37562 27
1391.18
5
Total 50217 29
Table 5
From the analysis of variance table results above, it can be observed that the p-value computed
(.02) is less than the level of significance (.05). The decision rule is therefore to reject the null
hypothesis and accept the alternative. It is therefore concluded that the mean time used to
complete the 400 meters walk is between the three weights categories is not equal; at least one or
more mean time is different.
QUESTION 4
Statistical analysis test to determine if there is a difference in 400m Walk Test times
between the three visits
Since the test involves comparison of means of three variables, an analysis of variance test is
employed. Since ANOVA is a parametric test which is very sensitive to normality, we must first
ensure that the data is normally distributed. From the descriptive statistics in the earlier question,
it has been established the data is normally distributed.

Data analysis 8
ANOVA usually tests the null hypothesis that the means of the three variables are the same. The
hypothesis is as below,
Hypothesis
Null hypothesis: There is no significance difference in 400m Walk Test times between the three
visits
Alternative hypothesis: At least one mean time is different
Anova: Single Factor
SUMMARY
Groups Count Sum
Averag
e Variance
Time to complete 400m Walk (s) 60
1815
4 302.57
1770.03
7
Time to complete 400m Walk
(s)_6mth 60
1728
8 288.13
2050.15
3
Time to complete 400m walk
(s)_12mths 60
1741
5 290.25
2537.29
8
ANOVA
Source of Variation SS df MS F P-value F crit
Between Groups 7298.2 2 3649.1 1.72196
0.1816
9
3.04701
2
Within Groups
37509
2 177 2119.2
Total
38239
0 179
Table 6
From the analysis of variance table results above, it can be observed that the p-value computed
(.2) is greater than the level of significance (.05). The decision rule is therefore to accept the null
hypothesis and reject the alternative. It is therefore concluded that there is no significance
difference in 400m Walk Test times between the three visits.

Data analysis 9
QUESTION 5
Test for correlation between KOOS pain score and KOOS function
To test the correlation between the two variables above, a Pearson correlation coefficient is
employed. A scatterplot was also employed to provide a graphical representation of the
relationship. The table below shows the results of the correlation test.
test for correlation results
Right knee: KOOS Pain
Score KOOS Function, Daily Activity
Right knee: KOOS Pain Score 1
KOOS Function, Daily Activity 0.59315549 1
Table 7
Scatterplot diagram
50 60 70 80 90 100 110
0
20
40
60
80
100
120
f(x) = 0.733257215329748 x + 21.111530780905
R² = 0.351833435215427
Scatterplot
KOOS pain score
KOOS function
Figure 1
From table of correlation above, it can be observed that the correlation coefficient is .6. This is a
strong correlation value. It can therefore be concluded that there is a strong correlation between

Data analysis 10
KOOS pain score and KOOS function. It can also be said that the relationship is positive. The R-
squared in the scatter plot also indicate that 35% of the dependent variable is explained by the
independent variable.
QUESTION 6
A regression analysis to determine if age is predictive of time taken to complete 400 meters
walk
To determine if age can predict properly the time taken to complete 400 meters walk, a
regression analysis is employed. The sample data are normally distributed from the previous
analyses. The results of the regression analysis are as in the table below;
SUMMARY
OUTPUT
Regression Statistics
Multiple R 0.544139951
R Square 0.296088286
Adjusted R Square 0.283738958
Standard Error 35.75032737
Observations 59
ANOVA
df SS MS F
Significanc
e F
Regression 1 30643.47
30643.4
7
23.9760
6 8.41E-06
Residual 57 72850.9
1278.08
6
Total 58 103494.4
Coefficients
Standard
Error t Stat P-value Lower 95%
Upper
95%
Lower
95.0%
Upp
95.0
Intercept 146.6972996 32.06811
4.57455
5 2.62E-05 82.48203
210.912
6 82.48203 210.9
68 2.516814751 0.513999 4.89653 8.41E-06 1.487549 3.54608 1.487549 3.54

Data analysis 11
6
Table 7
40 45 50 55 60 65 70 75 80 85
0
50
100
150
200
250
300
350
400
450
f(x) = 2.53590949483902 x + 145.768262902454
R² = 0.300279999340709
Scatter plot
AGE
Time (s)
Figure 2
From the regression results above, it can be seen that R-squared has a value of .3. This is to mean
that the independent variable age, can only explain 30% of the variation on dependent variable
time. Since 70% of the variation cannot be explained, we can conclude that age cannot be used to
predict or is not a predictor of time taken to complete 400 meters walk.