Test for the difference in BMI between OA and Control participants
Test for the difference in BMI between OA and Control participants
Test for the difference in the means of two variables requires the employment of a parametric or a nonparametric test. A nonparametric test is used where the data is not normally distributed while a parametric test is employed where the data is normally distributed. To determine normality, the measure of skewness is established. If the measure is zero, it means the data is perfectly normal. If it is close to zero then it indicates that the data is almost normal. In regard to this Question, if the data is normal and only two variables are involved then a paired sample ttest is employed to determine whether there exists a significant difference in BMI between the OA participants and control participants.
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
The descriptive analysis table above shows that the BMI data has got a skewness score of 0.2. This value is so close to zero and therefore we can conclude that the data is almost normally distributed. The data can then be analyzed using paired sample ttest since the sample size is also more than 30. The test hypothesis is as below;
Null hypothesis: There is no significant 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 results are below;
tTest: Paired Two Sample for Means 

control 
OA 

Mean 
28.258621 
29.61 
Variance 
24.511798 
29.33019 
Observations 
29 
29 
Pearson Correlation 
0.1128918 

Hypothesized Mean Difference 
0 

df 
28 

t Stat 
1.052729 

P(T<=t) onetail 
0.1507329 

t Critical onetail 
1.7011309 

P(T<=t) twotail 
0.3014657 

t Critical twotail 
2.0484071 
Table 2
For us to make a decision, then the pvalue computed must be compared to the level of significance in this test which is 0.05. If the pvalue is less than the level of significance, then the null hypothesis is not accepted. The converse also applies. From the test results above, it can be observed that the pvalue (0.3) calculated is indeed greater than the level of significance (.05). The decision rule is to accept the null hypothesis. It is therefore concluded that there is no significant difference in mean BMI between OA and Control participants.
Test for the difference in heart rate before and after walking for 400 meters
In order to test for the difference in heart rate between the two variables, then a parametric on a nonparametric test is used. A nonparametric test is used where the data is not normally distributed while a parametric test is employed where the data is normally distributed. To determine normality, the measure of skewness is established. If the measure is zero, it means the data is perfectly normal. If it is close to zero then it indicates that the data is almost normal. In regard to this Question, if the data is normal and only two variables are involved then a paired sample ttest is employed to determine whether there exists a significant difference in heart rate before and after walking for 400 meters.
Testing 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 
Table 3
The descriptive analysis table above shows that the both heart rates data have got skewness score close to zero. Before and after have the walk got skew scores of 0.55 and 0.3 respectively. These values are so close to zero and therefore we can conclude that the data is almost normally distributed. The data can then be analyzed using paired sample ttest since the sample size is also more than 30. The test hypothesis is as below;
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 results are as illustrated below;
tTest: Paired Two Sample for Means 

at rest 
after 400 walk 

Mean 
77.54237288 
99.79661017 
Variance 
166.1490357 
289.9234366 
Observations 
59 
59 
Pearson Correlation 
0.648444259 

Hypothesized Mean Difference 
0 

df 
58 

t Stat 
13.05539228 

P(T<=t) onetail 
3.26086E19 

t Critical onetail 
1.671552762 

P(T<=t) twotail 
6.52172E19 

t Critical twotail 
2.001717484 
Table 4
For us to make a decision, then the pvalue computed must be compared to the level of significance in this test which is 0.05. If the pvalue is less than the level of significance, then the null hypothesis is not accepted. The converse also applies. From the test results above, it can be observed that the pvalue (0.00) calculated is less than the level of significance (.05). The decision rule is to reject the null hypothesis and accept the alternative. It is therefore concluded that there is a significant difference between heart rate at rest and heart rate after walking for 400 metres.
Is there is any significant difference in the mean times to complete 400m Walk Test (s) between obese, overweight & heavyweight in OA participants
In order to test for any significant difference involving more than two variables, then an ANOVA test is appropriate. However, prior to using the ANOVA test, the data must be confirmed to be normally distributed as the test is a parametric test and hence very sensitive to normality. Earlier tests indicate that the data is normally distributed and hence we proceed to conduct an analysis of variance test.
Analysis of variance test usually test for equality of means in the variables involved. The hypothesis is as below,
Null hypothesis: There is no significant difference in mean time to complete 400m Walk Test (s) between obese, overweight and heavyweight in OA participants
Alternative hypothesis: At least one or more mean time is difference.
Anova: Single Factor 

SUMMARY 

Groups 
Count 
Sum 
Average 
Variance 

Overweight 
9 
2953.64 
328.1822 
2117.703 

Heavyweight 
7 
2366.9 
338.1286 
1403.176 

Obese 
14 
4088.05 
292.0036 
938.5632 

ANOVA 

Source of Variation 
SS 
df 
MS 
F 
Pvalue 
F crit 
Between Groups 
12655 
2 
6327.498 
4.548278 
0.019844 
3.354131 
Within Groups 
37562 
27 
1391.185 

Total 
50217 
29 
Table 5
To make a decision, then the pvalue computed must be compared to the level of significance in this test which is 0.05. If the pvalue is less than the level of significance, then the null hypothesis is not accepted. The converse also applies. From the test results above, it can be observed that the pvalue (0.02) calculated is less than the level of significance (.05). The decision rule is to reject the null hypothesis and accept the alternative. It is therefore concluded that at least one or more mean time is different. To establish the different variables then a further Duncan’s test is recommended.
Test for the difference in heart rate before and after walking for 400 meters
Is there is a difference in 400m Walk Test times between the three visits
In order to test for any significant difference involving more than two variables, then an ANOVA test is appropriate. However, prior to using the ANOVA test, the data must be confirmed to be normally distributed as the test is a parametric test and hence very sensitive to normality. Earlier tests indicate that the data is normally distributed and hence we proceed to conduct an analysis of variance test.
Analysis of variance test usually test for equality of means in the variables involved. The hypothesis is as below,
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 
Average 
Variance 

Time to complete 400m Walk (s) 
60 
18154 
302.57 
1770.037 

Time to complete 400m Walk (s)_6mth 
60 
17288 
288.13 
2050.153 

Time to complete 400m walk (s)_12mths 
60 
17415 
290.25 
2537.298 

ANOVA 

Source of Variation 
SS 
df 
MS 
F 
Pvalue 
F crit 
Between Groups 
7298.2 
2 
3649.1 
1.72196 
0.18169 
3.047012 
Within Groups 
375092 
177 
2119.2 

Total 
382390 
179 
Table 6
The decision rule in ANOVA test is based on pvalue computed and the level of significance (0.05). If the pvalue is less than the level of significance, then the null hypothesis is not accepted. The converse also applies. From the test results above, it can be observed that the pvalue (0.2) calculated is greater than the level of significance (.05). The decision rule is 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.
Test for correlation between KOOS pain score and KOOS function
Pearson correlation test was employed to test whether a correlation exists between KOOS pain score and KOOS function. A scatter plot was also used to give a graphical representation.
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
Figure 1
It can be seen that the Pearson correlation value computed is 0.6. This means that a positive and significant correlation exists between KOOS pain score and KOOS pain.
Simple regression analysis between time taken to complete 400 meters walk and age
Since there is only one independent variable (age), a simple regression analysis is appropriate to determine whether age can be a better determiner of the time taken to walk the 400 meters. The regression analysis results are as illustrated 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 
Significance F 

Regression 
1 
30643.47 
30643.47 
23.97606 
8.41E06 

Residual 
57 
72850.9 
1278.086 

Total 
58 
103494.4 

Coefficients 
Standard Error 
t Stat 
Pvalue 
Lower 95% 
Upper 95% 
Lower 95.0% 
Upper 95.0% 

Intercept 
146.6972996 
32.06811 
4.574555 
2.62E05 
82.48203 
210.9126 
82.48203 
210.9126 
68 
2.516814751 
0.513999 
4.896536 
8.41E06 
1.487549 
3.54608 
1.487549 
3.54608 
Table 7
Figure 2
It can be observed that the regression analysis above between age and time taken to complete 400 meters walk has Rsquared value of 0.3. This means that age as an independent variable can only explain 30% of the variations in time taken to finish 400 meters walk. The model is therefore not a fit predictor and hence age since it cannot explain 70% of the variation in the dependent variable, time.
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