Statistical Tests For BMI, Heart Rate, And Time To 400m Walk Test

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 non-parametric test. A non-parametric 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 t-test is employed to determine whether there exists a significant difference in BMI between the OA participants and control participants.

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 t-test 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;

Table 2

For us to make a decision, then the p-value computed must be compared to the level of significance in this test which is 0.05. If the p-value 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 p-value (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 non-parametric test is used. A non-parametric 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 t-test is employed to determine whether there exists a significant difference in heart rate before and after walking for 400 meters.

Testing normality

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 t-test 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;

Table 4

For us to make a decision, then the p-value computed must be compared to the level of significance in this test which is 0.05. If the p-value 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 p-value (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.

Table 5

To make a decision, then the p-value computed must be compared to the level of significance in this test which is 0.05. If the p-value 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 p-value (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

Table 6

The decision rule in ANOVA test is based on p-value computed and the level of significance (0.05). If the p-value 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 p-value (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.

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;

Table 7

Figure 2

It can be observed that the regression analysis above between age and time taken to complete 400 meters walk has R-squared 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.