Data analysis2 QUESTION 1 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. Testing normality summary statistics Mean28.8879661 Standard Error0.671148243 Median28.7 Mode34 Standard Deviation5.155187471 Sample Variance26.57595786 Kurtosis-0.570124636 Skewness0.219884827 Count59 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;
Data analysis3 Hypothesis 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; t-Test: Paired Two Sample for Means controlOA Mean28.25862 1 29.61 Variance24.51179 8 29.3301 9 Observations2929 Pearson Correlation0.112891 8 Hypothesized Mean Difference 0 df28 t Stat-1.052729 P(T<=t) one-tail0.150732 9 t Critical one-tail1.701130 9 P(T<=t) two-tail0.301465 7 t Critical two-tail2.048407 1 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
Data analysis4 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. Question 2 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. Normality test test for normality at resttest for normality for heart rate after 400m walk descriptive statistics of heart rate at rest descriptive statistics for heart rate after 400m walk Mean77.54237288Mean99.79661017 Standard Error1.678119588Standard Error2.216744584 Median75Median101 Mode70Mode107 Standard Deviation12.88988113Standard Deviation17.02713824 Sample Variance166.1490357Sample Variance289.9234366
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Data analysis5 Kurtosis-0.20900471Kurtosis0.738911335 Skewness0.556200949Skewness-0.368718759 Range53Range87 Minimum57Minimum50 Maximum110Maximum137 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; 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 results are as illustrated below; t-Test: Paired Two Sample for Means at restafter 400 walk Mean77.5423728 8 99.79661017 Variance166.149035 7 289.9234366 Observations5959 Pearson Correlation0.64844425 9 Hypothesized Mean Difference 0
Data analysis6 df58 t Stat- 13.0553922 8 P(T<=t) one-tail3.26086E-19 t Critical one-tail1.67155276 2 P(T<=t) two-tail6.52172E-19 t Critical two-tail2.00171748 4 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 differencebetween heart rate at rest and heart rate after walking for 400 metres. QUESTION 3 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.
Data analysis7 Analysis of variance test usually test for equality of means in the variables involved.The hypothesis is as below, Hypothesis 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 GroupsCountSumAverageVariance Overweight92953.6 4 328.182 2 2117.70 3 Heavyweight72366.9338.128 6 1403.17 6 Obese144088.0 5 292.003 6 938.563 2 ANOVA Source of VariationSSdfMSFP-valueF crit Between Groups1265526327.49 8 4.54827 8 0.01984 4 3.35413 1 Within Groups37562271391.18 5 Total5021729 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
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Data analysis8 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. QUESTION 4 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, 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 GroupsCountSumAverag e Variance Time to complete 400m Walk (s)601815 4 302.571770.03 7 Time to complete 400m Walk601728288.132050.15
Data analysis9 (s)_6mth83 Time to complete 400m walk (s)_12mths 601741 5 290.252537.29 8 ANOVA Source of VariationSSdfMSFP-valueF crit Between Groups7298.223649.11.721960.1816 9 3.04701 2 Within Groups37509 2 1772119.2 Total38239 0 179 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 thealternative. It is therefore concluded thatthereisno significance difference in 400m Walk Test times between the three visits. QUESTION 5 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 PainKOOS Function, Daily Activity
Data analysis10 Score Right knee: KOOS Pain Score1 KOOS Function, Daily Activity0.593155491 Table 7 Scatterplot diagram 5060708090100110 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 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. QUESTION 6 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
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Data analysis11 Regression Statistics Multiple R0.544139951 R Square0.296088286 Adjusted R Square0.283738958 Standard Error35.75032737 Observations59 ANOVA dfSSMSFSignificance F Regression130643.4730643.4723.976068.41E-06 Residual5772850.91278.086 Total58103494.4 CoefficientsStandard Error t StatP-valueLower 95%Upper 95% Lower 95.0% Up 95 Intercept146.697299632.068114.5745552.62E-0582.48203210.912682.48203210 682.5168147510.5139994.8965368.41E-061.4875493.546081.4875493.5 Table 7 40455055606570758085 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 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
Data analysis12 therefore not a fit predictor and hence age since it cannot explain 70% of the variation in the dependent variable, time.