### Data Analysis by Kurtosis and Skew Method

Added on - 10 May 2020

• 11

pages

• 1866

words

• 13

views

• 0

Showing pages 1 to 4 of 11 pages
Data analysis1Student Name:Student number:Lecturer:
Data analysis2QUESTION 1To examine if there is a difference in BMI (kg/m2) between the OA and Control participants, wedetermine first whether the data is normally distributed or not. If the data ids found to benormally distributed, then a parametric test is employed to determine whether there is asignificant difference in BMI between OA and Control participants. There are various methodsused 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. Thedirection can be positive or negative or positive. A skew value of 0 (zero) means perfectnormality.Test for normalitysummary statisticsMean28.8879661Standard Error0.671148243Median28.7Mode34Standard Deviation5.155187471Sample Variance26.57595786Kurtosis-0.570124636Skewness0.219884827Count59Table 1It can be observed from the table above that the value of skewness is .2 which means that thedata is relatively normally distributed. Therefore a parametric test can be employed to establishwhether there is a significant difference in BMI between OA and Control participants. Since thesample size of the data is greater than 30 and we are only testing difference between twovariables only, then a paired sample t-test which is a parametric test is appropriate. T-testnormally test the hypothesis as illustrated below,
Data analysis3HypothesisNull hypothesis:There is no difference in mean BMI between OA and Control participantsAlternative hypothesis:There is a significant difference in mean BMI between OA and ControlparticipantsThe t-test results are as in the table below;t-Test: Paired Two Sample for MeanscontrolOAMean28.25862129.61Variance24.51179829.33019Observations2929Pearson Correlation0.1128918Hypothesized MeanDifference0df28t Stat-1.052729P(T<=t) one-tail0.1507329t Critical one-tail1.7011309P(T<=t) two-tail0.3014657t Critical two-tail2.0484071Table 2From the t-test results in the table above, it can be seen that the p-value computed (.3) is greaterthan 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 Controlparticipants.
Data analysis4Question 2To examine if there is a difference in heart rate before and after walking, we determine firstwhether 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 inheart 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. Thedirection can be positive or negative or positive. A skew value of 0 (zero) means perfectnormality.Test for normalitytest for normality at resttest for normality for heart rate after 400m walkdescriptive statistics of heart rate at restdescriptive statistics for heart rate after 400mwalkMean77.54237288Mean99.79661017Standard Error1.678119588Standard Error2.216744584Median75Median101Mode70Mode107Standard Deviation12.88988113Standard Deviation17.02713824Sample Variance166.1490357Sample Variance289.9234366Kurtosis-0.20900471Kurtosis0.738911335Skewness0.556200949Skewness-0.368718759Range53Range87Minimum57Minimum50Maximum110Maximum137Sum4575Sum5888Count59Count59Table 3As can be observed from table 3 above, the two variables can be said to be normally distributedsince they have skewness values of close to zero. This means that a parametric test can be  