SPSS Data Analysis American Heart Term Paper
- Length: 9 pages
- Subject: Other
- Type: Term Paper
- Paper: #27218660
Excerpt from Term Paper :
Asymp. Sig. (2-tailed)
Exact Sig. [2*(1-tailed Sig.)]
a. Not corrected for ties.
b. Grouping Variable: fuel additives (per m)
With the two P. values being so far apart, as well as the variance of the two groups being of significant value, around 2 whole values, it is clear that there is a significant difference to be noted between the two sample groups. Through the analysis of both the variance and the computations worked out through the Mann-Whitney test, it is clear that Sample B. has a higher rate of miles per gallon than the vehicles tested in Sample a. Here, the significant difference can then be interpreted that the fuel additive used within the context of Sample B. is more effective in terms of increased mileage within its test vehicles.
B. Exercise and Calories Burnt
Combined Median of Ranks
Three separate exercises were observed three times a week for forty minutes each session. The data here shows the number of calories burnt by each different activity within that context of forty minute work outs three days a week. By using the Kruskal -- Wallis test, the data can help determine if there was a significant difference between the three activities and corresponding calorie burnt data. The test itself requires a measured independent variable, and one nominal variable with one measurement variable. In the contest of this analysis, the ranked data is the set being computed. It also depends on the fact that the K. samples are random and independent, coming specifically out of a larger sample population. Additionally, all populations within the two sample sets are expected to have normal distribution and similar variances. Here the equation for analysis is as follows, with a=0.05.
SSbg (R)=n (mean of the group -- combined mean)
H= SSbg (R)
calories burned swimming
Test Statisticsa, b calories burned
a. Kruskal Wallis Test
b. Grouping Variable: Activities
Within this data set, the sample sizes are at the 5 limit mark to create the notion that the distribution of H. is closely corresponding to the approximation of df, where df=k-1. Thus, with the computed analysis, it is clear that one sample population does show a significant difference the other two. It can be assumed that Cycling is significantly different in terms of how many calories it burns compared to the other two sample groups. It is significantly lower in terms of how many calories it burns within the context in comparison to the other sampled activities of swimming and tennis. Swimming and Tennis are much closer, with less of a significant difference between them, showing much more correlation in regards to the amount of calories burned within the workout regime setting. Based on the analysis, however, it is clear hat Tennis burns the most calories out of the two listed activities with less of a significant difference. .
Quality of Inpatient Treatment
In thus data set, forty patients represent the sample set to be used to determine the correlation between the number of visitations and perceived quality of the care based on the opinion of the patient. The patients were divided into visitor categories, in which 1=frequent, 2=occasional, and 3=rare. Then, treatment was valued between the scale of 1=good, 2=fair, and 3=poor. A Chi-square Test was then performed on the data set to determine if there was a significant difference between the number of visits and the perceived quality of care within the given set of surveyed patients.
Chi-Square Test Results
a. 0 cells (.0%) have expected frequencies less than 5. The minimum expected cell frequency is 13.3.
The data analysis using the Chi-Square Test clearly shows a significant difference between the three categorized groups of patients. The sample size for frequent patients and medium frequency patients was thirteen variables. The sample size for patients who rarely needed care within the hospital setting was set at fourteen variables. Each was analyzed in comparison to each other utilizing the principle equations within the Chi-Square Test. It is clear here that the null hypothesis was true in this case. Through the analysis, it was…