Univariate and multivariate analysis assignment help sample

Univariate simply means that your data has only a single variable. It is one of the simplest methods of analyzing data and its purpose is to describe, summarize, and find patterns in data. The univariate analysis differs from the multivariate analysis in that the latter is more complex and it is utilized when a data set has more than two variables. However, both are great statistical techniques and have been used in analyzing data for decades.

Theoretically, as age increases, the injury index in a woman increases. Also, as medical difficulty increases, the injury index increases.

Coefficientsa
Model		Unstandardized Coefficients		Standardized Coefficients	t	Sig.	95.0% Confidence Interval for B
		B	Std. Error	Beta			Lower Bound	Upper Bound
1	(Constant)	-244.230	104.141		-2.345	.021	-450.922	-37.538
	Age	3.628	1.541	.225	2.353	.021	.568	6.687
	Medical difficulty index	284.142	95.017	.286	2.990	.004	95.559	472.725

a. Dependent Variable: Overall injury index

As the p-value for both the demographic variables is <0.05, we can say that both variables are statistically significant. And, as R2 is 0.162, it means that 16.2% variation in body strength is explained by the demographic variables.

Overall injury index (injury) = β0 + β1 age +β2 Medical difficulty index (medindex) + ei 

One of the team members, Tom, selected a model retaining four independent variables. His SPSS output is as follows:

The coefficient of 5.668 means that for each extra unit of age (1 year), the overall injury index increases by 5.668. Or, as age increases by 1 unit, the overall injury index increases by 5.668.

The coefficient of -3.981 means that for each extra unit of gluts, the overall injury index decreases by 3.981. Or, as strength in gluteus/hamstrings increases by 1 unit, the overall injury index decreases by 3.981.  

For the predicted injury index of -107.321, profile of independent variables has to be

i) 5.668*age + 0.300*quads - 3.981*gluts - 0.620*arms = 0 OR

ii) age = quads = gluts = arms = 0

H0 = βarms = 0

     H1 = βarms ≠ 0

As the above table shows, the p-value for βarms is 0.291 (>0.05), i.e. we can reject the null hypothesis. Hence, we can say that βarms are not statistically significantly different than zero. Thus, the partial regression coefficient of -0.62 may differ from zero because of sampling error.

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Although Tom’s model may not be the one you picked, for some theoretical reasons, the team agreed that Tom’s model is the ultimate model to pick anyway. Recognizing that the correct model is never known, nonetheless, before reporting the result, you would like to check the aptness of Tom’s model. Using numbers and/or graphs, show the following diagnostics for Tom’s model. Specify your cut-off criterion when necessary. You do not need to re-run the analysis after checking the diagnostics.

We check the Variance Inflation Factor (VIF) to check multicollinearity.

The normality assumption of residuals can be checked using normal probability plots for both standardized and unstandardized residuals.

As the above graph shows, residuals are scattered randomly, hence there is no problem of heteroscedasticity in the model.

To check residual outliers, tests such as Mahalanobis distance, Cook’s distance, etc. can be calculated:

 

Residuals Statistics
	Minimum	Maximum	Mean	Std. Deviation	N
Predicted Value	56.25	209.17	145.80	28.151	100
Std. Predicted Value	-3.181	2.251	.000	1.000	100
Standard Error of Predicted Value	5.778	15.924	9.752	2.371	100
Adjusted Predicted Value	54.56	209.42	145.77	28.055	100
Residual	-103.802	87.387	.000	43.953	100
Std. Residual	-2.313	1.948	.000	.980	100
Stud. Residual	-2.342	2.023	.000	1.004	100
Deleted Residual	-106.349	94.677	.031	46.224	100
Stud. Deleted Residual	-2.400	2.057	.000	1.014	100
Mahal. Distance	.652	11.480	3.960	2.414	100
Cook's Distance	.000	.091	.010	.015	100
Centered Leverage Value	.007	.116	.040	.024	100
a. Dependent Variable: Overall injury index

As the above table shows, the maximum Mahalanobis distance in our model is 11.48 and the critical value for 4 degrees of freedom using the Chi-Sq test is 9.49. Hence, outliers are present in the dataset (Row 1,7,12 and 79 are the outliers).

Model Summary
Model	R	R Square	Adjusted R Square	Std. The error of the Estimate
1	.539a	.291	.261	44.869
a. Predictors: (Constant), Strength in arms/shoulders, Age, Strength in gluteus/hamstrings, Strength in quadriceps

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ANOVA
Model		Sum of Squares	df	Mean Square	F	Sig.
1	Regression	78456.806	4	19614.202	9.743	.000b
	Residual	191257.194	95	2013.234
	Total	269714.000	99
a. Dependent Variable: Overall injury index
b. Predictors: (Constant), Strength in arms/shoulders, Age, Strength in gluteus/hamstrings, Strength in quadriceps

Although the model R² is quite low (0.291). But the model F-statistic is 9.743 and the significance F is 0.000. i.e. H₀ (the model doesn’t fit the data)) is rejected. Hence, the model fits the data very well.