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  • Assistance with Biostatistics Assignments on Exploratory Data Analysis

Assistance with Biostatistics Assignments on Exploratory Data Analysis

A scatter plot and correlation matrix is one of the plots that are used in exploratory data analysis.  We are given data that we have to investigate if a relationship exists between the concentrations of two hormones. A scatter plot will make it easy to interpret the result visually, and the correlation matrix determines the strength of the relationship. Knowing how to plot a scatter plot can help the majority of students answer their biostatistics assignment questions with minimal hardship. Note that this is a pre-analysis stage. Later, we shall fit a linear regression model.

Output for the Scatter Plot for the Biostatistics Assignment

Look at the example that is provided by our experts below.

The above scatter plot shows that the trend for the slope seems to be upward sloping, suggesting there appears a positive correlation in the two variables. We can use the correlation matrix to compute the strength and magnitude of that relationship since both variables are scale variables.

Correlations
Hormone M (concentration) Hormone T (concentration)
Hormone M (concentration) Pearson Correlation 1 .787**
Sig. (2-tailed) .000
N 100 100
Hormone T (concentration) Pearson Correlation .787** 1
Sig. (2-tailed) .000
N 100 100
**. Correlation is significant at the 0.01 level (2-tailed).
The correlation matrix presented between Hormone M and Hormone T shows that the correlation value is 0.787, which is a  positive correlation coefficient values suggesting that an increase in one of the variables will increase the other variable and vice versa. A correlation value that is closer to one suggests that the correlation is strong between two variables.

Help with Biostatistics Homework on Prediction with Linear Regression

As a part of our biostatistics homework help service, the linear regression model was fitted on the data as we found that the data has met the linearity assumptions.

Linear Regression Output (H3) 

The linear regression model has been created in SPSS to create a model for individuals' amount of Enzyme E concentration based on their BMI. The output from SPSS is as follows:
Variables Entered/Removeda  
Model Variables Entered Variables Removed Method  
1 Body Mass Indexb . Enter  
a. Dependent Variable: Enzyme E (concentration)  
b. All requested variables entered.  

Model Summary

  
Model R R Square Adjusted R Square Std. Error of the Estimate  
1 .258a .066 .057 .8428  
a. Predictors: (Constant), Body Mass Index  
ANOVAa  
Model Sum of Squares Df Mean Square F Sig.  
1 Regression 4.944 1 4.944 6.961 .010b  
Residual 69.604 98 .710      
Total 74.548 99        
a. Dependent Variable: Enzyme E (concentration)  
b. Predictors: (Constant), Body Mass Index  
Coefficientsa
Model Unstandardized Coefficients Standardized Coefficients t Sig.
B Std. Error Beta
1 (Constant) 14.402 .470   30.656 .000
Body Mass Index -.043 .016 -.258 -2.638 .010
a. Dependent Variable: Enzyme E (concentration)

The Model obtained from the Analysis

This is the model that was obtained after the analysis.  The model that should be used to make the predictions is as follows: Enzyme E = 14.402 – 0.043*BMI

Checking for linearity between Enzymes E concentration and Body mass Index.

The p-value of the F-stat for the model under the null hypothesis, which states that there does not exist any linear relationship between Enzyme E concentration and BMI, is close to 0.01. At a 5% alpha level, because the p-value(0.01) is comparatively lower 0.05, its safe to reject the null hypothesis concluding that linearity exists between Enzyme E concentration and BMI.

Prediction

The predicted value of Enzyme E based on BMI of 25 is equal to ; Enzyme E = 14.402 – 0.043*BMI Enzyme E = 14.402 – 0.043*25 Enzyme E = 14.402 – 1.075 = 13.33 The actual value for Enzyme E for BMI of 25 in the sample data is 13.8. And the value predicted using the equation is 13.33. The predicted value is marginally below the actual data, but still very closer.

Chi-Square Application by our Online Biostatistics Tutor

For this scenario, our online biostatistics tutor wanted to analyses the data to answer the question, "In the population of adults between 56 and 65 years of age, is there a difference between the male gender andthe female gender in terms of their distributions of getting a medical Checkup within the last one year or 12 months?"Evidently, this is a Chi-square test since it's a test of independence. The results of the analysis are presented below.

Chi-square output

The chi-square statistics have been computed to see if the male gender and the female gender in terms of their distributions of getting a medical Checkup within the last year or 12 months differ. The output for the same is presented below.             
Gender * Medical Checkup (within last 12 months) Crosstabulation  
  Medical Checkup (within last 12 months) Total  
No Yes  
Gender Males Count 27 22 49  
% within Medical Checkup (within last 12 months) 60.0% 40.0% 49.0%  
Females Count 18 33 51  
% within Medical Checkup (within last 12 months) 40.0% 60.0% 51.0%  
Total Count 45 55 100  
% within Medical Checkup (within last 12 months) 100.0% 100.0% 100.0%  
chi-Square Tests
  Value df Asymp. Sig. (2-sided) Exact Sig. (2-sided) Exact Sig. (1-sided)
Pearson Chi-Square 3.962a 1 .047    
Continuity Correctionb 3.202 1 .074    
Likelihood Ratio 3.987 1 .046    
Fisher's Exact Test       .070 .037
Linear-by-Linear Association 3.922 1 .048    
N of Valid Cases 100        
a.       0 cells (0.0%) have expected count less than 5. The minimum expected count is 22.05.

Decision criteria
From the above results, the p-value for the test was 0.047 under the null hypothesis, which in our case states that the difference between the male gender and the female gender in terms of their distributions of getting a medical Checkup within the last one year or the last 12 months is lacking. At a 5% significance level, since the p-value(0.047) is is comparatively lower than 0.05, it's safe to accept the alternative hypothesis concluding that the male and the female gender in terms of their distributions of getting a medical Checkup within the last one year or 12 months differ.