## 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 |
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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). |

### 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/Removed^{a} |
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Model | Variables Entered | Variables Removed | Method | |||||||||||||||||

1 | Body Mass Index^{b} |
. | Enter | |||||||||||||||||

a. Dependent Variable: Enzyme E (concentration) | ||||||||||||||||||||

b. All requested variables entered. | ||||||||||||||||||||

## Model Summary |
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Model | R | R Square | Adjusted R Square | Std. Error of the Estimate | ||||||||||||||||

1 | .258^{a} |
.066 | .057 | .8428 | ||||||||||||||||

a. Predictors: (Constant), Body Mass Index | ||||||||||||||||||||

ANOVA^{a} |
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Model | Sum of Squares | Df | Mean Square | F | Sig. | |||||||||||||||

1 | Regression | 4.944 | 1 | 4.944 | 6.961 | .010^{b} |
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Residual | 69.604 | 98 | .710 | |||||||||||||||||

Total | 74.548 | 99 | ||||||||||||||||||

a. Dependent Variable: Enzyme E (concentration) | ||||||||||||||||||||

b. Predictors: (Constant), Body Mass Index | ||||||||||||||||||||

Coefficients^{a} |
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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 |
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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 |
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Value | df | Asymp. Sig. (2-sided) | Exact Sig. (2-sided) | Exact Sig. (1-sided) | |||||||

Pearson Chi-Square | 3.962^{a} |
1 | .047 | ||||||||

Continuity Correction^{b} |
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 |
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