Statistics Bivariate Analysis

1. Minutes Exercised Per Day vs. Weighted GPA Statistics Bivariate Analysis By: Student 1, 2, 3

2. Why did we choose this study? • Exercise is a vital part of staying healthy and living an active and accomplished lifestyle. • We believe that physical activity improves a student’s will to learn and may increase study habits. • Previous studies have concluded that children who live a more active lifestyle are more compelled to succeed in school. We want to see if this is true at our school. • We like to exercise, and we were curious to see if there is a correlation between these two variables.

3. Collected Data N=30

4. For X -X bar: 81.1 -Sx: 72.886 -5 # Summary: MinX: 0 Q1: 30 Med: 60 Q3: 120 MaxX: 240 For Y -Y bar: 3.494 -Sy: .3297 -5 # Summary: MinY: 2.6 Q1: 3.33 Med: 3.5 Q3: 3.7 MaxY: 4.3 Vital Stats

5. Outliers? In order to find outlier, we used the two formulas: • #<Q1-1.5(IQR) • #>Q3+1.5(IQR) • 0<30-1.5(90) • 240>120+1.5(90) • 0<-105 • 240>255 NO OUTLIERS • 2.6<3.33-1.5(.37) • 4.3>3.7=1.5(.37) • 2.6<-2.22 • 4.3>4.2554.3 is an OUTLIER

6. Histogram of X (exercise in min) The shape of the data is slightly right skewed.

7. Histogram of Y (Weighted GPA) The graph has a bell-shaped distribution. Outlier=4.5

8. Empirical Rule Test • Exercise (X) Mean=81.1 Standard Deviation=72.887 • 81.1 +/- 72.887= 153.986 & 8.213 • 81.1+/- 72.887(2)= 226.873 & -64.674 • 81.1 +/- 72.887(3)= 299.76 & -137.561 68% of the data falls between 153.986 & 8.213 95% of the data falls between 226.873 & -64.674 99.7% of the data falls between 299.76 & -137.561

9. Empirical Rule Test • GPA (Y) Mean= 3.494, Standard Deviation= .3297 • 3.494 +/- .3297 = 3.8237 & 3.1634 • 3.494 +/- .3297(2)= 4.1534 & 2.8346 • 3.494 +/- .3297(3)= 4.4831 & 2.5049 68% of the data falls between 3.8237 & 3.1634 95% of the data falls between 4.1534 & 2.8346 99.7% of the data falls between 4.4831 & 2.5049

10. Explanatory & Response Variable • The explanatory variable (X) in our data is the number of minuets exercised per day, it is used to predict changes in the response variable (Y) or GPA. • GPA is the response variable, and is dependent on the other data. This allows us to find a relationship between the two values.

11. Scatterplot

12. Analysis • The Scatterplot shows that there is no linear correlation between exercise and weighted GPA due to the graph. In order to receive that conclusion, we know that when a correlation graph has a pattern it is linear. When the correlation graph does not have a pattern it is not linear. • The coefficient of correlation is r = -0.038168. This also gives another reason why the scatter plot is not linear. If the r value is closer to 1 then it is linear. If the r value rounds close to zero it is not linear. If the r value was close to one, it would be very strong but in this case the r value is not strong at all because it is closer to zero. The outlier in this scatter plot is 4.3 which slightly altered our data.

13. Regression Line on Scatterplot Equation: y= 3.508 + -.0002x

14. The y-intercept of the regression line gives the predicted value of y for any given value of x. • The slope shows the relationship between x and y as the steepness of the regression line is analyzed. • Our data does not prove a correlation between weighted GPA and average minutes exercise performed in a day, so this equation should not be used to predict the response variable.

15. R & R Squared • The r-squared value is explained variation over total variation and will give the accuracy (in a percentage) for a given value. • R2= .00145681è .14% of the variation in Y is explained by the variation in x. • R measures the strenght and direction of a linear relationshop between two variables • R= -.038168 negative, with no correlation.

16. Total Variation: is the sum of the y values minus the mean of y values, squared • 362595.172 • Explained Variation: is the sum of the y-hat values minus the mean of y values, squared • 181283.8495 • Unexplained Variation: is the sum of the y values minus the y-hat values • 181311.3225 • 362595.172= 181283.8495 + 181311.3225

17. y2 –b0y – b1xy se = n – 2 Standard Error of Estimate • The standard error of estimate is a measure of how sample points deviate from the regression line. Se measures the difference between the observed y-values and the predicted y-values. One would take the unexplained variable, divide that by the degree of freedom and square the result. Se= .3353

18. 95% Prediction Interval • For X we choose: 70 • With wanting to find the possible GPA of a person with an average 70 minute workout, there will be a .3353 standard of error. The GPA would fall between 2.6889 and 4.0855.

19. Residual Plot

20. Interpretation • The Residual plot shows that it is not a good model for the LSRL. This is because the plot contains a pattern and is in the negative range. In other words, this graph is not linear. On the residual plot, the X-values equals GPA weighted and the Y-values is exercise in minutes.

21. Conclusion • In conclusion, we have found that there is no correlation between how many minuets a high school student exercises, and their GPA. • Our graphs and data values are not strong enough to draw conclusions based on our sample. • Despite the amount of time that a student does or does not spend working out, their grades will neither increase or decrease.

22. Possible Problems • If the sample had been larger, the results may have been more accurate. • It is possible that subjects may have lied either about the amount they exercise or their true GPA, thus hindering our results. • It is sometimes difficult to estimate how much you exercise each day because it varies depending on your changing daily activities.

23. The End.