Statistics Bivariate Analysis

By: Student 1, 2, 3

Minutes Exercised Per Day vs. Weighted GPA

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.

Exercise per day (minutes) X

Weighted GPA Y

30 3.7 30 3.5 0 3.7 0 3.5 60 3.2

120 3.12 120 3.67 30 3.2

120 3.5 30 3.6 90 3.7

180 2.6 0 3.33

150 4.3 180 3.7 120 3.6 15 3.52 60 3.5

240 3.33 180 3.7

0 3.65 0 4.0 23 3.5 60 3.0

240 3.7 40 3.9 60 3.0 60 3.2

160 3.5 35 3.4

Collected DataN=30

Vital Stats

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

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>4.255 4.3 is an OUTLIER

Histogram of X (exercise in min)

The shape of the data is slightly right skewed.

Histogram of Y (Weighted GPA)

The graph has a bell-shaped distribution. Outlier=4.5

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

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

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.

Scatterplot

GPA vs. Excercise

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Exercise (minutes)

GP

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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.

Regression Line on Scatterplot

Excersise Vs. GPAy = -0.0002x + 3.508R2 = 0.0015

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Equation: y= 3.508 + -.0002x

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.

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.

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

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 = y2 – b0 y – b1 xyn – 2

Se= .3353

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.

Residual Plot

Residual Plot

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GPA (weighted)

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.

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.

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.

The End.