In: Statistics and Probability
Aron | |||||
1 | 2 | 3 | 4 | Average | |
8/6/2017 |
90 | 138 | 118 | 105 | 112.8 |
8/19/2017 | 162 | 101 | 120 | 145 | 132 |
9/16/2017 | 101 | 129 | 132 | 111 | 118.3 |
Average | 117.7 | 122.7 | 123.3 | 120.3 | 121 |
Mjorgan | |||||
1 | 2 | 3 | 4 | Average | |
8/6/2017 | 115 | 88 | 94 | 102 | 99.8 |
8/19/2017 | 89 | 75 | 77 | 90 | 82.8 |
9/16/2017 | 74 | 110 | 117 | 90 | 97.8 |
Average | 92.7 | 91 | 96 | 94 | 93.4 |
Is there a correlation between the two bowlers themselves? In other words – does Aron’s score in each game depend on Mjorgan’s score at all? You might want to create a data table of each bowler’s score in each respective game to find out. Develop a statistical model to estimate Aron’s score based on Mjorgan’s score, then interpret its findings and its accuracy.
We first arrange the data in the below manner
Arons | Mjorgan |
90 | 115 |
138 | 88 |
118 | 94 |
105 | 102 |
162 | 89 |
101 | 75 |
120 | 77 |
145 | 90 |
101 | 74 |
129 | 110 |
132 | 117 |
111 | 90 |
Using the corellation functon in excel =CORREL () we find the correlation coefficeint
The correlation coefficient = 0.01048
Using regression we find the below results
SUMMARY OUTPUT | ||||||||
Regression Statistics | ||||||||
Multiple R | 0.01048 | |||||||
R Square | 0.00011 | |||||||
Adjusted R Square | -0.09988 | |||||||
Standard Error | 22.03966 | |||||||
Observations | 12 | |||||||
ANOVA | ||||||||
df | SS | MS | F | Significance F | ||||
Regression | 1 | 0.533571 | 0.533571 | 0.001098 | 0.974213 | |||
Residual | 10 | 4857.466 | 485.7466 | |||||
Total | 11 | 4858 | ||||||
Coefficients | Standard Error | t Stat | P-value | Lower 95% | Upper 95% | Lower 95.0% | Upper 95.0% | |
Intercept | 119.6154 | 42.25733 | 2.830644 | 0.017832 | 25.46024 | 213.7706 | 25.46024 | 213.7706269 |
Mjorgan | 0.014821 | 0.447197 | 0.033143 | 0.974213 | -0.98159 | 1.011238 | -0.98159 | 1.011237779 |
Here we see that the significance F = 0.97 which is more than 0.05 hence the model is imsignificant
Using the model we can say Arons score = 119.61 + 0.014821 * Mjorgan
As the significance is not good the equations accuracy will we extremely poor