5 out of 13 is roughly 1/7 but the chance of it being defective or not you have 5/13 of a chance of being defective or not
<u>Answer-</u>
<em>The </em><em>exponential model</em><em> best fits the data set.</em>
<u>Solution-</u>
x = input variable = number of practice throws
y = output variable = number of free throws
Using Excel, Linear, Quadratic and Exponential regression model were generated.
The best fit equation and co-efficient of determination R² are as follows,
<u>Linear Regression</u>
![y=2.155x+0.391,\ R^2=0.903](https://tex.z-dn.net/?f=y%3D2.155x%2B0.391%2C%5C%20R%5E2%3D0.903)
<u>Quadratic Regression</u>
![y=0.096x^2+0.713x+3.803,\ R^2=0.948](https://tex.z-dn.net/?f=y%3D0.096x%5E2%2B0.713x%2B3.803%2C%5C%20R%5E2%3D0.948)
<u>Exponential Regression</u>
![y=4.625e^{0.141x},\ R^2=0.951](https://tex.z-dn.net/?f=y%3D4.625e%5E%7B0.141x%7D%2C%5C%20R%5E2%3D0.951)
The value of co-efficient of determination R² ranges from 0 to 1, the more closer its value to 1 the better the regression model is.
Now,
![R^2_{Linear}< R^2_{Quadratic}< R^2_{Exponential}](https://tex.z-dn.net/?f=R%5E2_%7BLinear%7D%3C%20R%5E2_%7BQuadratic%7D%3C%20R%5E2_%7BExponential%7D)
Therefore, the Exponential Regression model must be followed.