Diameter of objects moved (mm)

Transcription

1 PRACTICE SAC on DATA TRANSFORMATIONS Rivers and streams carry small solid particles of rock downhill.solid particles are classified according to their mean diameter from smallest to largest as clay, silt, sand, pebble, cobble and boulder.the data below shows the necessary speed of a current to carry particles in suspension. This data is important for many reason. When the environment near a river is altered, the volume and speed of the current can be altered, resulting in the transport of more and larger particles that may change the ecology of the riverbed. Many species of fish have very specific requirements of sediment sizes on the river bed for successful reproduction. Diameter of objects moved (mm) Speed of current (m/sec) Classification of Objects Mud Sand Gravel Coarse Gravel Pebbles Small Stones Large stones (fist sized) Boulders a. There are three variables in this scatterplot. List them, and state whether they are numerical or categorical. Diameter of object (numerical) Speed of current (numerical) Classification of object (categorical) b. Scientists are interested in developing a least squares regression model in which the velocity of the current required to move a particle can be predicted from the diameter of the object. i. What is the explanatory variable in this model? Diameter ii. Create a scatterplot on CAS, and roughly sketch its shape in the space below. Inidcate the scale by including the lowest and largest value on each axis.

2 c. Would it be appropriate to fit a least squares regression line to this data? Justify your answer. No, it would not. The scatterplot of the raw data suggests strongly that the data is non-linear. This is confirmed by the residual plot, shown below: As can be seen, the residual plot displays a clear curved pattern, indicating that the original data is non-linear. It is now up to you to determine the best possible least squares regression line equation to model this data. You will have to: (1) Test out different possible models, and justify that the model you select is the best, by referring to appropriate statistical measures. Summarize your findings. (2) State the final equation for your least squares model, with the coefficients expressed to three significant figures. A local council has decided to build a car park which will lie close to a river. The planning authority wishes to establish how much the river ecology will be impacted by the construction of the car park. Ecologists have estimated that the construction of the car park will be ill-advised if particles of a diameter of more than 50mm will be moved by the current. (3) Use your regression model to predict the speed at which the current would have to be moving if particles of 50mm were being transported. (4) Comment on how reliable your prediction is likely to be by referring to appropriate statistical measures.

3 From the shape of the scatterplot, it will be appropriate to test out three possible transformations: a y 2 or (speed) 2 transformation (in order to stretch the y axis) a log(x) or log(diameter) transformation (in order to compress the x-axis) a 1 or 1 transformation (in order to compress the x x diameter axis)(: We will then have to decide which (if any) of the three succeeds best in linearizing the data. The (speed) 2 transformation: In the scatterplot below, we see that we have transformed the data value to (speed) 2 We can see that the transformed scatterplot now looks much more linear. The residual plot confirms this. As we see below, we see that the residuals are now much more randomly scattered around the horizontal axis than in the previous residual plot for the raw data.

4 The value for the coefficient of determination for the transformed data is: r 2 = The equation of the least squares regression line for the transformed data is: (speed) 2 = diameter Now we test out a log(diameter) transformation: We can see that the scatterplot of log(diameter) on the x-axis and speed on the y-axis looks non-linear. Likewise, the residual plot exhibits a clearcut curved pattern, further confirming that this transformation has not succeeded in linearizing the data.

5 1 Finally, we test out the transformation: diameter We can see that the scatterplot with 1 diameter on the x-axis has failed to linearize the data. This is likewise confirmed with by the residual plot The residual plot displays a clear curved pattern, confirming that the reciprocal x transformation has not succeeded in linearizing the data. SUMMARY Transformation Description of Residual plot Value of r 2 y 2 log 10 (x) Transformed scatterplot looked linearized. This was confirmed by the residual plot which exhibited a random scattering of the residuals. Transformed scatterplot was clearly non-linear. This was confirmed by the residual plot, which showed a clear pattern Did not calculate as this regression model was clearly inappropriate.

6 1 x transformation Transformed scatterplot was clearly non-linear. This was confirmed by the residual plot, which showed a clear pattern Did not calculate as this regression model was clearly inappropriate. CONCLUSION: From testing out these three transformations, we have established that only the y 2 transformation succeeds in linearizing the data. The value of r 2 for the y 2 transformation is equal to , indicating strong predictive power for the transformed regression model. The equation for the best possible regression model is: (3) Prediction: If diameter = 50, (speed) 2 = =3.3 speed = 3. 3 = 1. 8 (speed) 2 = diameter The predicted speed of the current would be 1.8 m/s. (4) This is a very reliable prediction because: a) The value of r 2 is , indicating a strong predictive power b) The prediction is an example of interpolation, since it falls within the range of the data set. Interpolation is a reliable process.