Exponential Regression. Suppose we have paired sample data {{ x 1, y 1 }, { x 2, y 2 },..., { x n, y n }}, with all x i > 0,

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1 MATH 482 More Regression Dr. Neal, WKU The least-squares regression technique for linear data can be adjusted to paired data that appears to be non-linear but instead exponential, logarithmic, or power-based. Here we explain three basic transformations that are used for these regression models. Exponential Regression Suppose we have paired sample data {{ x 1, y 1 }, { x 2, y 2 },..., { x n, y n }}, with all y i > 0, where it appears that y is an exponential function of x of the form y = a b x. Then ln y will be linear function of x. Indeed, we have ln y = ln (a b x ) = ln a + ln b x = ln a + x (ln b) = a x + b. y = a b x ln y = a x + b, where a = ln b and b = ln a So we can take the natural logarithm of each y i, then apply least-squares linear regression to the data {{ x 1, ln y 1 }, { x 2, ln y 2 },..., { x n, ln y n }} to obtain the function ln y = a x + b, where a = x ln y x ln y x 2 x 2 and b = ln y a x. Then we obtain the exponential regression function y = a b x using a = e b and b = e a. Logarithmic Regression Suppose we have paired sample data {{ x 1, y 1 }, { x 2, y 2 },..., { x n, y n }}, with all x i > 0, where it appears that y is a logarithmic function of x of the form y = a + b ln x with b 0. Then ln x will be linear function of y of the form ln x = 1 b y + a b = a y + b.

2 y ln x x y = a + b ln x y ln x = a y + b, where a = 1 / b and b = a / b Thus we take the natural logarithm of each x i, then apply least-squares linear regression to the data {{ y 1, ln x 1 }, { y 2, ln x 2 },..., { y n, ln x n }} to obtain the function ln x = a y + b, where a = y ln x y ln x y 2 y 2 and b = ln x a y. Then we obtain the logarithmic regression function y = a + b ln x using b = 1 / a and a = b b = b / a. Power Regression Suppose we have paired sample data {{ x 1, y 1 }, { x 2, y 2 },..., { x n, y n }}, with all x i > 0 and all y i > 0, where it appears that y is a power-based function of x of the form y = a x b with a 0. Then ln y will be linear function of ln x. Here we have ln y = ln a + b ln x = a ln x + b. y = a x b ln y = a ln x + b, where a = b and b = ln a

3 We now take the natural logarithm of each x i and of each y i, then apply leastsquares linear regression to the data {{ln x 1, ln y 1 }, {ln x 2, ln y 2 },..., {ln x n, ln y n }} to obtain the function ln y = a ln x + b, where ln x ln y ln x ln y a = (ln x ) 2 ln x 2 and b = ln y a ln x. Then we obtain the power regression function y = a x b using b = a b and a = e. Correlation and Coefficient of Determination In each case, the sample correlation r and the resulting coefficient of determination r 2 are obtained from whatever transformed data is being used. If we are using the data {{c 1, d 1 }, {c 2, d 2 },..., {c n, d n }}, then r is given by r = c d c d c 2 c 2 d 2 d 2. The interpretation of r 2 remains the same. It gives the proportion of y -data that is determined by the x -data when using the given regression model as a predictor. Fortunately, calculators and software have built-in functions that allow us to compute exponential, logarithmic, and power regression functions directly rather than having to make these transformations. We shall do both in the following example. Example 1. Compute the exponential and power regression models for the following data: {{1.4, 3.7}, {1.8, 5.8}, {2.5, 10.4}, {3, 14.5}, {3.6, 20.1}, {4, 24.2} }. Solution. We first plot the data: STAT EDIT STAT PLOT Adjust Settings ZOOM 9 We see that the exponential and power regression models both may be appropriate. We first evaluate the exponential regression by using transformation: Store ln of yi Re-adjust Settings ZOOM 9 LinReg(ax+b) L1,L3

4 Computing linear regression on L1 ( x ) and L3 (ln y ), we find ln y x = a x + b. Thus, y = a b x b, where a = e e a and b = e e We can verify this result by computing exponential regression directly on the original data: ExpReg L1, L2 We now evaluate the power regression by using transformation: Store ln of xi and ln of yi Re-adjust Settings ZOOM 9 LinReg(ax+b) L3,L4 Computing linear regression on L3 (ln x ) and L4 (ln y ), we obtain ln y 1.79 ln x = a ln x + b. Thus, y = a x b, where b = a and b a = e e We can verify this result by computing exponential regression directly on the original data: PwrReg L1, L2 We note that the power regression model has an r 2 of which means that it is a nearly perfect fit. We see that its plot of the transformed data is almost perfectly linear (unlike the plot of the exponentially transformed data). We also note that the r 2 values obtained via transformation are the same as those obtained through nontransformation. Other Regression Models Calculators and software have other built-in regression models that can be used to fit data. These regression models include quadratic, cubic, quartic, sine, and logistic functions. When data does not fit a clear pattern, we can use several models for comparison and then use the one that yields the highest r 2 value to obtain the best fit.

5 Example 2. (a) Plot the following set of paired data of engine size vs. mpg for certain models of cars. (b) Find a good regression fit. (c) Estimate the mpg for a cylinder volume of 110. (d) Find the cylinder volume that would give an mpg of 27. Cylinder Volume MPG VW Rabbit Nissan Chevette Omni Mazda Starfire Capri Celica Nissan Solution. (a) We first plot the data: STAT EDIT STAT PLOT Adjust Settings ZOOM 9 We observe the general tendency that mpg decreases as cylinder volume increases. Before computing a regression model, make sure that the calculator s diagnostics are turned on. Enter the CATALOG (2nd 0) and scroll down to the DiagnosticOn command. Press ENTER to bring the command to the Home screen, then press ENTER again. (b) We now compute various regression models: LinReg(ax+b) L1,L2 LnReg L1,L2 ExpReg L1,L2 CubicReg L1,L2 The cubic regression model y x x x gives a better fit than the linear, logarithmic, and exponential models due to the higher value of r Using this cubic function as a predictor, mpg is 89.15% determined by cylinder volume, and is about 10.85% determined by other factors. Now we will store the cubic regression function into Y1 and graph.

6 Y= VARS 5 EQ 1 GRAPH (c) After graphing, we easily can evaluate the cubic regression function for other x values. To evaluate mpg for a cylinder volume of 110, we use the value option from the CALC screen (2nd Trace). A cylinder volume of 110 should give mpg. (d) We also can solve for the x -value for which y equals a certain amount. For example, using the cubic model, find the x for which we get an mpg of 27. We wish to solve the equation Y1 = 27, or instead Y1-27 = 0. Look at the graph and make an initial guess for x, say x = 90. Then bring up the solve( command from the CATALOG and enter the command solve(y1 27, X, 90). We see that a cylinder volume of about should yield 27 mpg.

7 Exercises 1. For those who have long term exposure to loud machinery, hearing loss is a function of decibel level. Consider the following data: Decibel Level Pct loss (a) Find the linear regression function. (b) Find the logarithmic regression function. (c) Which function from (a) and (b) is a better fit and why? (d) Using the better fit, evaluate the pct hearing loss for long term exposure to a decibel level of 100. (e) According to this model, what decibel level would result in a 40% hearing loss? (i.e., solve for where y = 40, or y 40 = 0) 2. Chlorine residual in a pool is a function of the number of hours after treatment. Consider the following data: Hrs after treatment Chlorine residual Find the exponential regression fit. Using this fit, find the number of hours required for the residual to drop to 0.5 (i.e., solve for where y 0.5 = 0) 3. Small amounts of a chemical were introduced into the food supply of lab mice. The number of mice, per hundred, experiencing side effects is a function of the dosage. Here is some data: Dosage (mg) # mice Determine an appropriate functional fit. How many mice should suffer with 18 mg? What dosage would cause 30 mice to have side effects? 4. The growth of cactus (in inches) is a function of the time after grafting. Here are some data: Weeks Growth Determine an appropriate functional fit. Then find the number of weeks that would give a growth of 25 in.