Transforming Relationships Chapter 4.1: Exponential Growth and Power Law Models Part A: Day 1: Exponential Growth
Beyond Linear What if your data is clearly curved in some manner? Are there models we can use, and prediction equations we can develop? Of course … and we will examine two basic types … Exponential Growth & The Power Law But ARE we really beyond LINEAR?
The Exponential Model Y = abx “a” is the initial value when x = 0, it is the y-intercept. “a” is often unrealistically small depending on the manner in which the data is entered. Each subsequent Y value is obtained by multiplying by a factor “b”. Taking a Logarithm (LOG) of the y-value, and using the old x-value will linearize the data.
Transforming Data Enter the following data and observe the curved pattern. Do a LinReg on L1, L2 ŷ = -18081823.7 + 9083.3487(x); r = .96501 Check out the residuals, just to reinforce the non-linearity. Cell Phone Subscribers in the U.S., 1990-1999 Year 1990 1993 1994 1995 1996 1997 1998 1999 Subscribers (1000’s) 5283 16,009 24,134 33,784 44,043 55,312 69,209 86,047
Logarithmic Transformation Now, obtain the data from the Y-value list (L2), and “take the LOG” of each value. Place the resulting LOGGED DATA into L3 Re-plot the L1/L3 data. Are things perfectly linear? Explore with LinReg, r-value. Comment. Log (ŷ) = -263.203 + 0.13417(x); r = .99116 Year 1990 1993 1994 1995 1996 1997 1998 1999 Log of Subscribers (1000’s) 3.7229 4.2044 4.3826 4.5287 4.6439 4.7428 4.8402 4.9347
Still not perfect – Is it? Eliminate all data except the last 4 years Dump this data into L4/L5 Do a LinReg on L4/L5 … better? Log (ŷ) = -188.951 + 0.09699(x); r = .99995 How about that Residual Plot still? Grrrrrr. Year 1996 1997 1998 1999 Log of Subscribers (1000’s) 4.6439 4.7428 4.8402 4.9347
Predictions? OK, regardless of the suspicious Residual Plot … we move on. Let’s use the last Prediction Equation to Predict for the year 2000. Log (ŷ) = -188.951 + 0.09699(2000) Log (ŷ) = 5.032878574 ŷ =10^ 5.032878574 = 107, 864.5
Conclusions? If a variable grows exponentially, then its logarithm grows linearly High r and R-square values are not the total picture. Near perfect (.99999) r values and R-Square values still are incomplete. Residuals tell a big tale. But the magnitude of the error can still warrant usage, if we are simply trying to predict! Plug into the “Log Equation”, then raise answer to the 10th power.
Another Problem! The combined American Indian, Eskimo, Aleut, Asian and Pacific Islander population grew in the US from 1950 to 1990 …as shown below … When entering the year, enter 50 for 1950, 60 for 1960, etc. Perform the Regression after transforming the data. Make a prediction of this combined population in the year 2000 … Year 1950 1960 1970 1980 1990 Population(1000’s) 1131 1620 2557 5150 9534
So how’d ya do? Log (ŷ) = 1.8246 + 0.023539(x); r = .992025 Log (ŷ) = 1.8246 + 0.023539(100) … note we are using “100” to represent the year 2000. Log (ŷ) = 4.17853333477 ŷ = 10 ^ 4.17853333477 = 15084.5839 So … the population would be predicted to be : 15,084,584 people. Do you think your prediction (Extrapolation) is too high or too low as compared to the actual population in 2000? Why?
Gypsy Moths Enter the data into L1 and L2 for the year (x–List 1) and the Acres of land defoliated by the Gypsy Moth (y-List 2). Year Acres 1978 63,042 1979 226,260 1980 907,075 1981 2,826,095
Gypsy Moths P.212/#4.6 – A) Plot the number of acres defoliated (y) against the year (x). B) Check out the three consecutive ratios of the Acreage … to verify the approximate exponential growth. What is that approximate growth RATIO (to the nearest integer)? C) “Linearize” the data – i.e. Transform the y-values, and plot the results. D) Calculate the LSRL for the transformed data. Log (ŷ) = -1094.51 + 0.5558(x); r = .999293 E) Construct and interpret the residual plot. F) Perform an inverse transformation to express ŷ as an exponential function of year. G) Predict the number of acres defoliated in 1982. Log (ŷ) = -1094.51 + 0.5558(1982) = 7.0302 ŷ = 10 ^ 7.0302 = 10,719,964.92 acres.