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FITTING LINEAR MODELS TO DATA
Supplement to Unit 9B MATH 1001 In the handout we will learn how to find a linear model for data that is given and use it to make predictions. We will also learn how to measure how closely the model “fits” the given data. We will learn how to find the linear model that best fits a set of given data. Finding a Linear Model for Data and Making Predictions First, we consider the following table that gives the population for Spalding County, Georgia from 1960 to 2000. (Source: U.S. Census Bureau) Year Pop. (thousand) Change 1960 35. 1970 39.5 4. 1980 47.9 8. 1990 54.5 6. 2000 58.4 3. The third column of this table shows (for each decade year) the change in population during the preceding decade. We see the population of Spalding County increased by about 4 thousand people in the 1950s and 1990s. In the 1970s and 1980s the population increased by roughly 7 to 8 thousand people. We might wonder whether this qualifies as almost linear population growth. So, we will plot the data and look. We plot the year on the x -axis and the population on the y - axis. Figure 1 It appears from Figure 1 that these points do appear to lie on or near some straight line. But, how can we find a straight line that passes through or near each data point? One way is to simply pass a straight line through the first and the last data points. To make this easier, we will let t be the number of years after 1960. Thus, our data will look like:
Fitting Linear Models to Data t (years since 1960)
P
(in thousands) 0 35. 10 39. 20 47. 30 54. 40 58. To find the slope between the first data point (0, 35.4) and the last data point (40, 58.4), we use the formula for the slope:
- 575 40 23 40 0
- 4 35. 4 change in change in slope t P m Note that the P -intercept is (0, 35.4). Thus, a linear model for the population of Spalding County is P ( t ) = 0.575 t + 35. The graph of this line and the data points is shown in Figure 2 below. You can see that this line passes through or near each data point. Figure 2 One of the reasons to find a model for real-world is to use the model to make predictions. These predictions fall into two categories: (1) making predictions within the scope of the data and (2) make predictions beyond the scope of the data. As an example of the first type of prediction use the model we found for Spalding County population to predict the population for the year 1995. Notice that 1995 is 35 years after 1960. Thus, we substitute t = 35 into our equation: P (35) = 0.575×35 + 35.4 ≈ 55.
Fitting Linear Models to Data Definition: The phrase “ Sum of Squares of Errors ” is so common in data modeling that it is abbreviated SSE. Thus, the SSE associated with data modeling based on n data points is defined by 2 2 3 2 2 2 SSE E 1 E E En To find the SSE we first begin by finding the squares of the errors. t P (Actual) P ( t ) (Predicted) Error, Ei P − P ( t ) 2 E i 0 35.4 35.4 0 0 10 39.5 41.15 −1.65 2. 20 47.9 46.9 1 1 30 54.5 52.65 1.85 3. 40 58.4 58.4 0 0 Thus, the SSE is
- 145 SSE 0 2. 7225 1 3. 4225 0 The smaller the SSE is the better the model fits the data. This allows you to compare two or more different models to determine which one is the best. Another way to compare models is by finding the average error. Definition: The average error in a linear model fitting n given data points is defined by n SSE average error The average error for our model is
- 195
- 429 5
- 145 average error
Fitting Linear Models to Data Example: (a) Find a linear model for the population of Spalding County using the first and fourth data points; that is, (0, 35.4) and (30, 54,5). (b) Use your model to predict the Spalding County population in the years 1995 and 2010. (c) Find the SSE and average error. Use these to determine whether the model found in this example or the previous model is a better fit for the data. (a) (b)
Fitting Linear Models to Data Example: (a) Find the best-fit linear model for the population data for Spalding County. (b) Use your model to predict the population in 1995 and 2010. (c) Find the SSE and the average error for the model. (a) P ( t ) = 0.61 t + 34. (b) P (35) = 0.61×35 + 34.94 ≈ 56.3 thousand P (50) = 0.61×50 + 34.94 ≈ 65.4 thousand The population of Spalding County was about 56.3 thousand in 1995 and will be about 65. thousand in 2010. (c) t P (Actual) P ( t ) (Predicted) Error, Ei P − P ( t ) 2 E i 0 35.4 34.94 0.46 0. 10 39.5 41.04 −1.54 2. 20 47.9 47.14 0.76 0. 30 54.5 53.24 1.26 1. 40 58.4 59.34 −0.94 0. SSE = 0.2116 + 2.3716 + 0.5776 + 105876 + 0.8836 = 5. average error = 1. 061 5
632 Exercises: In each of problems 1 and 2 the population census data for a U.S. city is given. (a) Find a linear model for the data using the first and last data points. Let t = 0 in the year
Use it to predict the population in 2000. Calculate the average error of the model. (b) Find the linear model that best fits this census data. Let t be 0 in the year 1950. Use it to predict the population in 2000. Calculate the average error of the model.
San Diego, California: Year 1950 1960 1970 1980 1990 Pop. (thous) 334 573 697 876 1111
Riverside, California: Year 1950 1960 1970 1980 1990 Pop. (thous) 47 84 140 171 227 In each of problems 3 and 4 the population census data for a U.S. city is given.
Fitting Linear Models to Data (a) Find a linear model for the data using the second and fourth data points. Let t = 0 in the year 1950. Use it to predict the population in 2000. Calculate the average error of the model. (b) Find the linear model that best fits this census data. Let t be 0 in the year 1950. Use it to predict the population in 2000. Calculate the average error of the model.
- Garland, Texas: Year 1950 1960 1970 1980 1990 Pop. (thous) 11 39 81 139 181
- Santa Anna, California: Year 1950 1960 1970 1980 1990 Pop. (thous) 46 100 156 204 294
- The following table gives the number of compact discs (in millions) sold in the United States for the even-numbered years 1988 through 1996. Year 1988 1990 1992 1994 1996 Sales, S , (millions)
Source: The World Almanac and Book of Facts 1998. (a) Find the linear model S ( t ) = mt + b that best fits this data. Let t = 0 in 1988. (b) Compare the model’s prediction for the year 1995 with the actual 1995 CD sales of 722.9 million. (c) Use the model to predict the CD sales for the year 2002. (d) Which prediction, the one for 1995 or the one from 2002, is likely to be closer to actual sales? Why?
- The table below lists the number of passenger cars (in millions) in the United States for the years 1940 through 1990. Year 1940 1950 1960 1970 1980 1990 Number of Cars, N , (millions)
Source: Statistical Abstracts of the United States. (a) Find the best-fit linear model for the data. Let t = 0 in 1940. (b) Use your model to predict the number of passenger cars in the year 2000 and in the year 2010.
Fitting Linear Models to Data Answers:
- (a) P ( t ) = 19.425 t + 334; P (50) = 1305 thousand; average error ≈ 29. (b) P ( t ) = 18.57 t + 346.8; P (50) ≈ 1275 thousand; average error ≈ 26.
- (a) P ( t ) = 4.5 t + 47; P (50) = 272 thousand; average error ≈ 6. (b) P ( t ) = 4.47 t + 44.4; P (50) ≈ 268 thousand; average error ≈ 5.
- (a) P ( t ) = 5 t − 11; P (50) = 239 thousand; average error ≈ 11. (b) P ( t ) = 4.4 t + 2.2; P (50) ≈ 222 thousand; average error ≈ 7.
- (a) P ( t ) = 5.2 t + 48; P (50) = 308 thousand; average error ≈ 17. (b) P ( t ) = 6 t + 40; P (50) = 340 thousand; average error ≈ 10.
- (a) S ( t ) = 81.7 t + 130. (b) S (7) =702.04 million. This prediction is below what the actual sales were in 1995. (c) S (14) =1273.9 million (d) The prediction for 1995 is more accurate because it is between two known data points. The year 2002 is six years after the last data point.
- (a) N ( t ) = 2.29 t +21. (b) N(60) ≈ 159.3 million; N (70) ≈ 182.2 million
- (a) S ( v ) = 0.08 v +1. (b) S (18) ≈ 3.21 steps/sec; S (10) ≈ 2.57 steps/sec (c) We have more confidence in the prediction for the speed of 18 ft/sec since it is within the scope of the given data. The speed of 10 ft/sec is much lower than the lowest speed given.
- (b) W ( h ) = 7.33 h − 348. (c) W (72) = 179.33 pounds (d) No, a professional baseball player is not a typical American male.
