College Algebra 9th txtbk

SECTION 2–4 Linear Equations and Models 153
MATCHED PROBLEM 5
Table 3 Emerald-Shaped
Diamond Prices
Weight (Carats)
Price
0.5 $1,350
0.6 $1,740
0.7 $2,610
0.8 $3,320
0.9 $4,150
1.0 $4,850
Source: www.tradeshop.com
Prices for emerald-shaped diamonds taken from an online trader are given in Table 3. Repeat
Example 5 for this data with the linear model
p 7,270c 2,450
where p is the price of an emerald-shaped diamond weighing c carats.
The model we used in Example 5 was obtained by using a technique called linear
regression and the model is called the regression line. This technique produces a line that
is the best fit for a given data set. We will not discuss the theory behind this technique, nor
the meaning of “best fit.” Although you can find a linear regression line by hand, we prefer
to leave the calculations to a graphing calculator or a computer. Don’t be concerned if you
don’t have either of these electronic devices. We will supply the regression model in the
applications we discuss, as we did in Example 5.
Technology Connections
If you want to use a graphing calculator to construct regression
lines, you should consult your user’s manual.* The
process varies from one calculator to another. Figure 4
shows three of the screens related to the construction of the
model in Example 5 on a Texas Instruments TI-84 Plus.
8,000
0
1.5
(a) Entering the data. (b) Finding the model. (c) Graphing the data and the model.
Z Figure 4
1,000
*User’s manuals for the most popular graphing calculators are readily available on the Internet.
In Example 5, we used the regression line to approximate points that were not given in
Table 2, but would fit between points in the table. This process is called interpolation. In the
next example we use a regression model to approximate points outside the given data set. This
process is called extrapolation and the approximations are often referred to as predictions.
EXAMPLE 6 Telephone Expenditures
Table 4 gives information about expenditures for residential and cellular phone service. The
linear regression model for residential service is
r 722 33.1t
where r is the average annual expenditure (in dollars per consumer unit) on residential service
and t is time in years with t 0 corresponding to 2000.
(A) Interpret the slope of the regression line as a rate of change.
(B) Use the regression line to predict expenditures for residential service in 2018.