The table lists the cost in millions of dollars for a Year 1993 1997 2004 2009 2
ID: 3136999 • Letter: T
Question
The table lists the cost in millions of dollars for a Year 1993 1997 2004 2009 2012 30-second commercial for selected years a. Find a linear function f that models the data. b. Estimate the cost in 2010 and compare the estimate to the actual value of $23.6 million. Did your estimate involve interpolation or extrapolation? a. Use linear regression to determine the equation of the line that best fits the data. The linear function is f(x) Type an expression using x as the variable. Use integers or decimals for any numbers in the expression. Round to three decimal places as needed.) b. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. Cost 20.921.522.4 23.1 23.5 (Do not round until the final answer. Then round to the nearest tenth as needed.) O A. The estimated cost in the year 2010 is S million, which is outside $2 million of the actual cost. O B. The estimated cost in the year 2010 is S million, which is within $2 million of the actual cost. Did the estimate involve interpolation or extrapolation? interpolation O extrapolationExplanation / Answer
### By using R command
> y=c(20.9,21.5,22.4,23.1,23.5) ###Cost
> y
[1] 20.9 21.5 22.4 23.1 23.5
> x=c(1,5,12,17,20) ### Years taking 1993 =1 and so on
> x
[1] 1 5 12 17 20
> fit=lm(y~x)
> fit
Call:
lm(formula = y ~ x)
Coefficients:
(Intercept) x
20.7859 0.1358
a) The linear function is
f(x) = 20.7859 + 0.1358*x
b) The cost in year 2010 that is when x=18 is
f(x) = 20.7859 + 0.1358*18
=23.23
hence the estimated cost is $23.3 million which is within $2 million of actual cost.
c) The estimate involves interpolation. Since we are estimating the cost for the year which is within the data set.
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