Question

4. The table shows the average cost, in dollars, of a movie ticket in the United States for different years. Year 1948 1963 1974 1978 1985 1991 1997 2005 2010 Cost 0.36 0.86 1.89 2.34 3.55 4.21 4.59 6.41 7.89 Use technology to answer the following: (a) Find the equation of the least squares regression line. Round values of a and b to four decimal places. (b) Find r. Round to three decimal places. Interpret the r-value. (c) Find . Round to three decimal places. Tell what it means. (d) Use the equation of the least squares regression line to predict the average cost of a movie ticket in 1995.

Answer 1

Part A

Given the table which shows the average cost, in dollars, of a movie ticket in the United States for different years. 

$\begin{center} \begin{tabular} {|c|c|} Year&Cost\\[1ex] 1948&0.36\\ 1963&0.86\\ 1974&1.89\\ 1978&2.34\\ 1985&3.55\\ 1991&4.21\\ 1997&4.59\\ 2005&6.41\\ 2010&7.89 \end{tabular} \end{center}$

We can form our regression table by rewriting the year column with 1948 represented 0 and the other years represented by the number of years after 1948.

$\begin{center} \begin{tabular} {|c|c|c|c|} Year(X)&Cost(Y)&X^2&XY\\[1ex] 0&0.36&0&0\\ 15&0.86&225&12.9\\ 26&1.89&676&49.14\\ 30&2.34&900&70.2\\ 37&3.55&1,369&131.35\\ 43&4.21&1,849&181.03\\ 49&4.59&2,401&224.91\\ 57&6.41&3,249&365.37\\ 62&7.89&3,844&489.18\\ \sum X=319&\sum Y=32.1&\sum X^2=14,513&\sum XY=1,524.08\end{tabular} \end{center}$

The equation of the least square regression line is given by $y=a+bx$, where $a= \frac{(\sum Y)(\sum X^2)-(\sum X)(\sum XY)}{n(\sum X^2)-(\sum X)^2}$ and $b=\frac{n(\sum XY)-(\sum X)(\sum Y)}{n(\sum X^2)-(\sum X)^2}$

Thus, we have

$a=\frac{(32.1)(14,513)-(319)(1,524.08)}{9(14,513)-(319)^2} \\ \\ = \frac{465,867.3-486,181.52}{130,617-101,761} = \frac{-20,314.22}{28,856} \\ \\ =-0.7040$

and

$b=\frac{9(1,524.08)-(319)(32.1)}{9(14,513)-(319)^2} \\ \\ = \frac{13,716.72-10,239.9}{130,617-101,761} = \frac{3,476.82}{28,856} \\ \\ =0.1205$

Therefore, the equation of the least square regression line is given by $y=-0.7040+0.1205x$.


Part B

To find the correlation coefficient (r), we need an additional column to our table above, the column for Y^2.

$\begin{center} \begin{tabular} {|c|} Y^2\\ [1ex] 0.1296\\0.7396\\3.5721\\5.4756\\12.6025\\17.7241\\21.0681\\41.0881\\62.2521\\ \sum Y^2=164.6518 \end{tabular} \end{center}$

The correlation coefficient (r) is given by

$r= \frac{n(\sum XY)-(\sum X)(\sum Y)}{\sqrt{[n(\sum X^2)-(\sum X)^2][n(\sum Y^2)-(\sum Y)^2]}} \\ \\ = \frac{9(1,524.08)-(319)(32.1)}{\sqrt{[9(14,513)-(319)^2][9(164.6518-(32.1)^2]}} \\ \\ = \frac{13,716.72-10,239.9}{\sqrt{(130,617-101,761)(1,481.8662-1,030.41}} \\ \\ = \frac{3,476.82}{\sqrt{(28,856)(451.4562)}} = \frac{3,476.82}{\sqrt{13,027,220.11}} \\ \\ = \frac{3,476.82}{3,609.32} =0.963$

Since r = 0.963 which is very close to positive 1, thus, there exist a strong positive association between the two variables, i.e. between the year and the average cost of a movie ticket. In other words, as the year increases, the average cost of movie ticket increases.


Part 3

$r^2=0.963^2=0.928$

$r^2$ is called the coefficient of determination and is a measure of the closeness of the data points are to the regression line. The closer the value of $r^2$ to 1, the closer are the data points to the regression line.

Since we obtained $r^2=0.928$, which is very close to 1, this means that the data points are close to the regression line.


Part D

We obtained the equation of the least square regression line as $y=-0.7040+0.1205x$, where y is the average cost of movie ticket in x years after 1948.

1995 is 1995 - 1948 = 47 years after 1948, thus, the predicted average cost of movie ticket in 1995 is given by

$y=-0.7040+0.1205(47) \\ \\ =-0.7040+5.6635\approx\ 4.96$