Respuesta :
Part A
Given the table which shows the average cost, in dollars, of a movie ticket in the United States for different years.
[tex]\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}[/tex]
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.
[tex]\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}[/tex]
The equation of the least square regression line is given by [tex]y=a+bx[/tex], where [tex]a= \frac{(\sum Y)(\sum X^2)-(\sum X)(\sum XY)}{n(\sum X^2)-(\sum X)^2} [/tex] and [tex]b=\frac{n(\sum XY)-(\sum X)(\sum Y)}{n(\sum X^2)-(\sum X)^2}[/tex]
Thus, we have
[tex]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[/tex]
and
[tex]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[/tex]
Therefore, the equation of the least square regression line is given by [tex]y=-0.7040+0.1205x[/tex].
Part B
To find the correlation coefficient (r), we need an additional column to our table above, the column for Y^2.
[tex]\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}[/tex]
The correlation coefficient (r) is given by
[tex]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[/tex]
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
[tex]r^2=0.963^2=0.928[/tex]
[tex]r^2[/tex] 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 [tex]r^2[/tex] to 1, the closer are the data points to the regression line.
Since we obtained [tex]r^2=0.928[/tex], 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 [tex]y=-0.7040+0.1205x[/tex], 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
[tex]y=-0.7040+0.1205(47) \\ \\ =-0.7040+5.6635\approx\$4.96[/tex]
Given the table which shows the average cost, in dollars, of a movie ticket in the United States for different years.
[tex]\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}[/tex]
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.
[tex]\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}[/tex]
The equation of the least square regression line is given by [tex]y=a+bx[/tex], where [tex]a= \frac{(\sum Y)(\sum X^2)-(\sum X)(\sum XY)}{n(\sum X^2)-(\sum X)^2} [/tex] and [tex]b=\frac{n(\sum XY)-(\sum X)(\sum Y)}{n(\sum X^2)-(\sum X)^2}[/tex]
Thus, we have
[tex]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[/tex]
and
[tex]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[/tex]
Therefore, the equation of the least square regression line is given by [tex]y=-0.7040+0.1205x[/tex].
Part B
To find the correlation coefficient (r), we need an additional column to our table above, the column for Y^2.
[tex]\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}[/tex]
The correlation coefficient (r) is given by
[tex]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[/tex]
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
[tex]r^2=0.963^2=0.928[/tex]
[tex]r^2[/tex] 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 [tex]r^2[/tex] to 1, the closer are the data points to the regression line.
Since we obtained [tex]r^2=0.928[/tex], 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 [tex]y=-0.7040+0.1205x[/tex], 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
[tex]y=-0.7040+0.1205(47) \\ \\ =-0.7040+5.6635\approx\$4.96[/tex]
