Chapter 4.6, Problem 11PPS

a.

To determine

To draw: A scatter plot of the data ,where x =1 represents 2005.Then the equation for the best-fit line.

Answer to Problem 11PPS

  $y=0.0326x+\mathrm{1.598.}$

Explanation of Solution

Given: State fair : Refer to the beginning of the lesson, it gives the following table for the attendance at the state fair.

  

Calculation:

To graph a scatter plot of the data ,where x =1 represents 2005.then the table becomes,

Year x	1	2	3	4	5
Attendance (millions) y	1.633	1.681	1.682	1.693	1.790

To use the above data and a graphing calculator the linear regression equation is,

  $y=0.0326x+\mathrm{1.598.}$

Graph this equation on scatterplot.

  ..

b.

To determine

To draw: A graph and analyze the residual plot.

Answer to Problem 11PPS

The residuals are randomly scattered above and below the x -axis so the regression equation fits the data well. Regression model is a good fit for the data.

Explanation of Solution

Given: State fair :Refer to the beginning of the lesson, it gives the following table for the attendance at the state fair.

  

Calculation:

To graph a scatter plot of the data ,where x =1 represents 2005.then the table becomes,

Year x	1	2	3	4	5
Attendance (millions) y	1.633	1.681	1.682	1.693	1.790

To use the above data and a graphing calculator the linear regression equation is,

  $y=0.0326x+\mathrm{1.598.}$

Graph the residuals plot. The residuals are randomly scattered above and below the x -axis so the regression equation fits the data well.

  

c.

To determine

To predict: The total attendance in 2025.

Answer to Problem 11PPS

2.28 million people, Yes.

Explanation of Solution

Given: State fair :Refer to the beginning of the lesson, it gives the following table for the attendance at the state fair.

  

Calculation:

Conclusion:

2025 is 20 years after 2005 so we need to find y -value for x= 21(since x =1 represents 2005)on our graph of linear regression.

Put x= 21 in linear regression equation ,

  $\begin{array}{l}y=0.0326x+\mathrm{1.598.}\\ y=0.0326\times 21+1.598=2.28\\ \end{array}$

So, the predicted attendance in 2005 is 2.28 million people. This prediction is reasonable since it is not drastically larger than the attendance in the table. The correlation coefficient for the linear regression was also about 0.896. Since it is very close to 1 , the regression model is a very good fit for the data so it will gives accurate extrapolations , provided the trend in attendance increses.