Chapter 12.2, Problem 50E

(a)

To determine

To construct: an appropriate scatter plot predicting click through rate from position.

Answer to Problem 50E

There is negative, strong, curved relation between position and click through rate.

Explanation of Solution

Given:

Position	Click through rate
1	23.53
2	14.94
3	11.19
4	7.47
5	5.29
6	3.8
7	2.79
8	2.11
9	1.57
10	1.18

Graph:

Scatter plot

  

Form: curved, the reason is that the points are not lying on the straight line

Direction: Negative, the reason is that the pattern is downwards in the scatter plot.

Strength: strong, the reason is that all points lie don’t deviate much from the general pattern in the points.

(b)

To determine

To Explain: that this relationship follows an exponential model or a power model.

Answer to Problem 50E

Exponential Model

Explanation of Solution

Given:

Position	Click through rate
1	23.53
2	14.94
3	11.19
4	7.47
5	5.29
6	3.8
7	2.79
8	2.11
9	1.57
10	1.18

Graph:

  

  

The scatter plot of position v/s log (click through rate) represent no strong curved, that means that an exponential model would be appropriate.

(c)

To determine

To Explain: the equation of the regression line.

Answer to Problem 50E

  $\ln y=1.4674-0.1430\ln x$

Explanation of Solution

Given:

Position	Click through rate
1	23.53
2	14.94
3	11.19
4	7.47
5	5.29
6	3.8
7	2.79
8	2.11
9	1.57
10	1.18

Formula used:

  $\begin{array}{l}a=\frac{{\displaystyle \sum y}.{\displaystyle \sum x^2}-{\displaystyle \sum x}{\displaystyle \sum xy}}{n{\displaystyle \sum x^2}-{\left({\displaystyle \sum x}\right)}^2}\\ b=\frac{n{\displaystyle \sum xy}-{\displaystyle \sum x}.{\displaystyle \sum y}}{n{\displaystyle \sum x^2}-{\left({\displaystyle \sum x}\right)}^2}\\ y=a+b.x\end{array}$

Calculation:

X is the logarithm of length and Y is the logarithm of heart weight

Position	Click through rate	X	Y	XY	X2
1	23.53	0	3.158276203	0	0
2	14.94	0.693147181	2.70404218	1.874299213	0.480453
3	11.19	1.098612289	2.415020522	2.653171223	1.206949
4	7.47	1.386294361	2.010894999	2.787692398	1.921812
5	5.29	1.609437912	1.665818246	2.68103104	2.59029
6	3.8	1.791759469	1.335001067	2.392000803	3.210402
7	2.79	1.945910149	1.026041596	1.996584755	3.786566
8	2.11	2.079441542	0.746687947	1.552693937	4.324077
9	1.57	2.197224577	0.451075619	0.991114437	4.827796
10	1.18	2.302585093	0.165514438	0.381111079	5.301898
		${\displaystyle \sum x}$=15.10441257	${\displaystyle \sum y}$ =15.67837282	${\displaystyle \sum xy}$ =17.30969888	${\displaystyle \sum x^2}$ =27.65024
				XY	X2
				0	0

Using from the table find the value of a and b

  $a=\frac{{\displaystyle \sum y}.{\displaystyle \sum x^2}-{\displaystyle \sum x}{\displaystyle \sum xy}}{n{\displaystyle \sum x^2}-{\left({\displaystyle \sum x}\right)}^2}$ =1.4674

  $b=\frac{n{\displaystyle \sum xy}-{\displaystyle \sum x}.{\displaystyle \sum y}}{n{\displaystyle \sum x^2}-{\left({\displaystyle \sum x}\right)}^2}$ =-0.1430

Substituting the value of a and b in the regression equation formula

  $\begin{array}{l}y=a+b.x\\ y=1.4674-0.1430x\end{array}$

The least squares regression equation

  $\ln y=1.4674-0.1430\ln x$

Where x is the position and y the click through rate

(d)

To determine

To Explain: the prediction from the part (c) of the click through rate for a website in the 11^th position.

Answer to Problem 50E

0.784152%

Explanation of Solution

Given:

Position	Click through rate
1	23.53
2	14.94
3	11.19
4	7.47
5	5.29
6	3.8
7	2.79
8	2.11
9	1.57
10	1.18

Calculation:

Using the part (c)

  $\ln y=1.4674-0.1430\ln x$

Putting the value of x

  $\begin{array}{l}\ln y=1.4674-0.1430\ln x\\ \ln y=1.4674-0.1430\ln \left(11\right)\\ \ln y=-0.1056\end{array}$

Taking the exponential

  $\begin{array}{c}y=e^{\log y}=e^{-0.1056}\\ =0.784152\end{array}$

Therefore the expected click through rate is 0.784152%