Question

Click-through rates Companies work hard to have their website listed at the top of an Internet search. Is there a relationship between a website’s position in the results of an Internet search (1=top position,2=2nd position, etc.) and the percentage of people who click on the link for the website? Here are click-through rates for the top 10 positions in searches on a mobile device:

a. Make an appropriate scatterplot for predicting click-through rate from the position. Describe what you see.

b. Use transformations to linearize the relationship. Does the relationship between click-through rate and position seem to follow an exponential model or a power model? Justify your answer.

c. Perform least-squares regression on the transformed data. Give the equation of your regression line. Define any variables you use.

d. Use your model from part (c) to predict the click-through rate for a website in the 11th position.

Short answer

(a) The pattern in the scatter plot is downwards, hence the direction is negative.

Strength: strong

(b) The model is exponential

(c)$\ln y=1.4674-0.1430\ln x$

(d) The expected click-through rate is $0.784152\%$.

Step-by-step solution

Part (a) Step 1: Given information

The given data is

Part (a) Step 2: Explanation

The scatter plot for the given data is

Because the points do not lie on a straight line, the shape is curved.

The pattern in the scatter plot is downwards, hence the direction is negative.

Strength: strong, due to the fact that all points lie within a narrow range of deviations from the basic pattern in the points.

Part (b) Step 1: Given information

The given data is

Part (b) Step 2: Explanation

From the above scatter plot

Because there is no strong curvature in the scatter plot of position vs log (click-through rate), an exponential model would be acceptable.

Part (c) Step 1: Given information

The given data is

Part (c) Step 2: Explanation

The logarithm of length is $X$, and the logarithm of heart weight is $Y$.

Find the values from the above table

$$\begin{gathered} a=\frac{\sum y·\sum x^2-\sum x\sum xy}{n\sum x^2-{\left(\sum x\right)}^2} \\ a=1.4674 \end{gathered}$$

$$\begin{gathered} b=\frac{n\sum xy-\sum x·\sum y}{n\sum x^2-{\left(\sum x\right)}^2} \\ b=-0.1430 \end{gathered}$$

Substitute the values

$y=a+b.x$

$y=1.4674-0.1430x$

The least-square regression equation is

$\ln y=1.4674-0.1430\ln x$.

Part (d) Step 1: Given information

The given data is

Part (d) Step 2: Explanation

From part (c)

$\ln y=1.4674-0.1430\ln x$

Putting the value of x

$\ln y=1.4674-0.1430\ln \left(11\right)$

$=-0.1056$

Then take the exponential

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

The expected click-through rate is$0.784152\%.$