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.

The given data is

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.

The given data is

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.

The given data is

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

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

The given data is

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$0.784152\%.$