An advertising firm wishes to demonstrate to its clients the effectiveness of the advertising campaigns it has conducted. The following bivariate data on twelve recent campaigns, including the cost of each campaign (denoted by x, in millions of dollars) and the resulting percentage increase in sales (denoted by y) following the campaign, were presented by the firm. A scatter plot of the data is shown in Figure 1. Also given is the product of the campaign cost and the percentage increase in sales for each of the twelve campaigns. (These products, written in the column labelled "xy", may aid in calculations.)

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KensingtonD19834 KensingtonD19834

28-10-2022
Mathematics

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An advertising firm wishes to demonstrate to its clients the effectiveness of the advertising campaigns it has conducted. The following bivariate data on twelve recent campaigns, including the cost of each campaign (denoted by x, in millions of dollars) and the resulting percentage increase in sales (denoted by y) following the campaign, were presented by the firm. A scatter plot of the data is shown in Figure 1. Also given is the product of the campaign cost and the percentage increase in sales for each of the twelve campaigns. (These products, written in the column labelled "xy", may aid in calculations.)

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ClariceX478929 ClariceX478929 · Answer 1 · 2022-10-28T18:08:18-04:00

To find the correlation coefficient, we have to use the following formula

[tex]r=\frac{n\Sigma(xy)-\Sigma(x)\cdot\Sigma(y)}{\sqrt[]{\lbrack n\Sigma(x)^2-(\Sigma(x))^2\rbrack\lbrack n\Sigma(y)^2-(\Sigma(y))^2\rbrack}}[/tex]

So, we have to find the sum of all three columns.

[tex]\begin{gathered} \Sigma(xy)=211.8685 \\ \Sigma(x)=31.62 \\ \Sigma(y)=79.68 \end{gathered}[/tex]

The sum of all the x-values to the square power is

[tex]\begin{gathered} \Sigma(x)^2=92.5414 \\ \Sigma(y)^2=529.7792 \end{gathered}[/tex]

Now, we include all the numbers in the formula

[tex]\begin{gathered} r=\frac{12\cdot211.8685-31.62\cdot79.68}{\sqrt[]{\lbrack12\cdot92.5414-(31.62)^2\rbrack\lbrack12\cdot529.7792-(79.68)^2\rbrack}} \\ r=\frac{2542.422-2519.4816}{\sqrt[]{110.6724\cdot8.448}} \\ r=\frac{22.9404}{30.5771} \\ r\approx0.750 \end{gathered}[/tex]

Answer :

Hence, the correlation coefficient is r = 0.750, approximately.

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