The problem provides a table of data relating advertising budget $x$ and revenue $y$ for a company over four consecutive months. The regression line of $y$ on $x$ is given by the equation $y = 9x + 0.6$. We are asked to calculate the mean of $x$, the mean of $y$ in terms of $a$, show that $a = 20$, calculate the correlation coefficient and assess the strength of the correlation, and estimate $y$ for $x = 3.2$.

1. Problem Description

The problem provides a table of data relating advertising budget $x$ and revenue $y$ for a company over four consecutive months. The regression line of $y$ on $x$ is given by the equation $y = 9x + 0.6$. We are asked to calculate the mean of $x$, the mean of $y$ in terms of $a$, show that $a = 20$, calculate the correlation coefficient and assess the strength of the correlation, and estimate $y$ for $x = 3.2$.

2. Solution Steps

1. Calculate $\bar{x}$:

$\bar{x}$ is the mean of the $x$ values.

$$$$

\bar{x} = \frac{1.2 + 1.4 + 1.6 + 1.8 + 2}{5} = \frac{8}{5} = 1.6

$$$$

2. Calculate $\bar{y}$ in terms of $a$:

$\bar{y}$ is the mean of the $y$ values.

$$$$

\bar{y} = \frac{13 + 12 + 14 + 16 + a}{5} = \frac{55 + a}{5}

$$$$

3. Show that $a = 20$:

The regression line passes through the point $(\bar{x}, \bar{y})$. Therefore, we have

$$$$

\bar{y} = 9\bar{x} + 0.6

$$$$

Substitute $\bar{x} = 1.6$ and $\bar{y} = \frac{55+a}{5}$ into the equation:

$$$$

\frac{55+a}{5} = 9(1.6) + 0.6

$$$$

$$$$

\frac{55+a}{5} = 14.4 + 0.6 = 15

$$$$

$$$$

55 + a = 5(15) = 75

$$$$

$$$$

a = 75 - 55 = 20

$$$$

4. Calculate the correlation coefficient:

The correlation coefficient, $r$, can be calculated using the formula:

$r = \frac{n\sum xy - (\sum x)(\sum y)}{\sqrt{[n\sum x^2 - (\sum x)^2][n\sum y^2 - (\sum y)^2]}}$

Here $n=5$.

We have $\sum x = 8$

$\sum y = 13 + 12 + 14 + 16 + 20 = 75$

$\sum x^2 = 1.2^2 + 1.4^2 + 1.6^2 + 1.8^2 + 2^2 = 1.44 + 1.96 + 2.56 + 3.24 + 4 = 13.2$

$\sum y^2 = 13^2 + 12^2 + 14^2 + 16^2 + 20^2 = 169 + 144 + 196 + 256 + 400 = 1165$

$\sum xy = (1.2)(13) + (1.4)(12) + (1.6)(14) + (1.8)(16) + (2)(20) = 15.6 + 16.8 + 22.4 + 28.8 + 40 = 123.6$

$r = \frac{5(123.6) - (8)(75)}{\sqrt{[5(13.2) - (8)^2][5(1165) - (75)^2]}} = \frac{618 - 600}{\sqrt{[66-64][5825 - 5625]}} = \frac{18}{\sqrt{[2][200]}} = \frac{18}{\sqrt{400}} = \frac{18}{20} = 0.9$

Since $r = 0.9$, the correlation is strong.

5. Estimate $y$ for $x = 3.2$:

Use the regression equation to estimate $y$ for $x = 3.2$.

$$$$

y = 9x + 0.6

$$$$

$$$$

y = 9(3.2) + 0.6 = 28.8 + 0.6 = 29.4

$$$$

3. Final Answer

1. $\bar{x} = 1.6$

2. $\bar{y} = \frac{55 + a}{5}$

3. $a = 20$

4. The correlation coefficient is $0.9$. The correlation is strong.

5. Estimated $y$ for $x = 3.2$ is $29.4$.