A company invests continuously in advertising. The table provides the advertising budget $x$ and the turnover $y$, in millions of francs, for four consecutive months. The regression line of $y$ on $x$ is given by $y = 9x + 0.6$. We need to: 1. Calculate the mean of $x$, denoted as $\bar{x}$.

1. Problem Description

A company invests continuously in advertising. The table provides the advertising budget $x$ and the turnover $y$, in millions of francs, for four consecutive months. The regression line of $y$ on $x$ is given by $y = 9x + 0.6$. We need to:

1. Calculate the mean of $x$, denoted as $\bar{x}$.

2. Calculate the mean of $y$, denoted as $\bar{y}$, as a function of $a$.

3. Using the results from questions 1 and 2, show that $a = 20$.

4. Calculate the correlation coefficient and determine if the correlation is strong.

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

2. Solution Steps

1. Calculate $\bar{x}$:

$\bar{x}$ is the average of the $x$ values: $1.2, 1.4, 1.6, 1.8, 2$.

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

2. Calculate $\bar{y}$ as a function of $a$:

$\bar{y}$ is the average of the $y$ values: $13, 12, 14, 16, a$.

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

3. Show that $a = 20$:

Since the regression line is $y = 9x + 0.6$, it passes through the point $(\bar{x}, \bar{y})$.

Therefore, $\bar{y} = 9\bar{x} + 0.6$.

Substituting the values of $\bar{x}$ and $\bar{y}$, we get:

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

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

$\frac{55 + a}{5} = 15$

$55 + a = 15 \times 5$

$55 + a = 75$

$a = 75 - 55$

$a = 20$

4. Calculate the correlation coefficient:

The formula for the correlation coefficient $r$ is:

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

Where $n$ is the number of data points, which is

5. Let's calculate the necessary sums:

$\sum x = 1.2 + 1.4 + 1.6 + 1.8 + 2 = 8$

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

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

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

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

Plugging these values into the formula for $r$:

$r = \frac{5(123.6) - (8)(75)}{\sqrt{[5(13.2) - (8)^2][5(1165) - (75)^2]}}$

$r = \frac{618 - 600}{\sqrt{[66 - 64][5825 - 5625]}}$

$r = \frac{18}{\sqrt{(2)(200)}}$

$r = \frac{18}{\sqrt{400}}$

$r = \frac{18}{20} = 0.9$

Since $r = 0.9$, which is close to 1, the correlation is strong.

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

Using the regression line equation $y = 9x + 0.6$:

$y = 9(3.2) + 0.6$

$y = 28.8 + 0.6$

$y = 29.4$

3. Final Answer

1. $\bar{x} = 1.6$

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

3. $a = 20$

4. $r = 0.9$. The correlation is strong.

5. $y = 29.4$