A company invests continuously in advertising. The table gives the advertising budget $x$ and the turnover $y$, in millions of francs, for four consecutive months. A regression line of $y$ on $x$ is given by $y = 9x + 0.6$. We need to calculate the mean of $x$, the mean of $y$ in terms of $a$, show that $a=20$, calculate the correlation coefficient and estimate $y$ for $x=3.2$.

1. Problem Description

A company invests continuously in advertising. The table gives the advertising budget $x$ and the turnover $y$, in millions of francs, for four consecutive months. A regression line of $y$ on $x$ is given by $y = 9x + 0.6$. We need to calculate the mean of $x$, the mean of $y$ in terms of $a$, show that $a=20$, calculate the correlation coefficient and estimate $y$ for $x=3.2$.

2. Solution Steps

1. Calculate $\bar{x}$:

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

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

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

3. Show that $a = 20$:

The regression line is $y = 9x + 0.6$. The point $(\bar{x}, \bar{y})$ must lie on the regression line. Therefore:

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

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

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

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

$55 + a = 75$

$a = 75 - 55$

$a = 20$

4. Calculate the correlation coefficient:

The correlation coefficient $r$ can be calculated using the formula $b = r \frac{s_y}{s_x}$, where $b$ is the slope of the regression line. We also have the following formulas:

$s_x^2 = \frac{\sum x_i^2}{n} - \bar{x}^2$ and $s_y^2 = \frac{\sum y_i^2}{n} - \bar{y}^2$

$x_i$: 1.2, 1.4, 1.6, 1.8, 2

$y_i$: 13, 12, 14, 16, 20

$n = 5$

$\bar{x} = 1.6$

$\bar{y} = \frac{55+20}{5} = \frac{75}{5} = 15$

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

$s_x^2 = \frac{13.2}{5} - (1.6)^2 = 2.64 - 2.56 = 0.08$

$s_x = \sqrt{0.08} \approx 0.2828$

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

$s_y^2 = \frac{1165}{5} - (15)^2 = 233 - 225 = 8$

$s_y = \sqrt{8} \approx 2.828$

The slope of the regression line $b = 9$.

Therefore, $9 = r \frac{2.828}{0.2828}$

$9 = r \times 10$

$r = \frac{9}{10} = 0.9$

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

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

Using the regression line equation:

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

3. Final Answer

1. $\bar{x} = 1.6$

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

3. $a = 20$

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

5. $y = 29.4$