The problem describes a company's advertising budget $x$ and sales revenue $y$ (in millions of francs) over four consecutive months. A table provides the values of $x$ and $y$ for these months, with one sales revenue value, $a$, unknown. The problem also provides a regression line for $y$ in terms of $x$: $y = 9x + 0.6$. The questions ask us to calculate the mean of $x$ (denoted as $\overline{x}$), the mean of $y$ in terms of $a$ (denoted as $\overline{y}$), show that $a = 20$, calculate the correlation coefficient and assess its strength, and estimate $y$ for $x = 3.2$.

1. Problem Description

The problem describes a company's advertising budget $x$ and sales revenue $y$ (in millions of francs) over four consecutive months. A table provides the values of $x$ and $y$ for these months, with one sales revenue value, $a$, unknown. The problem also provides a regression line for $y$ in terms of $x$: $y = 9x + 0.6$. The questions ask us to calculate the mean of $x$ (denoted as $\overline{x}$), the mean of $y$ in terms of $a$ (denoted as $\overline{y}$), show that $a = 20$, calculate the correlation coefficient and assess its strength, and estimate $y$ for $x = 3.2$.

2. Solution Steps

1. Calculate $\overline{x}$:

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

$\overline{x} = \frac{8}{5}$

$\overline{x} = 1.6$

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

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

$\overline{y} = \frac{55 + a}{5}$

3. Show that $a = 20$:

The point $(\overline{x}, \overline{y})$ lies on the regression line $y = 9x + 0.6$. Therefore,

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

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

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

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

$55 + a = 75$

$a = 75 - 55$

$a = 20$

4. Calculate the correlation coefficient:

Given the regression line $y = 9x + 0.6$, we know that the slope $b = 9$.

We have the data points $(1.2, 13), (1.4, 12), (1.6, 14), (1.8, 16), (2, 20)$.

Also, $\overline{x} = 1.6$ and $\overline{y} = \frac{55 + 20}{5} = \frac{75}{5} = 15$.

The correlation coefficient $r$ is given by

$r = \frac{\sum_{i=1}^{n} (x_i - \overline{x})(y_i - \overline{y})}{\sqrt{\sum_{i=1}^{n} (x_i - \overline{x})^2 \sum_{i=1}^{n} (y_i - \overline{y})^2}}$

Calculate the numerator:

$(1.2 - 1.6)(13 - 15) = (-0.4)(-2) = 0.8$

$(1.4 - 1.6)(12 - 15) = (-0.2)(-3) = 0.6$

$(1.6 - 1.6)(14 - 15) = (0)(-1) = 0$

$(1.8 - 1.6)(16 - 15) = (0.2)(1) = 0.2$

$(2 - 1.6)(20 - 15) = (0.4)(5) = 2$

Numerator = $0.8 + 0.6 + 0 + 0.2 + 2 = 3.6$

Calculate the denominator:

$\sum_{i=1}^{n} (x_i - \overline{x})^2 = (-0.4)^2 + (-0.2)^2 + (0)^2 + (0.2)^2 + (0.4)^2 = 0.16 + 0.04 + 0 + 0.04 + 0.16 = 0.4$

$\sum_{i=1}^{n} (y_i - \overline{y})^2 = (-2)^2 + (-3)^2 + (-1)^2 + (1)^2 + (5)^2 = 4 + 9 + 1 + 1 + 25 = 40$

Denominator = $\sqrt{0.4 \times 40} = \sqrt{16} = 4$

$r = \frac{3.6}{4} = 0.9$

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

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

Using the regression line $y = 9x + 0.6$, we have

$y = 9(3.2) + 0.6$

$y = 28.8 + 0.6$

$y = 29.4$

3. Final Answer

1. $\overline{x} = 1.6$

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

3. $a = 20$

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

5. $y = 29.4$