The problem provides a table of observed prices ($Y$) and available quantities ($X$) of a product in a market. $Y$ is in hundreds of CFA francs, and $X$ is the number of products. The problem asks to: 1. Plot the scatter plot of the data.

1. Problem Description

The problem provides a table of observed prices ($Y$) and available quantities ($X$) of a product in a market. $Y$ is in hundreds of CFA francs, and $X$ is the number of products. The problem asks to:

1. Plot the scatter plot of the data.

2. Determine the equation of the regression line of $Y$ on $X$.

3. Calculate the linear correlation coefficient between $X$ and $Y$, and comment on the result.

4. Calculate the coefficient of determination $R^2$, and interpret the result.

5. Predict the price when the available quantity is 5 and

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2. Solution Steps

First, let's extract the data from the table:

$X = [2, 4, 6, 12, 15, 20, 22, 24, 28, 30]$

$Y = [98, 90, 84, 72, 64, 62, 55, 54, 45, 40]$

2. Establish the equation of the regression line of Y on X.

The regression line equation is given by:

$Y = a + bX$

Where:

$b = \frac{n\sum XY - \sum X \sum Y}{n\sum X^2 - (\sum X)^2}$

$a = \bar{Y} - b\bar{X}$

Calculate the required sums:

$n = 10$

$\sum X = 2+4+6+12+15+20+22+24+28+30 = 163$

$\sum Y = 98+90+84+72+64+62+55+54+45+40 = 664$

$\sum X^2 = 2^2+4^2+6^2+12^2+15^2+20^2+22^2+24^2+28^2+30^2 = 3369$

$\sum XY = (2*98)+(4*90)+(6*84)+(12*72)+(15*64)+(20*62)+(22*55)+(24*54)+(28*45)+(30*40) = 9358$

Calculate $\bar{X}$ and $\bar{Y}$:

$\bar{X} = \frac{\sum X}{n} = \frac{163}{10} = 16.3$

$\bar{Y} = \frac{\sum Y}{n} = \frac{664}{10} = 66.4$

Calculate $b$:

$b = \frac{10*9358 - 163*664}{10*3369 - 163^2} = \frac{93580 - 108232}{33690 - 26569} = \frac{-14652}{7121} \approx -2.0576$

Calculate $a$:

$a = 66.4 - (-2.0576)*16.3 = 66.4 + 33.539 = 99.939$

Therefore, the regression line equation is:

$Y = 99.939 - 2.0576X$

3. Calculate the linear correlation coefficient between the variables X and Y.

The correlation coefficient is given by:

$r = \frac{n\sum XY - \sum X \sum Y}{\sqrt{[n\sum X^2 - (\sum X)^2][n\sum Y^2 - (\sum Y)^2]}}$

We already have:

$n = 10$

$\sum X = 163$

$\sum Y = 664$

$\sum X^2 = 3369$

$\sum XY = 9358$

Now we need to calculate $\sum Y^2$:

$\sum Y^2 = 98^2+90^2+84^2+72^2+64^2+62^2+55^2+54^2+45^2+40^2 = 47290$

Calculate $r$:

$r = \frac{10*9358 - 163*664}{\sqrt{[10*3369 - 163^2][10*47290 - 664^2]}} = \frac{-14652}{\sqrt{[7121][472900 - 440896]}} = \frac{-14652}{\sqrt{7121 * 32004}} = \frac{-14652}{\sqrt{227890484}} = \frac{-14652}{15096.04} \approx -0.9706$

The correlation coefficient is approximately -0.

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6. This indicates a strong negative linear correlation between $X$ and $Y$. As $X$ increases, $Y$ tends to decrease significantly.

4. Calculate the coefficient of determination $R^2$.

$R^2 = r^2 = (-0.9706)^2 \approx 0.9421$

The coefficient of determination is approximately 0.

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1. This means that about 94.21% of the variance in $Y$ is explained by the variance in $X$. It confirms a strong linear relationship.

5. What price can be expected if the available quantity is 5, then if the available quantity is 26?

Using the regression equation $Y = 99.939 - 2.0576X$:

If $X = 5$:

$Y = 99.939 - 2.0576 * 5 = 99.939 - 10.288 = 89.651$

If $X = 26$:

$Y = 99.939 - 2.0576 * 26 = 99.939 - 53.4976 = 46.4414$

3. Final Answer

1. Scatter plot: Refer to the data points (X,Y) = (2,98), (4,90), (6,84), (12,72), (15,64), (20,62), (22,55), (24,54), (28,45), (30,40).

2. Regression line equation: $Y = 99.939 - 2.0576X$

3. Linear correlation coefficient: $r \approx -0.9706$. There is a strong negative correlation between X and Y.

4. Coefficient of determination: $R^2 \approx 0.9421$. About 94.21% of the variation in Y is explained by the variation in X.

5. Predicted price when $X = 5$: $Y \approx 89.65$ (hundreds of CFA francs). Predicted price when $X = 26$: $Y \approx 46.44$ (hundreds of CFA francs).