We are given data on the family size and the amount spent on food for ten families. We need to compute the correlation coefficient between these two variables and interpret the result. The family sizes are 3, 6, 5, 6, 6, 3, 4, 4, 5, 3. The corresponding amounts spent on food (GHS) are 99, 104, 151, 129, 142, 111, 74, 91, 119, 91.

1. Problem Description

We are given data on the family size and the amount spent on food for ten families. We need to compute the correlation coefficient between these two variables and interpret the result. The family sizes are 3, 6, 5, 6, 6, 3, 4, 4, 5,

3. The corresponding amounts spent on food (GHS) are 99, 104, 151, 129, 142, 111, 74, 91, 119,

9

1.

2. Solution Steps

Let $x$ be the family size and $y$ be the amount spent on food. We have $n = 10$ data points.

First, calculate the means of $x$ and $y$:

$\bar{x} = \frac{3+6+5+6+6+3+4+4+5+3}{10} = \frac{45}{10} = 4.5$

$\bar{y} = \frac{99+104+151+129+142+111+74+91+119+91}{10} = \frac{1111}{10} = 111.1$

Next, we calculate the standard deviations of $x$ and $y$:

$s_x = \sqrt{\frac{\sum_{i=1}^{n}(x_i - \bar{x})^2}{n-1}}$

$s_y = \sqrt{\frac{\sum_{i=1}^{n}(y_i - \bar{y})^2}{n-1}}$

To calculate $s_x$, we first calculate $(x_i - \bar{x})^2$ for each $x_i$:

$(3-4.5)^2 = 2.25$

$(6-4.5)^2 = 2.25$

$(5-4.5)^2 = 0.25$

$(6-4.5)^2 = 2.25$

$(6-4.5)^2 = 2.25$

$(3-4.5)^2 = 2.25$

$(4-4.5)^2 = 0.25$

$(4-4.5)^2 = 0.25$

$(5-4.5)^2 = 0.25$

$(3-4.5)^2 = 2.25$

$\sum_{i=1}^{10}(x_i - \bar{x})^2 = 2.25*4 + 0.25*4 + 2.25*2 = 9 + 1 + 4.5 = 14$

$s_x = \sqrt{\frac{14}{10-1}} = \sqrt{\frac{14}{9}} \approx 1.247$

To calculate $s_y$, we first calculate $(y_i - \bar{y})^2$ for each $y_i$:

$(99-111.1)^2 = (-12.1)^2 = 146.41$

$(104-111.1)^2 = (-7.1)^2 = 50.41$

$(151-111.1)^2 = (39.9)^2 = 1592.01$

$(129-111.1)^2 = (17.9)^2 = 320.41$

$(142-111.1)^2 = (30.9)^2 = 954.81$

$(111-111.1)^2 = (-0.1)^2 = 0.01$

$(74-111.1)^2 = (-37.1)^2 = 1376.41$

$(91-111.1)^2 = (-20.1)^2 = 404.01$

$(119-111.1)^2 = (7.9)^2 = 62.41$

$(91-111.1)^2 = (-20.1)^2 = 404.01$

$\sum_{i=1}^{10}(y_i - \bar{y})^2 = 146.41+50.41+1592.01+320.41+954.81+0.01+1376.41+404.01+62.41+404.01 = 5210.9$

$s_y = \sqrt{\frac{5210.9}{10-1}} = \sqrt{\frac{5210.9}{9}} \approx \sqrt{578.9889} \approx 24.062$

Now, we calculate the Pearson correlation coefficient $r$:

$r = \frac{\sum_{i=1}^{n}(x_i - \bar{x})(y_i - \bar{y})}{(n-1)s_x s_y}$

We need to calculate $(x_i - \bar{x})(y_i - \bar{y})$ for each data point:

$(3-4.5)(99-111.1) = (-1.5)(-12.1) = 18.15$

$(6-4.5)(104-111.1) = (1.5)(-7.1) = -10.65$

$(5-4.5)(151-111.1) = (0.5)(39.9) = 19.95$

$(6-4.5)(129-111.1) = (1.5)(17.9) = 26.85$

$(6-4.5)(142-111.1) = (1.5)(30.9) = 46.35$

$(3-4.5)(111-111.1) = (-1.5)(-0.1) = 0.15$

$(4-4.5)(74-111.1) = (-0.5)(-37.1) = 18.55$

$(4-4.5)(91-111.1) = (-0.5)(-20.1) = 10.05$

$(5-4.5)(119-111.1) = (0.5)(7.9) = 3.95$

$(3-4.5)(91-111.1) = (-1.5)(-20.1) = 30.15$

$\sum_{i=1}^{10}(x_i - \bar{x})(y_i - \bar{y}) = 18.15 - 10.65 + 19.95 + 26.85 + 46.35 + 0.15 + 18.55 + 10.05 + 3.95 + 30.15 = 163.5$

$r = \frac{163.5}{(10-1)s_x s_y} = \frac{163.5}{9 * 1.247 * 24.062} \approx \frac{163.5}{269.44} \approx 0.607$

Interpretation: The correlation coefficient is approximately 0.

6

0

7. Since it is a positive value, it indicates a positive correlation between family size and amount spent on food. This means that as family size increases, the amount spent on food tends to increase as well. The value of 0.607 suggests a moderate positive correlation.

3. Final Answer

The correlation coefficient is approximately 0.

6

0

7. This indicates a moderate positive correlation between family size and the amount spent on food, meaning that larger families tend to spend more on food.