Question

Most restaurants offer free refills for their customers who purchase a fountain soft drink. A certain buffet restaurant currently doesn’t. But, it has been argued that customers who drink more soft drinks tend to eat less and, thus, cost the restaurant less for the food the customer eats. To test this argument, this restaurant offered free refills to a sample of 6 of its customers. The total number of glasses each customer drank, X, and the $ cost of the food each ate, Y, were recorded. The findings along with some additional calculations appear in the following table.

       

Customer X Y X-X y-y (x - )? (y-7) 1 2 2 1 10 8 -0.5 - 1.5 4 0 2 0 -2 5 3 6 4 3 0 0.5 - 2.5 (x-7)(y-7) - 1.0 0.0 - 1.0 - 12.5 - 3.0 - 7.5 - 25.0 4 0.25 2.25 0.25 6.25 2.25 6.25 17.50 13 6 25 5 6 4 5 15 1.5 2.5 - 2 - 3 0 4 9 5 48 46

Using the above results, determine the following. (Show your work and highlight your final answers either with a highlighter or by placing a box around it. Calculate all values to 4 decimal places.)

1. The standard deviation of Y

2. The standard deviation of X

3. The correlation coefficient between X and Y

4. The slope of the regression line expressing the linear relationship between cost and number of glasses

5. The y-intercept of this regression line

6. R²

Answer 1

Solution :

a) The standard deviation of X is given by,

1 1 2 _Sr Σ (r; – ) η – 1 i=1

From the given table we have,

η = 6, Σ (ε; – )2 = 17.50 i=1

..Se 17.50 6 – 1

$\large \therefore \mathbf{s_{x} = 1.8708}$

The standard deviation of X is 1.8708.

b) The standard deviation of Y is given by,

n 2 Sy 1 η – 1 Σ(9: – 9) i=1

From the given table we have,

$\large n = 6,\ \ \sum_{i=1}^{n}\left ( y_{i}-\bar{y} \right )^{2} = 46$

$\large \therefore s_{y} = \sqrt{\frac{46}{6-1}}$

$\large \therefore \mathbf{s_{y} = 3.0332}$

The standard deviation of Y is 3.0332.

c) The correlation coefficient between X and Y is given as follows :

$\large r = \frac{\sum_{i=1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})}{\sqrt{\sum_{i=1}^{n}(x_{i}-\bar{x})^{2}}\sqrt{\sum_{i=1}^{n}(y_{i}-\bar{y})^{2}} }$

From the given table we have,

$\large \sum_{i=1}^{n}\left ( x_{i}-\bar{x} \right )^{2} = 17.50,\ \ \sum_{i=1}^{n}\left (y_{i}-\bar{y} \right )^{2} = 46\ \ and$

$\large \sum_{i=1}^{n}\left ( x_{i}-\bar{x} \right )\left (y_{i}-\bar{y} \right ) = -25.0$

$\large \therefore r = \frac{-25.0}{(\sqrt{17.50})(\sqrt{46})}$

$\large \therefore r = -0.8811$

The correlation coefficient between X and Y is -0.8811.

d) Let y = a + bx be the regression equation expressing the linear relationship between cost and number of glasses.

According to principle of least square, the estimate of the slope of regression line (b) is given as follows :

$\large b = \frac{\sum_{i=1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})}{\sum_{i=1}^{n}(x_{i}-\bar{x})^{2}}$

From the given table we have,

$\large \sum_{i=1}^{n}\left ( x_{i}-\bar{x} \right )\left (y_{i}-\bar{y} \right ) = -25.0\ \ \&\ \ \sum_{i=1}^{n}\left ( x_{i}-\bar{x} \right )^{2}=17.50$

$\large \therefore b = \frac{-25.0}{17.50} = -1.4286$

The slope of the regression line expressing the linear relationship between cost and number of glasses is -1.4286.

e) The y-intercept of regression line is given as follows :

$\large a = \bar{y}-b\bar{x}$

$\large Where,\ \bar{x} = \frac{1}{n}\sum_{i=1}^{n}x_{i},\ \ \bar{y} = \frac{1}{n}\sum_{i=1}^{n}y_{i}$

From the given table we have,

$\large n = 6,\ \ \sum_{i=1}^{n}x_{i} = 15,\ \ \sum_{i=1}^{n}y_{i} = 48$

$\large \bar{x} = \frac{15}{6} = 2.5\ \ \&\ \ \bar{y} = \frac{48}{6} = 8$

Also we have, b = -1.4286

$\large \therefore a = 8 -(-1.4286\times2.5)$

$\large \therefore a = 11.5714$

The y-intercept of the regression line is 11.5714.

f) We have, R = -0.8811

Hence, R² = (-0.8811)² = 0.7764

R² = 0.7764

Please rate the answer. Thank you.