Answer to Question #180963 in Statistics and Probability for AYESHA

Question

How much does ten pounds of weight change the estimated number of cups of dog

food consumed? Dog 1 2 3 4 5 6 7 8 9

Weight 0.41 1.48 0.79 0.41 0.85 1.11 0.37 1.11 0.41

Consumption in cups 3 8 5 4 5 6 3 6 3

Dog 10 11 12 13 14 15 16 17 18

Weight 0.91 1.09 2.07 0.49 1.13 0.84 0.95 0.57 1.68

Consumption in cups 5 6 10 3 6 5 5 4 9

Dog 1 2 3 4 5 6 7 8 9

Weight 0.41 1.48 0.79 0.41 0.85 1.11 0.37 1.11 0.41

Consumption in cups 3 8 5 4 5 6 3 6 3

Dog 10 11 12 13 14 15 16 17 18

Weight 0.91 1.09 2.07 0.49 1.13 0.84 0.95 0.57 1.68

Consumption in cups 5 6 10 3 6 5 5 4 9

Answer 1

We denote the weight as X and the consumption as Y. The means and standard deviations are

$\bar{X} = \frac{41+...+168}{18}=92.61 \\ s_x = \sqrt{\frac{1}{18-1}((41-92.61)^2+...+(168-92.61)^2)} = 47.139 \\ \bar{Y}= \frac{3+...+9}{18}=5.33 \\ s_y = \sqrt{\frac{1}{18-1}((3-5.33)^2...(9-5.33)^2)}=2.029 \\ r = \frac{\sum (X-\bar{X})(Y-\bar{Y})}{(n-1)s_xs_y} \\ = \frac{(41-92.61)(3-5.33)+…+(168-92.61)(9-5.33)}{(18-1) \times 47.139 \times 2.029} \\ = 0.9872$

The parameters of the regression line are

$b = r \frac{s_y}{s_x} \\ = 0.9872 \frac{2.029}{47.139} \\ = 0.0425 \\ a = \bar{Y} -b \bar{X} \\ = 5.33 -0.0425 \times 92.61 \\ = 1.3973$

The regression equation is Y = 1.3973 + 0.0425X

If the weight increases by 10 pounds the consumption increases by 0.425 cups.