Answer to Question #180963 in Statistics and Probability for AYESHA

Question #180963

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


1
Expert's answer
2021-04-15T06:51:03-0400


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 \\\\\n\ns_x = \\sqrt{\\frac{1}{18-1}((41-92.61)^2+...+(168-92.61)^2)} = 47.139 \\\\\n\n\\bar{Y}= \\frac{3+...+9}{18}=5.33 \\\\\n\ns_y = \\sqrt{\\frac{1}{18-1}((3-5.33)^2...(9-5.33)^2)}=2.029 \\\\\n\nr = \\frac{\\sum (X-\\bar{X})(Y-\\bar{Y})}{(n-1)s_xs_y} \\\\\n\n= \\frac{(41-92.61)(3-5.33)+\u2026+(168-92.61)(9-5.33)}{(18-1) \\times 47.139 \\times 2.029} \\\\\n\n= 0.9872"

The parameters of the regression line are

"b = r \\frac{s_y}{s_x} \\\\\n\n= 0.9872 \\frac{2.029}{47.139} \\\\\n\n= 0.0425 \\\\\n\na = \\bar{Y} -b \\bar{X} \\\\\n\n= 5.33 -0.0425 \\times 92.61 \\\\\n\n= 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.


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