Answer to Question #161821 in Statistics and Probability for Lemchi Divine

Question #161821

The table below shows the average body weight 'W' and food consumption 'F'


W- 5.1 4.6 5.1. 4.8 4.4 5.9 4.7 5.1

F- 87.1|93.1|89.8|91.4|95.5|92.1|95.5|99.3

W- 5.2 4.9

F- 93.4|94.4


1. Plot the scatter diagram for the data

2. Fit the least squares regression equation of W on F.

3. Using the fitted regression equation, estimate the average body weights 'W' given that food 'F' is 6.0

4. Compute the Spearman Rank Correlation Coefficient and comment on your results.


1
Expert's answer
2021-02-24T06:40:46-0500

1.




2.

"\\bar{x}=\\dfrac{931.6}{10}=93.16, \\bar{y}=\\dfrac{49.8}{10}=4.98"


"SS_{xx}=\\sum x_i^2-n\\bar{x}^2=102.484"

"SS_{yy}=\\sum y_i^2-n\\bar{y}^2=1.536"

"SS_{xy}=\\sum x_iy_i-n\\bar{x}\\bar{y}=-3.088"

"b=\\dfrac{SS_{xy}}{SS_{xx}}=\\dfrac{-3.088}{102.484}=-0.03013"

"a=\\bar{y}-b\\bar{x}=4.98-(-0.03013)(93.16)=7.78705"

"W=a+bF"

"W=7.78705-0.03013F"

3. Given "F=6"


"W(6)=7.78705-0.03013(6)=7.6"

4. Rank the two data sets. Ranking is achieved by giving the ranking '1' to the biggest number in a column, '2' to the second biggest value and so on. The smallest value in the column will get the lowest ranking. This should be done for both sets of measurements.


"\\begin{matrix}\n Rank(x) & Rank(y)\\\\\n 1 & 7\\\\\n5 & 2 \\\\\n2 &7 \\\\\n3 & 4 \\\\\n8.5 & 1 \\\\\n4 & 10 \\\\\n8.5 & 3 \\\\\n10 & 7 \\\\\n6 & 9 \\\\\n7 & 5\n\n\\end{matrix}"

"\\sum d^2=36+9+25+1+56.25+36+30.25"

"+9+9+4=215.5"

"r_s=1-\\dfrac{6\\sum d^2}{n^3-n}=1-\\dfrac{6(215.5)}{10^3-10}=-0.3061"

Or

"SS_{\\bar{x}\\bar{x}}=\\sum Rank(x_i)^2-\\dfrac{1}{n}(\\sum Rank(x_i))^2"

"=384.5-\\dfrac{1}{10}(55)^2=82"

"SS_{\\bar{y}\\bar{y}}=\\sum Rank(y_i)^2-\\dfrac{1}{n}(\\sum Rank(y_i))^2"

"=383-\\dfrac{1}{10}(55)^2=80.5"

"SS_{\\bar{x}\\bar{y}}=\\sum Rank(x_i) Rank(y_i)"

"-\\dfrac{1}{n}(\\sum Rank(x_i))(\\sum Rank(y_i))"

"=276-\\dfrac{1}{10}(55)^2=-26.5"


"r_s=\\dfrac{SS_{\\bar{x}\\bar{y}}}{\\sqrt{SS_{\\bar{x}\\bar{x}}}\\sqrt{S_{\\bar{y}\\bar{y}}}}=\\dfrac{-26.5}{\\sqrt{82}\\sqrt{80.5}}=-0.3262"

Negative correlation. The correlation is too weak to be thought significant.



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