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

Question

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.

Answer 1

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.

$\begin{matrix} Rank(x) & Rank(y)\\ 1 & 7\\ 5 & 2 \\ 2 &7 \\ 3 & 4 \\ 8.5 & 1 \\ 4 & 10 \\ 8.5 & 3 \\ 10 & 7 \\ 6 & 9 \\ 7 & 5 \end{matrix}$

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

$+9+9+4=215.5$

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

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.