Q. 4.101 - Elementary Statistics

Question

Plant Emissions. Following are the data on plant weight and quantity of volatile emissions from Exercise $4.61$

a. Compute $SST,SSR,SSE$

b. Compute the coefficient of determination 

c. Determine the percentage of variation in the observed values of the response variable explained by the regression, and interpret your answer.

d. State how useful the regression equation appears to be for making predictions.

Step-by-Step Solution

Answer

(a) $$\begin{gathered} SST=296.6818 \\ SSR=32.5179 \\ SSE=264.1639 \end{gathered}$$

(b) $0.1096$

(c) $10.96\%$

(d) Utilising the regression equation to generate predictions is useless, as the regression can only explain roughly $11\%$ of the variation.

Part (a) Step 1: Given information

The given data is 

Part (a) Step 2: Explanation

The amount of volatile chemicals emitted and the weight of $11$ potato plants are shown in the table below. The plant's weight is expressed in grammes, while the amount of volatile chemical released is expressed in hundreds of nanograms.

The formulas to calculate the sum of squares is  

$SST=\sum y_i^2-{\left(\sum y_i\right)}^2/n$

$SSR=\frac{{\left[\sum x_i{y}_i-\left(\sum x_i\right)\left(\sum y_i\right)/n\right]}^2}{\sum x_i^2-{\left(\sum x_i\right)}^2/n}$

$SSE=SST-SSR$

As shown in the table below, the relevant sums can be determined. 

$$\begin{gathered} SST=2523.25-\frac{156.5^2}{11} \\ SST=294.6818 \end{gathered}$$

$$\begin{gathered} SSR=\frac{|10486-723\times 156.5÷11{|}^2}{48747-{723}^2÷11} \\ SSR=32.5179 \end{gathered}$$

$$\begin{gathered} SSE=296.6818-32.5179 \\ SSE=264.1639 \end{gathered}$$

Part (b) Step 1: Given information

The given data is 

Part (b) Step 2: Explanation

The coefficient of determination is 

$r^2=\frac{SSR}{SST}$

    $$\begin{gathered} =\frac{32.5179}{296.6818} \\ =0.1096 \end{gathered}$$

Part (c) Step 1: Given information

The given data is

Part (c) Step 2: Explanation

The coefficient of determination restated as a percentage is the percentage of variation: 

$0.1096=10.96\%$

Part (d) Step 1: Given information

The given data is

Part (d) Step 2: Explanation

The regression equation can be used to generate predictions if the estimated $r^2$ is near to 1.

The computed $r^2=0.1096$, which is far from $1$.

As a result, utilising the regression equation to generate predictions is useless, as the regression can only explain roughly $11\%$ of the variation.