Question

In an area of the Midwest, records were kept on the relationship between the

rainfall (in inches) and the yield of wheat (bushels per acre).

Rain (inches)	10.5	8.8	13.4	12.5	18.8	10.3	7.0	15.6	16.0
Yield (bushels/acre)	50.5	46.2	58.8	59.0	82.4	49.2	31.9	76.0	78.8

Which is the residual for y given x = 16.0? Assume that the variables x and y have a significant correlation.

Answer 1

Solution:

	X	Y	XY	X^2	Y^2
	10.5	50.5	530.25	110.25	2550.25
	8.8	46.2	406.56	77.44	2134.44
	13.4	58.8	787.92	179.56	3457.44
	12.5	59	737.5	156.25	3481
	18.8	82.4	1549.12	353.44	6789.76
	10.3	49.2	506.76	106.09	2420.64
	7	31.9	223.3	49	1017.61
	15.6	76	1185.6	243.36	5776
	16	78.8	1260.8	256	6209.44
SUM	112.90	532.80	7187.81	1531.39	33836.58

n = 9

Now ,

Slope of the regression line is

$$ \begin{array}{l} \mathrm{b}=\frac{\mathrm{n} \sum \mathrm{xy}-(\Sigma \mathrm{x})(\Sigma \mathrm{y})}{\mathrm{n} \sum \mathrm{x}^{2}-(\Sigma \mathrm{x})^{2}} \\ \therefore \mathrm{b}=4.3791 \end{array} $$

Now, y intercept of the line is

$$ a=\frac{\sum y-b \Sigma x}{n} $$

$$ \therefore a=4.2668 $$

The equation of the regression line is

\(\hat{y}=a+b x\)

$$ \hat{y}=4.2668+(4.3791) \mathrm{x} $$

For \(x=16.0,\) find the predicted value of \(y\).

Put \(x=16.0\) in the regression line equation.

$$ \therefore \hat{y}=4.2668+\left[(4.3791)^{*} 16.0\right]=74.3324 $$

So , at x = 16.0 , predicted value is 74.3324

But , at x = 16.0 , actual value is 78.8

We know that ,

Residual = Actual value - Predicted value = 78.8 - 74.3324 =  4.4676

Answer :

Residual = 4.4676