Question

The data of an experiment to supply sufficient quantities of fertilizer to optimize vegetation growth(grass yield) and avoid excessive application that could lead to a runoff and nutrient enrichment of a nearby lake is shown in the data below. determine the regression of y on x.
y - 25- 50- 75- 100- 125- 150- 175- 200- 225- 250.
x - 84- 9- 90- 154- 148- 169- 206- 244- 212- 248.

Answer 1

Answer on Question #62757 – Math – Statistics and Probability

Question

The data of an experiment to supply sufficient quantities of fertilizer to optimize vegetation growth (grass yield) and avoid excessive application that could lead to a runoff and nutrient enrichment of a nearby lake is shown in the data below. Determine the regression of $y$ on $x$.

Solution

We shall derive the linear regression equation, i.e. we must calculate the coefficients $a$ and $b$ in the equation $y = a + bx$. Now we calculate the next values:

$$\bar{x} = \frac{84+9+90+154+148+169+206+244+212+248}{10} = 156.4;$$ $$\bar{y} = \frac{25+50+75+100+125+150+175+200+225+250}{10} = 137.5;$$ $$\overline{xy} = \frac{84\cdot25+9\cdot50+90\cdot75+154\cdot100+148\cdot125+169\cdot150+206\cdot175+244\cdot200+212\cdot225+248\cdot250}{10} = 26310;$$ $$\sigma_{x}^{2} = \frac{(84-156.4)^2+(9-156.4)^2+(90-156.4)^2+(154-156.4)^2+(148-156.4)^2+(169-156.4)^2+(206-156.4)^2+(244-156.4)^2+(212-156.4)^2+(248-156.4)^2}{10} = 5322.84.$$

Now we can obtain the coefficients using the next formulas:

$$b = \frac{\bar{x}\bar{y} - \bar{x}\cdot\bar{y}}{\sigma_{x}^{2}} = 0.90; \quad a = \bar{y} - b\bar{x} = -3.68.$$

The regression equation for the given data is $y = -3.68 + 0.90 \cdot x$.

Answer: $y = -3.68 + 0.90 \cdot x$.

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