Question

The Agricultural Extension Agent's Office has tracked fertilizer application and crop yields for a variety of chickpea and has recorded the data shown in the following table. Their staff statistician developed the regression model and computed the performance statistics displayed below the data.
Fertilizer 10 15 Bushels
10 80
15 115
20 220
25 250
30 225
FERTILIZER APPLICATION VERSUS YEILD Summary
Measure Value
Error Measures
Cumulative forecast error 0
MAD (Me an Absolute Deviation) 28.6
MSE (Mean Squared Error) 933.5
Standard Error (denom=n-23) 39.44
MAPE (Me an Absolute Percent Error) 16
Regression line BUSHELS = 8
+ 8.5* FERTILIZER
Statistics
Correlation coefficient .89
Coefficient of determination (12) .79
1. For every unit of fertilizer applied, the crop yield increases by:_________.
A. 8.0 bushels.
B. 8.9 bushels.
C. 8.5 bushels.
D. 7.9 bushels
2) What percent in the variation of the variable Bushels is explained by the value of the variable Fertilizer?
3) The value of Bushels when Fertilizer is 60 is:__________.
4) The value of Fertilizer required to generate 100 bushels yield must be:___________.
5) If the correlation coefficient were negative, what would also be true?

Answer 1

Answer:

Following are the solution to the given points:

Explanation:

In point 1:

The yield added by the regression equation increases 8.5 times of per each unit of fertilizer.

In point 2:

Definition i.e. $R^2$ describes the percentage of variation. Consequently, the value of Fertilizer variable describes 0.79 percent throughout the variance of the Bushels variable.

In point 3:

At $60 \times 8.5 +8 = 518$, Bushels has an fertilizer of 60.

In point 4:

The fertilizer should be 100 when bushels are:

$\to 100= 8+ Fertilizer \times 8.5 \\\\ \to Fertilizer= 10.82 units$

In point 5:

Increased that amount of fertilizer will reduce the amount of bushels unless the value of determination was negative.