Question 261439: 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 fall, x 10.5 8.8 13.4 12.5 18.8 10.3 7.0 15.6 16.0
Yield, y 50.5 46.2 58.8 59.0 82.4 49.2 31.9 76.0 78.8
Find the equation of the regression line for the given data. Round the line values to the nearest two decimal places.
Answer: 9(7187.77)-(112.9)(532.8)/square root 9 (1521.39)-112.92 square root 9 (33,836.58)-532.8*532.8=453.6/square root 9.46 square root 2.06 =0.962
Using the equation found in part a, predict the bushel yield when the rainfall is 11 inches
Answer: 11(7187.77)-(112.9)(532.8)/square root (1521.39-112.92*112.92 square root 11 (33,836.58-532.8*532.8=189.1/square root 1.28 square root 1.82=1.440
This is what I have come up with.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 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 fall, x 10.5 8.8 13.4 12.5 18.8 10.3 7.0 15.6 16.0
Yield, y 50.5 46.2 58.8 59.0 82.4 49.2 31.9 76.0 78.8
Find the equation of the regression line for the given data. Round the line values to the nearest two decimal places.
Answer: 9(7187.77)-(112.9)(532.8)/square root 9 (1521.39)-112.92 square root 9 (33,836.58)-532.8*532.8=453.6/square root 9.46 square root 2.06 =0.962
Using the equation found in part a, predict the bushel yield when the rainfall is 11 inches
Answer: 11(7187.77)-(112.9)(532.8)/square root (1521.39-112.92*112.92 square root 11 (33,836.58-532.8*532.8=189.1/square root 1.28 square root 1.82=1.440
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Linear Regression Equation:
y = 4.3791x+4.267
r = 0.9808...
Positive linear correlation
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Cheers,
Stan H.
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