Question #138990

The following table shows the amount of converted sugar in a chemical process at different temperatures.
Temperature, x Converted Sugar, y
1.2 8.2
1.4 8.5
1.6 8.4
1.8 9.3
2.0 8.9
2.2 10.5
2.4 9.3


a. Compute for the correlation coefficient
b. Estimate the linear regression line.
c. Estimate the mean amount of sugar produced when the temperature recorded is 1.7.

Expert's answer

Solution

a. Correlation coefficient, r.


r=n∑(xy)−(∑x)(∑y)[n∑x2−(∑x)2][n∑y2−(∑y)2]r={n \sum (xy)- (\sum x)( \sum y) \over \sqrt{[n\sum x^2 - ({\sum x})^2][n\sum y^2 - ({\sum y})^2]}}=(7∗115.14)−(12.6∗63.1)[7∗23.8−(12.6)2][7∗572.49−(63.1)2]={(7*115.14)-(12.6*63.1) \over \sqrt{[7*23.8 - {(12.6)}^2][7*572.49-{(63.1)}^2]}}=10.9214.2277=0.76751={10.92 \over 14.2277} = 0.76751

Ans: 0.76751


b. Linear regression line

y=a+bxy=a+bx


b=n∑(xy)−(∑x)(∑y)[n∑x2−(∑x)2]b= {n \sum (xy)- (\sum x)( \sum y) \over [n\sum x^2 - ({\sum x})^2]}=7∗115.14−(12.6∗63.1)(7∗23.8−(12.6)2=10.927.84={7*115.14 - (12.6*63.1) \over (7*23.8 - {(12.6)}^2} = {10.92 \over 7.84 }=1.3929=1.3929

a=∑y−b∑xna= {\sum y - b\sum x \over n}=63.1−1.3929(12.6)7={63.1 - 1.3929(12.6) \over 7}=6.5071=6.5071

∴y=6.5071+1.3929x\therefore y=6.5071+1.3929x


c. mean amount of sugar when temp. is 1.7


y=6.5071+1.3929(1.7)y=6.5071+1.3929(1.7)=8.87503=8.87503


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