Question #79272

1. Given the data,
x 1 2 3 4 5 6 7 8 9
y 9 8 10 12 11 13 14 16 15
(a) Calculate the coefficient of correlation?
(b) Obtain the line of regression.
(c) Estimate the value of y which should correspond to x = 6.2
1

Expert's answer

2018-07-23T14:10:01-0400

Answer on Question #79272 – Math – Statistics and Probability

Question

Given the data:

x 1 2 3 4 5 6 7 8 9

y 9 8 10 12 11 13 14 16 15

(a) Calculate the coefficient of correlation

(b) Obtain the line of regression

(c) Estimate the value of yy which should correspond to x=6.2x = 6.2

Solution

(a)

A correlation coefficient is given by a formula:


rxy=i=1N(xixˉ)(yiyˉ)i=1N(xixˉ)2i=1N(yiyˉ)2,r_{xy} = \frac{\sum_{i=1}^{N} (x_i - \bar{x})(y_i - \bar{y})}{\sqrt{\sum_{i=1}^{N} (x_i - \bar{x})^2 \sum_{i=1}^{N} (y_i - \bar{y})^2}},


where xx and yˉ\bar{y} are the sample means of xx and yy

Calculate the sample means (N=9):


xˉ=1Ni=1Nxi=5\bar{x} = \frac{1}{N} \sum_{i=1}^{N} x_i = 5yˉ=1Ni=1Nyi=12\bar{y} = \frac{1}{N} \sum_{i=1}^{N} y_i = 12


Thus, the correlation coefficient rxy=0.95r_{xy} = 0.95.

(b)

Consider the simple linear regression model and find the linear function y=Ax+By = Ax + B that best fits the data. Use the method of least squares to find optimal values of parameters A and B.

It is necessary to minimize the sum:


S=i=1N(yi(Axi+B))2S = \sum_{i=1}^{N} (y_i - (A x_i + B))^2


The minimum of the sum can be found by setting the partial derivatives to zero.


SA=2i=1Nxi(yiAxiB)=2N(xˉyˉAxˉ2Bxˉ)=0\frac{\partial S}{\partial A} = -2 \sum_{i=1}^{N} x_i (y_i - A x_i - B) = -2 N (\bar{x} \bar{y} - A \bar{x}^2 - B \bar{x}) = 0SB=2i=1N(yiAxiB)=2N(yˉAxˉB)=0,\frac{\partial S}{\partial B} = -2 \sum_{i=1}^{N} (y_i - A x_i - B) = -2 N (\bar{y} - A \bar{x} - B) = 0,


where xˉyˉ=1Ni=1Nxiyi=1993\bar{x} \bar{y} = \frac{1}{N} \sum_{i=1}^{N} x_i y_i = \frac{199}{3}, xˉ2=1Ni=1Nxi2=953\bar{x}^2 = \frac{1}{N} \sum_{i=1}^{N} x_i^2 = \frac{95}{3}.

Thus, we receive the system of equations:


{Ax2Bxˉ+xy=0AxˉB+yˉ=0\left\{ \begin{array}{l} - A \overline {{x ^ {2}}} - B \bar {x} + \overline {{x y}} = 0 \\ - A \bar {x} - B + \bar {y} = 0 \end{array} \right.


Solving this system, we obtain the values of the parameters A and B:


A=xyxˉyˉx2xˉ=0.95A = \frac {\overline {{x y}} - \bar {x} \bar {y}}{\overline {{x ^ {2}}} - \bar {x}} = 0. 9 5B=x2yˉxˉxyˉx2xˉ=7,25B = \frac {\overline {{x ^ {2}}} \bar {y} - \bar {x} \bar {x y}}{\overline {{x ^ {2}}} - \bar {x}} = 7, 2 5


The equation of the regression line is y=0.95x+7.25y = 0.95x + 7.25

(c)

To estimate the value of yy , which correspond to x=6.5x = 6.5 we should substitute the value of xx into the equation of the line.


y=0.95×6.5+7.25=13.425y = 0. 9 5 \times 6. 5 + 7. 2 5 = 1 3. 4 2 5

Comment

The data used in the calculations are given in table.



Answer:

(a) rxy=0.95r_{xy} = 0.95

(b) y=0.95x+7.25y = 0.95x + 7.25

(c) x=6.5,y=13.425x = 6.5, y = 13.425

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