Answer to Question #236472 in Statistics and Probability for XL Jacobs

Question #236472

Suppose that you are assigned a task to investigate the relationship between selling price and valuation of plots sold by a local municipality. Data was obtained for a random sample of ten plots. Plot Selling price (N$ 000) Valuation (N$ 000) 1 120 72 2 100 68 3 140 72 4 150 70 5 155 75 6 100 50 7 150 58 8 200 90 9 80 56 10 145 70 a) Discuss the difference between correlation analysis and regression analysis. (10) b) Use the method of least squares and estimate the regression equation between selling price and valuation. (15) c) Provide an interpretation for the slope coefficient. (5) d) Use the estimated regression equation and make a prediction for a value of the dependent variable when the independent variable is N$ 85 000. (5) e) Calculate and interpret the coefficient of determination. (10) f) Calculate and interpret the correlation coefficient. 


1
Expert's answer
2021-09-14T06:08:58-0400

a.

Correlation is a statistical tool which determines the association of two quantities whereas Regression describes how an independent variable is related numerically to the dependent variable.

Correlation is used to represent the linear relationship between two variables but regression is used to fit the best line and estimate one variable on the basis of another variable.

Correlation shows the strength of association between variables opposed to regression which mirrors the impact of the unit change in the independent variable on the dependent variable.

The main objective of correlation is to find a numerical value that expresses the relationship between variables. Unlike regression whose aim is to forecast values of the random variable on the basis of the values of the fixed variable.

There is no difference between dependent and independent variables in correlation. For example, correlation between X and Y is similar to Y and X. However, regression of the dependent variable on the independent variable differs with regression from the independent variable on the dependent variable.

b.

In this case the selling price is the dependent variable(y) and valuation is the independent variable(x).

Using the method of least squares, we estimate the regression equation given by

"y(hat)=(a*x)+ b" , where a is the slope coefficient and b the intercept.

In order to find this equation, we determine the slope coefficient and the intercept using the following formula.

"a=(n*summation(xy)-(summation(x)*summation(y)))\/(n*summation(x^2)-(summation(x))^2)"

"b=mean(y)-a*mean(x)"

From the data provided,


"summation(x)=681\nsummation(y)=1340\nsummation(xy)=93975\nsummation(x^2)=47517\nsummation(y^2)=190450\nmean(y)=summation(y)\/10=134\nmean(x)=summation(x)\/10=68.1"

therefore,

"a=((10*93975)-(1340*681))\/((10*47517)-(463761))\n =2.385(3 decimal places)\n\nb=134-(68.1*2.385)\n=-28.416(3 decimal places)"

the estimated regression equation is

"y(hat)=2.385x-28.416"


c.

the slope coefficient of 2.385 shows that a unit increase in valuation(x) results to an estimated increase of 2.385 N$ in selling price(y).

d.

When the independent variable(x) is N$ 85000, the predicted value of the dependent variable is found by substituting this value in the regression equation,

=(2.385*85000)-28.416

=N$ 202693.12


e.

The coefficient of determination is found using the following formula

"R^2=1-SSE\/SST\nwhere SSE=summation(y^2)-b*summation(y)-a*summation(xy)\nand \nSST=summation(y^2)-(summation(y))^2\/10"

SSE=190450-(1340*-28.416) -2.385*93975

=4400.526

SST=190450-(179560/10)

=10890

therefore,

"R^2=1-(4400.526\/1890)\n=0.5959" which is approximately equal to 0.60

interpretation

multiplying by 100, we get 60%

this means that the values of x account for 60% of variability observed in y.


f.

The correlation coefficient is obtained using the formula below

"r=(10*summation(xy)-summation(x)*summaton(y))\/squareroot((10*summation(x^2)-(summation(x))^2)*(10*summation(y^2)-(summation(y))^2)"

=(10*93975-1340*681)/"squareroot" ((475170-463761)*(1904500-1795600))

=0.77(2 decimal places)

this value shows that there is a significant and positive relationship between selling price and valuation.



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