Plot Selling price (N$ 000) Valuation (N$ 000)

1 120 72

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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

a.

Correlation is a statistical measure which determines the co-relationship 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. On the contrary, regression is used to fit the best line and estimate one variable on the basis of another variable.

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

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

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

b.

In this case, plot selling price is the dependent variable(Y) while valuation is the independent variable(X).

Using the method of least squares requires an equation of the form “\hat{y}=m*x+n”, where “m” is the slope coefficient and “n” is the y-intercept to be determined. To obtain this equation, values of “m” and “n” need to be found using the formulas below.

“m=(n*\displaystyle\sum(xy)-\displaystyle\sum(x)*\displaystyle\sum(y))/(n*\displaystyle\sum(x^2)-(\displaystyle\sum(x))^2)”

“n=\bar{y}-m*\bar{x}”

where n=10, “\bar{y}=\displaystyle\sum(y)/n” and “\bar{x}=\displaystyle\sum(x)/n”

Data above can be summarized as follows.

“\displaystyle\sum(xy)=93975\displaystyle\space,n\displaystyle\sum(y)=1340\displaystyle\space,n\displaystyle\sum(x)=681\displaystyle\space,n\displaystyle\sum(x^2)=47517\displaystyle\space.”

“\bar{y}=1340/10=134\displaystyle\space,n\bar{x}=681/10=68.1\displaystyle\space.”

Therefore,

“m=(10*(93975)-(1340)*(681))/(10*(47517)-681^2)=2.385(3\displaystyle\space decimal\displaystyle\space places)”

“n=134-(2.385*68.1)=-28.416(3\displaystyle\space decimal\displaystyle\space places)”

Hence the regression equation is given by,

“\hat{y}=2.385x-28.416”

c.

A slope coefficient of 2.385 represents the estimated increase in plot selling price(Y) for every one unit increase in valuation(X).

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