Answer to Question #350612 in Statistics and Probability for joy

Question

 The following data relate to advertising expenditure in. Sh. ten thousands and their corresponding sales (in Shs. millions). Advertising expenditure 10 12 15 23 20 Sales 14 17 23 25 21 i.) Fit a regression line of sales against advertising expenditure? (5mrks) ii) Estimate the sales when the advertising expenditure is of sh. 300,000? (2 marks)

Answer 1

In order to compute the regression coefficients, the following table needs to be used:

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} & X & Y & XY & X^2 & Y^2 \\ \hline & 10 & 14 & 140 & 100 & 196 \\ \hdashline & 12 & 17 & 204 & 144 & 289 \\ \hdashline & 15 & 23 & 345 & 225 & 529 \\ \hdashline & 23 & 25 & 575 & 529 & 625 \\ \hdashline & 20 & 21 & 420 & 400 & 441 \\ \hdashline Sum= & 80 & 100 & 1684 & 1398 & 2080 \\ \hdashline \end{array}$$

$$\bar{X}=\dfrac{1}{n}\sum _{i}X_i=\dfrac{80}{5}=16$$

$$\bar{Y}=\dfrac{1}{n}\sum _{i}Y_i=\dfrac{100}{5}=20$$

$$SS_{XX}=\sum_iX_i^2-\dfrac{1}{n}(\sum _{i}X_i)^2$$ $$=1398-\dfrac{80^2}{5}=118$$

$$SS_{YY}=\sum_iY_i^2-\dfrac{1}{n}(\sum _{i}Y_i)^2$$ $$=2080-\dfrac{(100)^2}{5}=80$$

$$SS_{XY}=\sum_iX_iY_i-\dfrac{1}{n}(\sum _{i}X_i)(\sum _{i}Y_i)$$ $$=1684-\dfrac{80(100)}{5}=84$$

$$r=\dfrac{SS_{XY}}{\sqrt{SS_{XX}SS_{YY}}}=\dfrac{84}{\sqrt{118(80)}}$$

$$=0.8646>0.7$$

Strong positive correlation

$$m=slope=\dfrac{SS_{XY}}{SS_{XX}}=\dfrac{84}{118}=0.7119$$ $$n=\bar{Y}-m\bar{X}=20-\dfrac{84}{118}(16)=8.6102$$

i) The regression equation is:

$$y=8.6102+0.7119x$$

ii)

$x=30$

$$y=8.6102+0.7119(30)$$

$$y=29.9672\ Shs. millions$$