Answer to Question #218189 in Statistics and Probability for Shivi

Question

https://www.chegg.com/homework-help/questions-and-answers/question-08-subjective-question-hence-write-answer-text-field-given--8-marks-retail-mercha-q66655636

Answer 1

(a)

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} & X & Y & XY & X^2 & Y^2 \\ \hline & 40 & 35 & 1400 & 1600 & 1225 \\ \hdashline & 20 & 40 & 800 & 400 & 1600 \\ \hdashline & 25 & 40 & 1000 & 625 & 1600 \\ \hdashline & 20 & 35 & 700 & 400 & 1225 \\ \hdashline & 30 & 45 & 1350 & 900 & 2025 \\ \hdashline & 50 & 50 & 2500 & 2500 & 2500 \\ \hdashline & 40 & 45 & 1800 & 1600 & 2025 \\ \hdashline & 20 & 40 & 800 & 400 & 1600 \\ \hdashline & 50 & 55 & 2750 & 2500 & 3025 \\ \hdashline & 40 & 55 & 1100 & 1600 & 3025 \\ \hdashline Sum= & 335 & 440 & 15300 & 12525 & 19850 \\ \hdashline \end{array}$$

$$\bar{X}=\dfrac{1}{n}\sum _iX_i=\dfrac{335}{10}=33.5$$

$$\bar{Y}=\dfrac{1}{n}\sum _iY_i=\dfrac{440}{10}=44$$

$$SS_{XX}=\sum _iX_i^2-\dfrac{1}{n}(\sum _iX_i)^2=12525-\dfrac{335^2}{10}$$

$$=1302.5$$

$$SS_{YY}=\sum _iY_i^2-\dfrac{1}{n}(\sum _iY_i)^2=19850-\dfrac{440^2}{10}$$

$$=490$$

$$SS_{XY}=\sum _iX_iY_i-\dfrac{1}{n}(\sum _iX_i)(\sum _iY_i)$$

$$=15300-\dfrac{335(440)}{10}=560$$

$$slope=m=\dfrac{SS_{XY}}{SS_{XX}}=\dfrac{560}{1302.5}=0.42994242$$

$$n=\bar{Y}-m\bar{X}=44-0.42994242(33.5)$$

$$=29.59692893$$

We find that the regression equation is:

$$y=29.596929+0.429942x$$

(b)

$$R^2=\dfrac{(SS_{XY})^2}{SS_{XX}SS_{YY}}=\dfrac{560^2}{1302.5(490)}=0.49136276$$

$$r=\sqrt{R^2}=\sqrt{0.49136276}=0.7010>0.7$$

Strong positive correlation.

(c)

$x=35$

$$y=29.596929+0.429942(35)=45$$

Weekly advertising expenditure 35000 rupees will give 45000 rupees sales.