Question

cquation The actu aur Cnot avcilab ut the folloui よata : 22 너 a) what i b.? b) what is bo

Answer 1

(a) We have $\dpi{80} b_1 = \frac{\sum (X-\bar{X})(Y-\bar{Y})}{\sum (X-\bar{X})^2}$ or $\dpi{80} b_1 = \frac{\sum xy}{\sum x^2}$ , for x be the mean deviation of X values and y be the mean deviation of Y values. Putting the values, we have $\dpi{80} b_1 = \frac{-654}{60}$ or $\dpi{80} b_1 = -10.9$ .

(b) As the mean is not given, neither is the number of observation, we must find the number of observations n.
We have

$\dpi{80} \sum (X - \bar{X})^2 = \sum x^2$

or $\dpi{80} \sum (X^2 - 2X\bar{X} + \bar{X}^2) = 60$

or $\dpi{80} \sum X^2 - 2\bar{X}\sum X + \bar{X}^2\sum 1 = 60$

or $\dpi{80} \sum X^2 - 2\bar{X}\sum X + n\bar{X}^2 = 60$

or $\dpi{80} \sum X^2 - 2(\frac{\sum X}{n})\sum X + n(\frac{\sum X}{n})^2 = 60$

or $\dpi{80} \sum X^2 - \frac{2(\sum X)^2}{n} + \frac{n(\sum X)^2}{n^2} = 60$

or $\dpi{80} \sum X^2 - \frac{2(\sum X)^2}{n} + \frac{(\sum X)^2}{n} = 60$

or $\dpi{80} \sum X^2 - \frac{(\sum X)^2}{n} = 60$

or $\dpi{80} 310 - \frac{(50)^2}{n} = 60$

or $\dpi{80} 310 - 60 = \frac{(50)^2}{n}$

or $\dpi{80} \frac{2500}{n} = 250$

or $\dpi{80} n = 10$ .

The intercept coefficient would be hence $\dpi{80} b_0 = \bar{Y} - b_1 \bar{X}$ or $\dpi{80} b_0 = \frac{\sum Y}{n} - b_1 \frac{\sum X}{n}$ or $\dpi{80} b_0 = \frac{580}{10} - 10.9*\frac{50}{10}$ or $\dpi{80} b_0 = 3.5$ .