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Answer the following on regression and correlation.
From the given data: Variable Mean 6 8 Standard Deviation 4 6 and correlation coefficient: . Find…
From the given data:
and correlation coefficient: $\frac{2}{3}$. Find:
| Variable | $x$ | $y$ |
|---|---|---|
| Mean | 6 | 8 |
| Standard Deviation | 4 | 6 |
(i)[1.3333333333333333]
Regression coefficients $b_{yx}$ and $b_{xy}$
(ii)[1.3333333333333333]
Regression line $x$ on $y$
(iii)[1.3333333333333333]
Most likely value of $x$ when $y = 14$
Answer
Answer (i)
AIRegression coefficient of y on x: 1
Regression coefficient of x on y: 4/9
$b_{yx} = r\frac{\sigma_y}{\sigma_x} = \frac23\times\frac64 = 1$
$b_{xy} = r\frac{\sigma_x}{\sigma_y} = \frac23\times\frac46 = \frac49$
Hence $b_{yx} = 1$ and $b_{xy} = \frac49$.
Answer (ii)
AILine of x on y: x = (4/9)y + 22/9
$x - \bar x = b_{xy}(y - \bar y)$: $x - 6 = \frac49(y - 8)$
$9x - 54 = 4y - 32$
Hence the line of $x$ on $y$ is $9x - 4y - 22 = 0$, i.e. $x = \frac49y + \frac{22}{9}$.
Answer (iii)
AILine of x on y: x = (4/9)y + 22/9
Estimate: x = 26/3 when y = 14
Put $y = 14$ in $x = \frac49y + \frac{22}{9}$: $x = \frac{56}{9} + \frac{22}{9} = \frac{78}{9} = \frac{26}{3}$
Hence the most likely value of $x$ is $\frac{26}{3}\approx 8.67$.
From ISC 2018 Mathematics Paper 1, question 20(b).
Check your working with the Regression lines calculator: both regression lines, b_yx, b_xy and r from data, summary sums or the two lines; with estimates.