Answer to Question #130557 in Statistics and Probability for Samuel kassapa

Question

As part of a conservation project, Kojo Nyamekye was asked to measure the circumference of trees that were growing at different distances from the Winneba beach. His results are shown in the table below.

Distance (x)m
6
14
20
25
35
48
46
48
52

Circumference (y)cm
52
57
57
68
65
70
75
80
82

Given Σy2 = 41,700 Σxy = 21, 139

On graph paper, draw a scatter diagram to show Kojo’s results. Use a scale of 1 cm to represent 5m on the x-axis and 1cm to represent 10cm on the y-axis.

Calculate the mean distance, , of the trees from the beach.

Work out the mean circumference, , of the trees.

Plot and label the point M(, ) on your graph.

Find the equation of the regression line y on x, for Akosua’s results.

Draw the regression line y on x on your graph.

Use the equation of the regression line y on x to estimate the circumference of a tree that is 42 m from the beach.

Answer 1

$$\begin{matrix} & x & y & xy & x^2 & y^2 \\ & 6 & 52 & 312 & 36 & 2704 \\ & 14 & 57 & 798 & 196 & 3249 \\ & 20 & 57 & 1140 & 400 & 3249 \\ & 25 & 68 & 1700 & 625 & 4624 \\ & 35 & 65 & 2275 & 1225 & 4225 \\ & 48 & 70 & 3360 & 2304 & 4900 \\ & 46 &75 & 3450 & 2116 & 5625 \\ & 48 & 80 & 3840 & 2304 & 6400 \\ & 52 & 82 & 4264 & 2704 & 6724 \\ Sum= & 294 & 606 & 21139 & 11910 & 41700 \\ \end{matrix}$$

$$\bar{x}={1\over n}\sum_ix_i=\dfrac{294}{9}=\dfrac{98}{3}$$

$$\bar{y}={1\over n}\sum_iy_i=\dfrac{606}{9}=\dfrac{202}{3}$$

$Point \ M\big(\dfrac{98}{3},\dfrac{202}{3}\big)$

$$SS_{xx}=\sum_ix_i^2-{1\over n}(\sum_ix_i)^2=$$

$$=11910-\dfrac{294^2}{9}=2306$$

$$SS_{yy}=\sum_iy_i^2-{1\over n}(\sum_iy_i)^2=$$

$$=41700-\dfrac{606^2}{9}=896$$

$$SS_{xy}=\sum_ix_iy_i-{1\over n}(\sum_ix_i)(\sum_iy_i)=$$

$$=21139-\dfrac{294\cdot606}{9}=1343$$

Therefore, based on the above calculations, the regression coefficients (the slope $m,$

and the y-intercept $n$) are obtained as follows:

$$m=\dfrac{SS_{xy}}{SS_{xx}}=\dfrac{1343}{2306}\approx0.582394$$

$$n=\bar{y}-m\cdot\bar{x}=\dfrac{202}{3}-\dfrac{1343}{2306}\cdot\dfrac{98}{3}=\dfrac{334198}{6918}\approx$$

$$\approx48.308471$$

Therefore, we find that the regression equation is:

$$y=\dfrac{1343}{2306}x+\dfrac{167099}{3459}$$

$$y=0.582394x+48.308471$$

$$y(42)=\dfrac{1343}{2306}\cdot42+\dfrac{167099}{3459}=72.769(m)$$

The circumference of a tree that is 42 m from the beach is 72.769 m.