Question

a Wksx 3 2 4 10. Five People are on diet reduction plan. Their results, are captured in the following table 5 6 5 lbsy 4 9 The fust pair of observations (X=3, 9=6) indicates the fust person lost 6 pounds in 3 weeks; the second par (x=2, Y =5) indicates that the second person lost s pounds in 2 weeks and soon, a) Graph the data and show that weeks of Diet and pounds lost have a positive correlatio of 97.69. show that the Regression - Predection - Expectation line is = 1.6+1.8.wks .. Using this equation draw the entire regression picture along with comments . If you diet for 10 wks what is your expected weirlit loss. Is the weight loss experienced at 2 weeks within one standard deviation you expected? d) Show that the second regression /prediction line of lbs versus weeks is Wies - 0.71+ 0.53 lbs Use 10 lbs to get another point ħ alone with the Average point to graph this line. b) Tos c) of what gº

Answer 1

a). We draw the scatter plot of the given data and is presented below:

Scatter Plot of the Data y = 1.8x + 1.6 R=0.9529 12 10 8 Weight Loss 6 4 2 0 0 2 3 4 5 6 Weeks of Diet

We shall calculate the correlation coefficient of data by calculating the squares and cross products.

	X	y	x^2	y^2	x*y
	3	6	9	36	18
	2	5	4	25	10
	1	4	1	16	4
	4	9	16	81	36
	5	11	25	121	55
Total	15	35	55	279	123

Calculate the following:

=1 TS ci η

$\bar{x}=\frac{15}{5}=3$

$\bar{y}=\frac{\sum_{i=1} ^{n}y_{i}}{n}$

35 y 5

η 1 SSxx = Σ? Ti n i=1 i=1

SSxx = 55 - (15)

$SS_{XX}=55-\frac{225}{5}$

$SS_{XX}=55-45$

$SS_{XX}=10$

$SS_{YY}=\sum_{i=1}^{n}y_{i}^{2}-\frac{1}{n}\left ( \sum_{i=1}^{n}y_{i} \right )^{2}$

$SS_{YY}=279-\frac{1}{5}\left (35 \right )^{2}$

$SS_{YY}=279-\frac{1225}{5}$

$SS_{YY}=279-245$

$SS_{YY}=34$

The variance of the weight loss is $s^{2}=\frac{1}{n-1}SS_{YY}$

$s^{2}=\frac{34}{4}=8.5$

  $s=2.9155$

$SS_{XY}=\sum_{i=1}^{n}x_{i}*y_{i}-\frac{1}{n}\left ( \sum_{i=1}^{n}x_{i} \right )\left ( \sum_{i=1}^{n}y_{i} \right )$

SSXY 123 (15) (35)

$SS_{XY}=123-\frac{525}{5}$

$SS_{XY}=123-105$

$SS_{XY}=18$

The correlation coefficient between the weeks of diet and the weight loss is

$r=\frac{SS_{xy}}{\sqrt{SS_{xx}*SS_{yy}}}$

$r=\frac{18}{\sqrt{10*34}}$

$r=\frac{18}{\sqrt{340}}$

$r=\frac{18}{18.4391}$

$r=0.9762$

We can see that the correlation is positive and is 0.9762 or 97.62%.

(b). The regression euation:

The slope is given by  $\hat{\beta }_{1}=\frac{SS_{xy}}{SS_{xx}}$

$\hat{\beta }_{1}=\frac{18}{10}=1.8$

The intercept is given by $\hat{\beta }_{0}=\bar{y}-\hat{\beta }_{1}*\bar{x}$

$\hat{\beta }_{0}=7-1.8*3$

  $\hat{\beta }_{0}=7-5.4=1.6$

Hence the least squares regression of weight loss and the weeks is

$\hat{wl}=1.6+1.8*weeks$

(c). The weight loss for 10 weeks of exercise is

$\hat{wl}=1.6+1.8*10$

$\hat{wl}=1.6+18$

The weight loss for 10 weeks of exercise is   $\hat{wl}=19.6$ lbs.

The weight loss at 2 weeks observed = 5 lbs and the estimated loss is 5.2. 1 standard deviation limit is $7\pm 2.91$ ie (4.09,9.91). Therefore this weight loss is within 1 sd of the loss.

(d). The second regression line of weeks versus weight loss:

The regression coefficient $\hat{\beta }_{1}=\frac{SS_{xy}}{SS_{yy}}$

  $\hat{\beta }_{1}=\frac{18}{34}=0.5294$

The intercept is $\hat{\beta }_{0}=\bar{x}-\hat{\beta }_{1}*\bar{y}$

$\hat{\beta }_{0}=3-0.5294*7$

  $\hat{\beta }_{0}=3-3.7059=-0.7059$

The equation is $\hat{Wks}=-0.71+0.53*wl.$

Now for a 10 lb, weight loss,

  $\hat{Wks}=-0.71+0.53*10=5.3-.071=4.59$.