Question

Modern medical practice tells us not to encourage babies to become too fat. Is there a positive correlation between the weight x of a 1-year old baby and the weight y of the mature adult (30 years old)? A random sample of medical files produced the following information for 14 females. x (lb) 23 23 21 26 20 15 25 21 17 24 26 22 18 19 y (lb) 127 122 119 125 130 120 145 130 130 130 130 140 110 115 In this setting we have Σx = 300, Σy = 1773, Σx2 = 6576, Σy2 = 225,649, and Σxy = 38,186.

Answer 1

Answer:

a) Figure attached

b) $y=1.31 x +98.57$

c) The correlation coefficient would be r =0.47719

d) $y=1.31 x +98.57=1.31*21 + 98.57 =126.08$

Step-by-step explanation:

(a) Draw a scatter diagram for the data.

See the figure attached

(b) Find x, y, b, and the equation of the least-squares line. (Round your answers to three decimal places.) x =__ y =__ b =__ y =__ + __x

$m=\frac{S_{xy}}{S_{xx}}$

Where:

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}$

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}$

With these we can find the sums:

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=6576-\frac{300^2}{14}=147.429$

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i){n}}=38186-\frac{300*1773}{14}=193.143$

And the slope would be:

$m=\frac{193.143}{147.429}=1.31$

Nowe we can find the means for x and y like this:

$\bar x= \frac{\sum x_i}{n}=\frac{300}{14}=21.429$

$\bar y= \frac{\sum y_i}{n}=\frac{1773}{14}=126.643$

And we can find the intercept using this:

$b=\bar y -m \bar x=126.643-(1.31*21.429)=98.571$

So the line would be given by:

$y=1.31 x +98.57$

(c) Find the sample correlation coefficient r and the coefficient of determination r?2. (Round your answers to three decimal places.)

n=14 $\sum x = 300, \sum y = 1773, \sum xy=38186, \sum x^2 =6576, \sum y^2 =225649$  

And in order to calculate the correlation coefficient we can use this formula:

$r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}$

$r=\frac{14(38186)-(300)(1773)}{\sqrt{[14(6576) -(300)^2][14(225649) -(1773)^2]}}=0.9534$

So then the correlation coefficient would be r =0.47719

What percentage of variation in y is explained by the least-squares model? (Round your answer to one decimal place.)

The % of variation is given by the determination coefficient given by $r^2$ and on this case $0.47719^2 =0.2277$, so then the % of variation explaines is 22.8%.

(d) If a female baby weighs 21 pounds at 1 year, what do you predict she will weigh at 30 years of age? (Round your answer to two decimal places.) ___ lb

So we can replace in the linear model like this:

$y=1.31 x +98.57=1.31*21 + 98.57 =126.08$