Question

https://www.chegg.com/homework-help/questions-and-answers/prob-2-data-collected-mass-kitten-kilograms-age-weeks-5-age-weeks-1-mass-kg-08-11-12-14-15-q36517894

Answer 1

Data were collected on the mass of a kitten, in kilograms, and its age in weeks.

$$\def\arraystretch{1.5} \begin{array}{c: c: c: c:c: c: c: c} Age(weeks) & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline Mass(kg) & 0.8 & 1.1 & 1.2 & 1.4 & 1.5 & 1.5 & 1.7 \end{array}$$

$$n=7,\sum_ix_i=28, \bar{x}=\dfrac{\sum_ix_i}{n}=4$$

$$n=7,\sum_iy_i=9.3, \bar{y}=\dfrac{\sum_ix_i}{n}=\dfrac{9.2}{7}$$

$$S_{xx}=\sum_i(x_i-\bar{x})^2=\sum_ix_i^2-n\bar{x}^2=$$ $$=140-7(4)^2=28$$

$$S_{yy}=\sum_i(y_i-\bar{y})^2=\sum_iy_i^2-n\bar{y}^2=$$ $$=12.64-7(\dfrac{9.2}{7})^2=0.548571$$

$$S_{xy}=\sum_i(x_i-\bar{x})(y_i-\bar{y})=\sum_ix_iy_i-n\bar{x}\bar{y}=$$ $$=40.6-7(4)(\dfrac{9.2}{7})=3.8$$

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

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

$$m=\dfrac{SS_{xy}}{SS_{xx}}=\dfrac{3.8}{28}=0.135714$$

$$n=\bar{y}-m\cdot\bar{x}=\dfrac{9.2}{7}-\dfrac{3.8}{28}\cdot 4=\dfrac{5.4}{7}\approx0.771429$$

Therefore, we find that the regression equation is:

$$y=0.771429+0.135714x$$

$$y(10)=0.771429+0.135714(10)=2.13(kg)$$

$$r=\dfrac{SS_{xy}}{\sqrt{SS_{xx}}\sqrt{SS_{yy}}}$$

$$r=\dfrac{3.8}{\sqrt{28}\sqrt{0.548571}}=0.9696$$

$0.7<r\leq1$ strong correlation