Question

1)

The data show the chest size and weight of several bears. Find the regression equation, letting chest size be the independent (x) variable. Then find the best predicted weight of a bear with a chest size of 58 inches. Is the result close to the actual weight of 632 pounds? Use a significance level of 0.05. Chest size (inches) 46 57 53 41 Weight (pounds) 384 580 542 358 306 320 Click the icon to view the critical values of the Pearson correlation coefficient r. 40 40 What is the regression equation? 9= x (Round to one decimal place as needed.)

2)

Answer 1

x	y	Ꮖ - Ꮖ	y Y	$(x-\bar{x})^2$	$(x-\bar{x})(y-\bar{y})$
46 57 53 41 40 40	384 580 542 358 306 320	-0.1667 10.8333 6.8333 -5.1667 -6.1667 -6.1667	-31 165 127 -57 -109 -95	0.0278 117.3611 46.6944 26.6944 38.0278 38.0278	5.1667 1787.5 867.8333 294.5 672.1667 585.8333
$\sum x = 277$	$\sum y = 2490$			$\sum(x-\bar{x})^2= 266.8333$	$\sum(x-\bar{x})(y-\bar{y})= 4213$

mean $\bar{x} = \frac{\sum{x}}{n}$ = 277/6 =46.1667

mean$\bar{y} = \frac{\sum{y}}{n}$ = 2490/6 =415

Regression Equation = ŷ = bx + a

where b= $\frac{\sum(x-\bar{x})(y-\bar{y})}{\sum(x-\bar{x})^2}$ =$\frac{4213}{266.8333}$ = 15.7888

a = $\bar{y}- b\bar{x}$ = 415 - (15.788*46.17) = -313.92005

Thus regression equation is
ŷ = -313.9 + 15.8 x

(b) when x = 51 inches
   ŷ = -313.9 + 15.8 (51)

= -313.9 +805.8

= 491.9 = 492

the best predicted weight for a bear with chect size 51 inches is 492 pounds.

(c) option (b) the result is not very close to actual weight of bear is correct answer.

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(2)

x	y	Ꮖ - Ꮖ	y Y	$(x-\bar{x})^2$	$(x-\bar{x})(y-\bar{y})$
29 30 29 62 32 33 47 28 64 21 44 53	44 38 39 46 49 49 62 52 38 54 46 32	-10.3333 -9.3333 -10.3333 22.6667 -7.3333 -6.3333 7.6667 -11.3333 24.6667 -18.3333 4.6667 13.6667	-1.75 -7.75 -6.75 0.25 3.25 3.25 16.25 6.25 -7.75 8.25 0.25 -13.75	106.7778 87.1111 106.7778 513.7778 53.7778 40.1111 58.7778 128.4444 608.4444 336.1111 21.7778 186.7778	18.0833 72.3333 69.75 5.6667 -23.8333 -20.5833 124.5833 -70.8333 -191.1667 -151.25 1.1667 -187.9167
$\sum x = 472$	$\sum y = 549$			$\sum(x-\bar{x})^2= 2248.6667$	$\sum(x-\bar{x})(y-\bar{y})= -354$

mean $\bar{x} = \frac{\sum{x}}{n}$ = 472 / 12 = 39.3333

mean$\bar{y} = \frac{\sum{y}}{n}$ = 549 / 12 = 45.75

Regression Equation = ŷ = bx + a

where b= $\frac{\sum(x-\bar{x})(y-\bar{y})}{\sum(x-\bar{x})^2}$ = -354 / 2248.67 = -0.15743

a = $\bar{y}- b\bar{x}$ = 45.75 - (-0.15743*39.33) = 51.94211

Thus regression equation is
ŷ =  51.94211 - 0.15743 x

   ŷ =  51.9 - 0.157 x

(b) when x = 30

      ŷ =  51.9 - 0.157 (30) = 51.9 - 4.71 = 47.19 = 47

(c) no the predited age is 47 which is 5 year more than the actual winner age which is 38 years.