Question

A Question Help O 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 39 inches. Is the result close to the actual weight of 126 pounds? Use a significance level of 0.05. Critical Values of the Pearson Correlation coefficient Х Chest size (inches) 44 41 41 55 51 42 Weight (pounds) 213 206 176 309 300 178 Click the icon to view the critical values of the Pearson correlation coefficient r. value in the table. What is the regression equation? ỹ=O+Nx (Round to one decimal place as needed.) What is the best predicted weight of a bear with a chest size of 39 inches? 7 8 9 10 11 12 13 14 15 16 pounds The best predicted weight for a bear with a chest size of 39 inches is (Round to one decimal place as needed.) Is the result close to the actual weight of 126 pounds? 17 O A. This result is not very close to the actual weight of the bear. O B. This result is exactly the same as the actual weight of the bear. O C. This result is close to the actual weight of the bear. OD. This result is very close to the actual weight of the bear. 18 19 20 25 30 35 40 45 50 60 70 80 90 100 0.754 0.707 0.666 0.632 0.602 0.576 0.553 0.532 0.514 0.497 0.482 0.468 0.456 0.444 0.396 0.361 0.335 0.312 0.294 0.279 0.254 0.236 0.220 0.207 0.196 Q = 0.05 0.875 0.834 0.798 0.765 0.735 0.708 0.684 0.661 0.641 0.623 0.606 0.590 0.575 0.561 0.505 0.463 0.430 0.402 0.378 0.361 0.330 0.305 0.286 0.269 0.256 a = 0.01 Click to select your answer(s). n

Answer 1

This is a regression model. The regression equation which is calculated is used to predict values for the dependent variable for a given value of the indpendent variable.

	X	Y	X2	Y2	XY
1	44	213	1936	45369	9372
2	41	206	1681	42436	8446
3	41	176	1681	30976	7216
4	55	309	3025	95481	16995
5	51	300	2601	90000	15300
6	42	178	1764	31684	7476
Total	274	1382	12688	335946	64805

Regression eq of Y on X

y = a + b 2

slope =

β - η(Στy) - (Σ) (Σ9) ηΣ2 – (Στ)2

=9.6597

   intercept

a = y B.T

= -210.793

Therefore the regression equation is

$\mathbf{\hat{y}=-210.793+9.6597x}$

y: weight x: chest size.

To find the predicted weight when chest size = 39 so we sub x = 39 in the reg equation

$\hat{y}$=-210.793+9.6597x

=-210.793+9.6597 * 39

$\hat{y}$ = 165.9354

Correlation coefficient

Y = η(Στy) - (Στ) (Σ9) η Σ2 - (Σκ) νηΣ 92 – (Σ9)?

r = 0.9634

n = 6

Critical value at 0.05 = 0.754 at 7 so at 6 it would be lower.

Since r > C.v.

we conclude that there is significant relationship.

Residual = Actual - Predicted

= 166 - 126

= 40

We can say that this is not very close to the actual value

Option A

Q2

plot for r

The plot for 'r' can be determined by the direction of the points and how far or close the points are to each other. If the points are close to each then 'r' will be close to -1 or 1. If the points slope upwards then the relation is positive and vice versa. If the points are far away then 'r' is close to 0

r	Plot no	Reason
-0.993	3	negative so downward, close to 1 so points are close to each other.
0.363	5	positve and close to 0
1	2	positve and perfect linear so points from a straight line
0.708	1	positive and moderate
-0.708	4	negative and moderate

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