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

Answer to 1^st Question:
The general way of obtaining a regression equation (least square) for two variables is given below –

Where b is the Slope of the regression equation and a is the Y – Intercept

Here, x represents the chest size of the bear (in inches) and y represents the weight (pounds)

The following tables shows all the calculations –

	Chest Size(inches) x	Weight(pounds) y	x^2	xy
	44	213	1936	9372
	41	206	1681	8446
	41	176	1681	7216
	55	309	3025	16995
	51	300	2601	15300
	42	178	1764	7476
Total	274	1382	12688	64805

Total number of observations, n = 6

Mean of x, = 274/6 = 45.667

T

Mean of y, $\bar{y}$ = 1382/6 = 230.333

Variance of x, Var.(x) = ($\sum$(x^2)/n) - (^2) = (12688/6) – (45.667^2) = 29.192

T

Covariance between x and y, Cov.(x, y) = ($\sum$xy / n) - (
T
​​​​​​​$\bar{y}$)  = (64805/6) – (45.667 x 230.333) = 282.216

Therefore, slope b = Cov.(x, y) / Var.(x) = 282.216 / 29.192 = 9.4 (approximately)

Y – Intercept, a = $\bar{y}$ - b = 230.333 – (9.4 x 45.667) = -198.9

T

The regression equation is -

$\hat{y}$ (predicted value) = -198.9 + 9.4x

when x = 39 inches

$\hat{y}$ = -198.9 + (9.4 x 39) = 167.7

Therefore, The best predicted weight for a bear with a chest size of 38 inches is 167.7 pounds

Given the actual weight is 126 pounds

Therefore, the result is not very close to the actual weight of the bear (Option A)

Answer to 2^nd Question:
Answer a:
The correct scatter – plot to the above data is given Option C

Answer b:
The following table shows all the calculations –

	x	y	x^2	y^2	xy
	10	7.47	100	55.8009	74.7
	8	6.78	64	45.9864	54.24
	13	12.73	169	162.0529	165.49
	9	7.12	81	50.6994	64.08
	11	7.82	121	61.1524	86.02
	14	8.84	196	78.1456	123.76
	6	6.08	36	36.9664	36.48
	4	5.38	16	28.9444	21.52
	12	8.15	144	66.4225	97.8
	7	6.41	49	41.0881	44.87
	5	5.73	25	32.8329	28.65
Total	99	82.51	1001	660.0919	797.61

Total number of observations, n = 11

Mean of x, = 99/11 = 9

T

Mean of y, $\bar{y}$ = 82.51/11 = 7.5009

Standard Deviation of x, Sx = {($\sum$(x^2)/n) - (^2)}^0.5 = {(1001/11) – (9^2)}^0.5 = 3.1623

T

Standard Deviation of y, Sy = {($\sum$(y^2)/n) - ($\bar{y}$^2)}^0.5 = {(660.0919/11) – (7.5009^2)}^0.5 = 1.9352

Covariance between x and y, Cov.(x, y) = ($\sum$xy / n) - (
T
$\bar{y}$)  = (797.61/11) – (9 x 7.5009) = 5.0019

Correlation Coefficient, r = Cov.(x, y) / (Sx.Sy) = 5.0019 / (1.9352 x 3.1623) = 0.817

Answer c:
To test for significance of the linear correlation, we can perform a hypothesis test,

The null hypothesis being the two variables have 0 correlation, H0: $\rho$ = 0 against

the alternative hypothesis the two variables are correlated, Ha: $\rho$   ≠ 0

$\rho$ - Population Correlation Coefficient

This is a two – sided test

We assume that the data is a random sample taken from a Bivariate Normal Distribution

The test statistic for this test would be t = [r x (n - 2) ^0.5]/ (1 – (r^2)) ^0.5, the test statistic t follows the t – distribution with (n - 2) degrees of freedom under the null hypothesis

Substituting all values, we get t = 4.250

The Significance Level, $\alpha$ = 5% = 0.05

Since, it is a 2 – sided test, we use the value of ($\alpha$/2) = 0.025 to get the critical value(s)

The critical value, t(0.025, 9) = 2.262

Now, if the value of the test statistic is greater than the critical value, we reject the null hypothesis else we fail to reject the null hypothesis

Here, the value of Test Statistic > The Critical Value hence we reject the null

So, there is sufficient evidence to support the claim of a linear correlation between the two variables (Option B)