Question

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 44 inches. Is the result close to the actual weight of 213 pounds? Use a significance level of 0.05. Chest size (inches) 50 41 45 52 45 45 Weight (pounds) 321 221 265 335 307 265 Critical Values of the Pearson Correlation coefficient Х 5 Click the icon to view the critical values of the Pearson correlation coefficient r. n 4 What is the regression equation? ŷ=0+ Ox (Round to one decimal place as needed.) What is the best predicted weight of a bear with a chest size of 44 inches? The best predicted weight for a bear with a chest size of 44 inches is pounds. (Round to one decimal place as needed.) Is the result close to the actual weight of 213 pounds? 6 O A. This result is close to the actual weight of the bear. O B. This result is very close to the actual weight of the bear. OC. This result is not very close to the actual weight of the bear. OD. This result is exactly the same as the actual weight of the bear 8 9 10 11 12 13 14 15 16 17 18 19 20 25 30 35 40 45 50 60 70 80 90 100 n Critical Values of the Pearson Correlation Coefficient Q = 0.05 a = 0.01 NOTE: To test Ho: p=0 0.950 0.990 against H, p70, reject Ho 0.878 0.959 if the absolute value of r is 0.811 0.917 Igreater than the critical 0.754 0.875 value in the table. 0.707 0.834 0.666 0.798 0.632 0.765 0.602 0.735 0.576 0.708 0.553 0.684 0.532 0.661 0.514 0.641 0.497 0.623 0.482 0.606 0.468 0.590 0.456 0.575 0.444 0.561 0.396 0.505 0.361 0.463 0.335 0.430 0.312 0.402 0.294 0.378 0.279 0.361 0.330 0.236 0.305 0.220 0.286 0.207 0.269 0.196 0.256 Q=0.05 a = 0.01 0.254

Answer 1

solation - of several bearl :- The data (show the chest Gite and weight Rezuired information i@given bog, 59 50 41 005 45 (15 x » Y = 3.91 021 265 365 Bot 265 - we have to draw the tabular form by asing given data is, х Y 42 KY 50 321 9500 103041 16050 gel 160) CEB841 9061 41 2005 265 toas5 I 1995 45 52 0335 Hoy 17420 (1 2295 45 13815 Bo7 2005 quocuq L995 45 70995 965 2025 Exe 210 Y = 14 [x=12960 y?- U9880 EXY = BoLab 6 Calculate Find the Regression - Ecaltion Line: we have to obtained the Regression Equation line by using given data is Elere, Regressions Ezciation is Y = atbf

ΣΥ- IXTY cohere, ba Real a Ex² (Ext > 6.4 Bolab 60-18-41414) (6 V12960) - (218) 466192 401176 67760 1-1284) 4689 4116 b 9.840 -> a= [y- b Ex #161–69.8407 278 ) 6 CHILI - 9785.59) 6 -1021.52 6 -170.253 From En ②, Weget Ý : -110.253 +9.8410 X Therefore Regresion Ezciation is 170.0531+ 840 x

3 o what is the best predicted colight at x=(160:- he have to obtained the best pre best predicted height of a bear ak x=44 is here, - from Ech we get -170.053 + 9.8410 X X=44 =-170.233+ [q.QuoV(14) =E170-993)+ 402.96] 269.7 ro Therefore predicted value is 1969. as the result close to the actual weight:- he hale to obtained rebult close to the official Cueight at Stais Eer, close to the actual weight of the bear. Arhis result is not very Therefore, -> option © il comiect answer