Find the regression equation, letting overhead width be the predictor (x) variable. Find the best predicted weight of a
Posted: Wed May 04, 2022 11:54 am
Find the regression equation, letting overhead width be the predictor (x) variable. Find the best predicted weight of a seal if the overhead width measured from a photograph is 2.4 cm. Can the prediction be correct? What is wrong with predicting the weight in this case? Use a significance level of 0.05. Overhead Width (cm) 7.6 9.5 D 7.5 157 8.3 223 9.6 273 8.6 203 X Weight (kg) 182 271 Critical Values of the Pearson Correlation Coefficient r Click the icon to view the critical values of the Pearson correlation coefficient r Critical Values of the Pearson Correlation Coefficient r n x=0.05 x=0.01 0.950 0.990 The regression equation is y=+x. NOTE: To test Ho: p=0 against H₁: p= 0, reject Ho if the absolute value of ris greater than the critical 0.878 0.959 (Round to one decimal place as needed.) 0.811 0.917 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 0.254 0.330 0.236 0.305 0.220 0.286 0.207 0.269 0.196 0.256 x=0.05 α = 0.01 4 15 16 7 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
The regression equation is y=-215.9 + 49.4 x. (Round to one decimal place as needed.) The best predicted weight for an overhead width of 2.2 cm is - 107.2 kg. (Round to one decimal place as needed.) Can the prediction be correct? What is wrong with predicting the weight in this case? A. The prediction cannot be correct because there is not sufficient evidence of a linear correlation. The width in this case is beyond the scope of the available sample data. B. The prediction cannot be correct because a negative weight does not make sense. The width in this case is beyond the scope of the available sample data. C. The prediction cannot be correct because a negative weight does not make sense and because there is not sufficient evidence of a linear correlation. D. The prediction can be correct. There is nothing wrong with predicting the weight in this case.
The regression equation is y=-215.9 + 49.4 x. (Round to one decimal place as needed.) The best predicted weight for an overhead width of 2.2 cm is - 107.2 kg. (Round to one decimal place as needed.) Can the prediction be correct? What is wrong with predicting the weight in this case? A. The prediction cannot be correct because there is not sufficient evidence of a linear correlation. The width in this case is beyond the scope of the available sample data. B. The prediction cannot be correct because a negative weight does not make sense. The width in this case is beyond the scope of the available sample data. C. The prediction cannot be correct because a negative weight does not make sense and because there is not sufficient evidence of a linear correlation. D. The prediction can be correct. There is nothing wrong with predicting the weight in this case.