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 2.3 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) 8.3 7.8 8.6 9.2 9.3 9.6 Weight (kg) 199 207 238 237 263 284 9 Click the icon to view the critical values of the Pearson correlation coefficient r. - X Critical Values of the Pearson Correlation Coefficient r %3D The regression equation is y =.x. (Round to one decimal place as needed.) Critical Values of the Pearson Correlation Coefficient r a-0.05 0.950 0.878 0.811 0.754 NOTE: To test Ho p=0 against H, p0, reject He Jf the absolute value of r is greater than the critical value in the table. In - 0.01 The best predicted weight for an overhead width of 2.3 cm is kg. (Round to one decimal place as needed.) 0.990 0.959 14 15 p.959 0.917 0.875 0.834 Can the prediction be correct? What is wrong with predicting the weight in this case? 17 18 19 10 11 12 13 14 15 16 17 18 19 20 25 30 35 40 0.707 0.666 0.632 l0,602 O A. The prediction cannot be correct because a negative weight does not make sense and because there is not sufficient evidence of a linear correlation. 0.798 0.765 0.735 0.708 0.684 0.661 0 641 0.623 0.606 l0.590 l0 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 OB. 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. OC. 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. OD. The prediction can be correct. There is nothing wrong with predicting the weight in this case. 0.576 0.553 0.532 0.514 0.497 0.482 0.468 0.456 0.444 0.396 10.361 0.335 0.312 0.294 45 50 60 70 80 90 100 In 0.279 0 254 0.236 0.220 0.207 0.196 -0.05 Print Done

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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 2.3 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)
8.3
7.8
8.6
9.2
9.3
9.6
Weight (kg)
199
207
238
237
263
284
Click the icon to view the critical values of the Pearson correlation coefficient r.
Critical Values of the Pearson Correlation Coefficient r
The regression equation is y =+ x.
(Round to one decimal place as needed.)
Critical Values of the Pearson Correlation Coefficient r
INOTE: To test Ho: p=0
a = 0.05
0.950
0.878
0.811
|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
l0.444
0.396
|0.361
0.335
0.312
0.294
0.279
0.254
0.236
0.220
0.207
0.196
|α=D 0.01
0.990
0.959
0.917
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
In
The best predicted weight for an overhead width of 2.3 cm is kg.
against H;: p+0, reject Ho
Lif the absolute value of r is
greater than the critical
value in the table.
(Round to one decimal place as needed.)
Can the prediction be correct? What is wrong with predicting the weight in this case?
17
18
O A. The prediction cannot be correct because a negative weight does not make sense and because there is not sufficient evidence of a linear correlation.
19
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.
10
11
O C. 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.
12
O D. The prediction can be correct. There is nothing wrong with predicting the weight in this case.
13
14
15
16
17
18
19
20
25
30
35
40
45
50
60
70
80
90
100
n
a = 0.05
Print
Done
Transcribed Image Text: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 2.3 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) 8.3 7.8 8.6 9.2 9.3 9.6 Weight (kg) 199 207 238 237 263 284 Click the icon to view the critical values of the Pearson correlation coefficient r. Critical Values of the Pearson Correlation Coefficient r The regression equation is y =+ x. (Round to one decimal place as needed.) Critical Values of the Pearson Correlation Coefficient r INOTE: To test Ho: p=0 a = 0.05 0.950 0.878 0.811 |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 l0.444 0.396 |0.361 0.335 0.312 0.294 0.279 0.254 0.236 0.220 0.207 0.196 |α=D 0.01 0.990 0.959 0.917 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 In The best predicted weight for an overhead width of 2.3 cm is kg. against H;: p+0, reject Ho Lif the absolute value of r is greater than the critical value in the table. (Round to one decimal place as needed.) Can the prediction be correct? What is wrong with predicting the weight in this case? 17 18 O A. The prediction cannot be correct because a negative weight does not make sense and because there is not sufficient evidence of a linear correlation. 19 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. 10 11 O C. 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. 12 O D. The prediction can be correct. There is nothing wrong with predicting the weight in this case. 13 14 15 16 17 18 19 20 25 30 35 40 45 50 60 70 80 90 100 n a = 0.05 Print Done
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