MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN: 9781119256830
Author: Amos Gilat
Publisher: John Wiley & Sons Inc
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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.

The regression equation is
y=___+___x
​(Round to one decimal place as​ needed.)
The best predicted weight for an overhead width of
2.3cm is ___kg. 
​(Round to one decimal place as​ needed.)
 
Can the prediction be​ correct? What is wrong with predicting the weight in this​ case?
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)
7.5
7.4
8.7
8.2
9.6
9.9
Weight (kg)
144
170
219
171
252
272
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
NOTE: To test Ho: p=0
against H,: p# 0, reject H
if the absolute value of r is
greater than the critical
value in the table.
In
a = 0.05
0.950
0.878
0.811
0.754
0.707
0.668
0.632
0.602
0.576
0.553
0.532
0.514
0.497
0.482
0.488
0.456
0.444
0.398
0.361
0.335
0.312
0.294
0.279
0.254
0.236
0.220
0.207
0.196
a = 0.05
a = 0.01
0.990
0.950
0.917
0.875
0.834
0.708
0.765
0.735
0.708
0.684
0.661
0.641
0.623
0.806
0.590
0.575
0.561
0.505
0.483
0.430
0.402
0.378
0.361
0.330
0.305
0.288
0.269
0.256
a = 0.01
The best predicted weight for an overhead width of 2.3 cm is kg.
14
(Round to one decimal place as needed.)
15
Can the prediction be correct? What is wrong with predicting the weight in this case?
18
17
O A. 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.
O B. The prediction cannot be correct because a negative weight does not make sense and because there is not sufficient evidence of a linear correlation.
10
11
12
13
14
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.
O D. The prediction can be correct. There is nothing wrong with predicting the weight in this case
15
16
17
18
19
20
25
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35
40
45
50
80
70
80
90
100
In
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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) 7.5 7.4 8.7 8.2 9.6 9.9 Weight (kg) 144 170 219 171 252 272 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 NOTE: To test Ho: p=0 against H,: p# 0, reject H if the absolute value of r is greater than the critical value in the table. In a = 0.05 0.950 0.878 0.811 0.754 0.707 0.668 0.632 0.602 0.576 0.553 0.532 0.514 0.497 0.482 0.488 0.456 0.444 0.398 0.361 0.335 0.312 0.294 0.279 0.254 0.236 0.220 0.207 0.196 a = 0.05 a = 0.01 0.990 0.950 0.917 0.875 0.834 0.708 0.765 0.735 0.708 0.684 0.661 0.641 0.623 0.806 0.590 0.575 0.561 0.505 0.483 0.430 0.402 0.378 0.361 0.330 0.305 0.288 0.269 0.256 a = 0.01 The best predicted weight for an overhead width of 2.3 cm is kg. 14 (Round to one decimal place as needed.) 15 Can the prediction be correct? What is wrong with predicting the weight in this case? 18 17 O A. 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. O B. The prediction cannot be correct because a negative weight does not make sense and because there is not sufficient evidence of a linear correlation. 10 11 12 13 14 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. O D. The prediction can be correct. There is nothing wrong with predicting the weight in this case 15 16 17 18 19 20 25 30 35 40 45 50 80 70 80 90 100 In
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