Answered step-by-stepFind the regression equation, letting overhead width be the...
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)
| 8.4 | 7.8 | 9.4 | 8.8 | 8.2 | 8.9 |
|
|---|---|---|---|---|---|---|---|
Weight (kg)
| 178 | 183 | 245 | 193 | 190 | 225 |
Critical Values of the Parson Correlation Coefficient r
n
| α=0.05
| α=0.01
| NOTE: To test H0: p=0 against H1: ρ≠0, reject H0 if the absolute value of r is greater than the critical value in the table.
|
|---|---|---|---|
4
| 0.950
| 0.990
| |
5
| 0.878
| 0.959
| |
6
| 0.811
| 0.917
| |
7
| 0.754
| 0.875
| |
8
| 0.707
| 0.834
| |
9
| 0.666
| 0.798
| |
10
| 0.632
| 0.765
| |
11
| 0.602
| 0.735
| |
12
| 0.576
| 0.708
| |
13
| 0.553
| 0.684
| |
14
| 0.532
| 0.661
| |
15
| 0.514
| 0.641
| |
16
| 0.497
| 0.623
| |
17
| 0.482
| 0.606
| |
18
| 0.468
| 0.590
| |
19
| 0.456
| 0.575
| |
20
| 0.444
| 0.561
| |
25
| 0.396
| 0.505
| |
30
| 0.361
| 0.463
| |
35
| 0.335
| 0.430
| |
40
| 0.312
| 0.402
| |
45
| 0.294
| 0.378
| |
50
| 0.279
| 0.361
| |
60
| 0.254
| 0.330
| |
70
| 0.236
| 0.305
| |
80
| 0.220
| 0.286
| |
90
| 0.207
| 0.269
| |
100
| 0.196
| 0.256
| |
n
| α=0.05
| α=0.01
|
The regression equation is
y= -141.0 + 40.0x
(Round to one decimal place as needed.)
Part 2
The best predicted weight for an overhead width of
2.4 cm is ______ kg.
(Round to one decimal place as needed.)
Part 3
Can the prediction be correct? What is wrong with predicting the weight in this case?
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.
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 there is not sufficient evidence of a linear correlation. The width in this case is beyond the scope of the available sample data.
D.
The prediction can be correct. There is nothing wrong with predicting the weight in this case.
Answer & Explanation
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