Question
In: Statistics and Probability
Fifty-four wild bears were anesthetized, and then their weights and chest sizes were measured and listed...
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Fifty-four wild bears were anesthetized, and then their weights and chest sizes were measured and listed in a data set. Results are shown in the accompanying display. Is there sufficient evidence to support the claim that there is a linear correlation between the weights of bears and their chest sizes? When measuring an anesthetized bear, is it easier to measure chest size thanweight? If so, does it appear that a measured chest size can be used to predict the weight? Use a significance level of α=0.05 |
Correlation Results |
|
|
Correlation coeff, r: |
0.9572230 |
|
|---|---|---|
|
Critical r: |
±0.26808550 |
|
|
P-value (two tailed): |
0.000 |
|
Solutions
Expert Solution
Yes, it is easier to measure a
chest size than a weight because measuring weight would require
lifting the bear onto the scale. The chest size could be used to
predict weight because there is a linear correlation between the
two
orchestra answered 1 year agoRelated Solutions
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64
65
49 47
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356
316
94 86
Find the value of the linear correlation coefficient.
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40
64
65
49 47
Weight
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356
316
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Find the value of the linear correlation coefficient.
Letting y represent weights of bears and letting
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192192
241241
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Listed below are the overhead widths (in cm) of seals measured from photographs and the weights...
Listed below are the overhead widths (in cm) of seals measured
from photographs and the weights (in kg) of the seals. Construct
a scatterplot, find the value of the linear correlation
coefficient r, and find the critical values of r using
α=0.05. Is there sufficient evidence to conclude that there is a
linear correlation between overhead widths of seals from
photographs and the weights of the seals?
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7.6
9.9
9.3
8.7
8.2
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250
199...
Listed below are the overhead widths (in cm) of seals measured from photographs and the weights...
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from photographs and the weights (in kg) of the seals. Construct
a scatterplot, find the value of the linear correlation
coefficient r, and find the critical values of r using
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9.69.6
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Weight
112112
197197
241241
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