Question

Bighorn sheep are beautiful wild animals found throughout the western United States. Let x be the...

Bighorn sheep are beautiful wild animals found throughout the western United States. Let x be the age of a bighorn sheep (in years), and let y be the mortality rate (percent that die) for this age group. For example, x = 1, y = 14 means that 14% of the bighorn sheep between 1 and 2 years old died. A random sample of Arizona bighorn sheep gave the following information:

x 1 2 3 4 5
y 13.0 20.5 14.4 19.6 20.0

Σx = 15; Σy = 87.5; Σx2 = 55; Σy2 = 1580.77; Σxy = 275.6

Conclusion

Reject the null hypothesis, there is sufficient evidence that ρ > 0.Reject the null hypothesis, there is insufficient evidence that ρ > 0.    Fail to reject the null hypothesis, there is sufficient evidence that ρ > 0.Fail to reject the null hypothesis, there is insufficient evidence that ρ > 0.


(e) Given the result from part (c), is it practical to find estimates of y for a given x value based on the least-squares line model? Explain.

Given the significance of r, prediction from the least-squares model is practical.Given the lack of significance of r, prediction from the least-squares model might be misleading.    Given the lack of significance of r, prediction from the least-squares model is practical.Given the significance of r, prediction from the least-squares model might be misleading

Homework Answers

Answer #1

To Test :-

H0 :- ρ = 0
H1 :- ρ > 0

Test Statistic :-
t = (r * √(n - 2) / (√(1 - r2))
t = ( 0.5887 * √(5 - 2) ) / (√(1 - 0.3465) )
t = 1.2613


Test Criteria :-
Reject null hypothesis if t > t(α)
t(α,n-2) = t(0.05 , 5 - 2 ) = 2.3534
t < t(α, n-2) = 1.2613 < 2.3534
Result :- We Accept H1


Decision based on P value
P - value = P ( t > 1.2613 ) = 0.1482
Reject null hypothesis if P value < α = 0.05 level of significance
P - value = 0.1482 > 0.05 ,hence we fail to reject null hypothesis
Conclusion :- We Accept H1

Fail to reject the null hypothesis, there is insufficient evidence that ρ > 0.

Part e)

Given the lack of significance of r, prediction from the least-squares model might be misleading.

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