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 12.2 20.9 14.4 19.6 20.0

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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 12.2 20.9 14.4 19.6 20.0

Σx = 15; Σy = 87.1 ; Σx2 = 55; Σy2 =1577.17; Σxy = 275.6

d) Test the claim that the population correlation coefficient is positive at the 1% level of significance. (Round your test statistic to three decimal places.)

t =



e) Find or estimate the P-value of the test statistic.

P-value > 0.250

0.125 < P-value < 0.250   

0.100 < P-value < 0.125

0.075 < P-value < 0.100

0.050 < P-value < 0.075

0.025 < P-value < 0.050

0.010 < P-value < 0.025

0.005 < P-value < 0.010

0.0005 < P-value < 0.005

P-value < 0.0005


Conclusion

Reject the null hypothesis, there is sufficient evidence that ρ> 0.

Reject the null hypothesis, there is insufficient evidence thatρ > 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 lack of significance of r, prediction from the least-squares model might be misleading.

Given the 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.

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

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