Question

The health of the bear population in Yellowstone National Park is monitored by periodic measurements taken from anesthetized bears. A sample of 54 bears has a mean weight of 182.9 lbs. Assume that the standard deviation is known to be 121.8 lbs

a. Use a 0.05 significance level to test the claim that the population mean of all such a bear weight is greater than 150 pounds. Use the usual critical value method to perform this test.

b. Find the P value for this test and explain how the P value would lead you to the same conclusion that your rejection region method did.

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Answer 1

Given the health of the bear population in Yellowstone National Park is monitored by periodic measurements taken from anesthetized bears. A sample of n = 54 bears has a mean weight of $\bar{x}$ = 182.9 lbs. Assume that the standard deviation is known to be $\sigma$ = 121.8 lbs.

a)  The claim is that the population mean of all such a bear weight is greater than $\mu$ =150 pounds. Thus the hypotheses are:

$Ho: \mu = 150$

$Ha: \mu > 150$

Based on the hypothesis it will be a right-tailed test Since the sample size is greater than 30, and the population standard deviation is known hence Z statistic is applicable for hypothesis testing.

Test statistic:

- μo 2= X o/vn 182.9 – 150 121.8/54 = 1.985

Reejction region:

Based on the significance level and the type of hypothesis the critical score for rejection region is calclated using the excel formula for normal distribution which is =NORM.S.INV(0.95), thus the Zc is computed as 1.645. Now reject Ho if Z > 1.645.

Conclusion:

Since the test statistic is greater than 1.645 hence we can reject the null hypothesis and conclude that at 0.05 level of significance there is enough evidence to support the claim.

b) P-value:

Rejection region:

The rejection region for p-value method is the significance level, reject Ho if P-value is less than 0.05.

P-value:

The P_value is computed using the excel formula for normal distribution which is =1-NORM.S.DIST(1.985, TRUE), thud the p-value is computed as 0.0236.

Conclusion:

Since P-value is less than 0.05 hence we can reject the null hypothesis and conclude that at 0.05 level of significance there is enough evidence to support the claim.

Thus the P value would lead you to the same conclusion that your rejection region method did.