Question

2. 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 lb. Assume that o is known to be 121.8 lb. (a) Use a 0.05 significance level to test the claim that the population mean of all such bear weights is greater than 150 lb. 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. (Note that you don't have to re-do the whole test, but show your work for finding the P-value.)

Answer 1

Sample size, n= 54

Sample mean, = 182.9 lb

T

Population standard dev, $\sigma$ = 121.8 lb

Null hypothesis, H0 : $\mu$ = 150

Alternate hypothesis, Ha : $\mu$ > 150

a) Test statistic , z=
u/ 11
= $\frac{182.9- 150}{121.8 / \sqrt{54}}$ = 1.9849

Critical value, Z_(0.05/2) = $\pm$ 1.96

Now , since the test stistic is greater than the critical value, that means that the test statistic lie in the critical region .Hence we will reject the null hypothesis.

Hence we have enough evidence to support the claim that the population mean of all such bear is greater than 150 .

b) P-value = P( z > 1.98) = 1- P(z < 1.98) = 1- 0.97615 = 0.02385

   Now, since the P-value < significance leve (0.05) , so we will reject the null hypothesis.

Hence we have enough evidence to support the claim that the population mean of all such bear is greater than 150.