Question

The health of the bear population in Yellowstone National Park is monitored by periodic measurements taken from anesthetized bears. A sample of 53 bears has a mean weight of 189.9 lb. At α = .06, can it be concluded that the average weight of a bear in Yellowstone National Park is different from 187 lb? Note that the standard deviation of the weight of a bear is known to be 8.9 lb. (a) Find the value of the test statistic for the above hypothesis. (b) Find the critical value. (c) Find the p-value. (d) What is the correct way to draw a conclusion regarding the above hypothesis test?

Answer 1

Given: = 187 lbs, = 8.9 lbs, = 189.9 lbs, n = 53, = 0.06

The Hypothesis:

H0: = 187

Ha: 187

This is a 2 tailed test

(a) The Test Statistic: Since the population standard deviation is known, and n > 30, we use the z test statistic.

The test statistic is given by the equation:

(b) The Critical Value:   The critical value (2 Tail) at = 0.06, Zcritical = +1.881 and -1.881

(c) The p Value:    The p value (2 Tail) for Z = 2.37, is; p value = 0.0178

(d) The Decision Rule:   If Zobserved is > Zcritical or if Zobserved is < -Zcritical, Then reject H0.

Also if P value is < , Then Reject H0.

The Decision:   Since Zobserved (2.37) is > Zcritical (1.881), We Reject H0.

Also since P value (0.0178) is < (0.06) , We Reject H0.

The Conclusion: There is sufficient evidence at the 94% significance level to conclude that the average weight of a bear in Yellowstone National Park is different from 187 lbs.

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