Dear beloved readers, welcome to our website! We hope your visit here brings you valuable insights and meaningful inspiration. Thank you for taking the time to stop by and explore the content we've prepared for you.
------------------------------------------------ The health of the bear population in Yellowstone National Park is monitored by periodic measurements taken from anesthetized bears. A sample of 39 bears has a mean weight of 189.2 lb. At [tex]\alpha = 0.01[/tex], 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 8.3 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?

The test statistic result is 1.63, which is less than the critical value of ±2.712. Therefore, we can't reject the null hypothesis and so we cannot conclude that the mean weight of bears in Yellowstone National Park is significantly different from 187lb.

Explanation:

In this scenario, we are performing a hypothesis test for the mean weight of bears in Yellowstone National Park using periodic measurements. We will use a two-tailed t-test in this case because we want to know if the average weight is significantly different from 187lb, either higher or lower.

(a) Test Statistic: The formula for the test statistic in a one sample t-test is: (sample mean - population mean) / (standard deviation/sqrt(number of observations)). Substituting the given values, we get: (189.2 - 187) / (8.3/sqrt(39)) = 1.63.
(b) Critical Value: At α=.01 for a two-tailed test, the degrees of freedom is n - 1, which equals 38. Checking the t-distribution table, the critical value for df=38 at α=.01 is ±2.712.
(c) P-Value: Since our calculated test statistic (1.63) falls within the critical region (-2.712,+2.712), we cannot reject the null hypothesis. However, the exact p-value needs to be looked up in the t-distribution table or calculated with statistical software.
(d) Conclusion: Since we cannot reject the null hypothesis, statistically we conclude that there is no significant evidence to say that the mean weight of bears in Yellowstone National Park is significantly different from 187lb at the 1% level of significance.

Learn more about Hypothesis Testing here:

https://brainly.com/question/31665727

#SPJ11