To test the hypothesis that the population mean weight of bears in Yellowstone National Park is greater than 150 lb, we can set up the following hypotheses:
- Null hypothesis (\(H_0\)): \(\mu \leq 150\) lb
- Alternative hypothesis (\(H_a\): \(\mu > 150\) lb
We are given the following data:
- Sample size (\(n\)) = 54
- Sample mean (\(\bar{x}\)) = 182.9 lb
- Population standard deviation (\(\sigma\)) = 121.8 lb
- Significance level (\(\alpha\)) = 0.05
Step 1: Calculate the test statistic
We use the Z-test for this scenario, given that we know the population standard deviation. The test statistic \(Z\) can be calculated using the formula:
\[ Z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}} \]
Where:
- \(\bar{x} = 182.9\)
- \(\mu_0 = 150\)
- \(\sigma = 121.8\)
- \(n = 54\)
Substituting the values into the formula:
\[ Z = \frac{182.9 - 150}{121.8 / \sqrt{54}} \]
Calculating the denominator:
\[ \sigma / \sqrt{n} = \frac{121.8}{\sqrt{54}} \approx \frac{121.8}{7.348} \approx 16.56 \]
Now calculate \(Z\):
\[ Z = \frac{182.9 - 150}{16.56} \approx \frac{32.9}{16.56} \approx 1.988 \]
Step 2: Find the P-value
To find the P-value for our test statistic, we look up the Z-score of 1.988 in the standard normal distribution table, or use a calculator.
For \(Z = 1.988\):
The P-value associated with \(Z = 1.988\) is approximately 0.0239.
Step 3: Conclusion
Now we compare the P-value to the significance level (\(α = 0.05\)):
- If P-value < \(α\), we reject the null hypothesis.
- If P-value ≥ \(α\), we fail to reject the null hypothesis.
Since \(0.0239 < 0.05\), we reject the null hypothesis.
Answer
The P-value is 0.0239 and we reject the null hypothesis. Therefore, the correct answer is:
D) 0.0239; reject the null hypothesis.