To solve this problem, we need to conduct a hypothesis test. Let's define our hypotheses:
- Null Hypothesis (\(H_0\)): The mean age of state senators is equal to or older than the mean age of U.S. Senators (\(\mu \geq 60.35\)).
- Alternative Hypothesis (\(H_1\)): The mean age of state senators is younger than the mean age of U.S. Senators (\(\mu < 60.35\)).
Given:
- Sample mean age of state senators (\(\bar{x}\)) = 55.4 years
- Population mean age of U.S. Senators (\(\mu_0\)) = 60.35 years
- Population standard deviation (\(\sigma\)) = 6.5 years
- Sample size (\(n\)) = 40
- Significance level (\(\alpha\)) = 0.05
We can perform a one-sample z-test to determine if we can reject the null hypothesis in favor of the alternative:
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Calculate the z-score: \[ z = \frac{\bar{x} - \mu_0}{\sigma/\sqrt{n}} = \frac{55.4 - 60.35}{6.5/\sqrt{40}} \]
First, calculate the standard error (SE): \[ SE = \frac{6.5}{\sqrt{40}} \approx \frac{6.5}{6.32} \approx 1.03 \]
Now, plug this into the z-score formula: \[ z = \frac{55.4 - 60.35}{1.03} \approx \frac{-4.95}{1.03} \approx -4.80 \]
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Determine the critical z-value for a one-tailed test at \(\alpha = 0.05\). The critical z-value is approximately -1.645.
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Compare the calculated z-score with the critical z-value:
- Calculated z-score ≈ -4.80
- Critical z-value ≈ -1.645
Since -4.80 is less than -1.645, we reject the null hypothesis.
- Conclusion: Since we rejected the null hypothesis, we have sufficient evidence to support the claim that state senators are on average younger than the Senators in Washington.
The correct conclusion is: c) The sample data support the claim that state senators are on average younger than the Senators in Washington.