SOLUTION: 500 people randomly polled from a city showed that 274 supported gun control. Test the claim that a majority in the city supported gun control. Use a significance level of 0.05.

Question 1198294: 500 people randomly polled from a city showed that 274 supported gun control. Test the claim
that a majority in the city supported gun control. Use a significance level of 0.05.

Answer by proyaop(69) (Show Source): You can put this solution on YOUR website!
**1. Set up Hypotheses**
* **Null Hypothesis (H0):** p ≤ 0.50 (Proportion of people in the city who support gun control is less than or equal to 50%)
* **Alternative Hypothesis (H1):** p > 0.50 (Proportion of people in the city who support gun control is greater than 50%)
**2. Calculate Sample Proportion**
* Sample Proportion (p̂) = Number of successes / Sample size
* p̂ = 274 / 500 = 0.548
**3. Calculate the Standard Error**
* Standard Error (SE) = √(p̂ * (1 - p̂) / n)
* SE = √(0.548 * (1 - 0.548) / 500)
* SE ≈ 0.022
**4. Calculate the Test Statistic (z-score)**
* z = (p̂ - p0) / SE
* where p0 is the hypothesized population proportion under the null hypothesis (0.50)
* z = (0.548 - 0.50) / 0.022
* z ≈ 2.18
**5. Determine the Critical Value**
* Since this is a one-tailed test (we're testing if the proportion is *greater* than 0.50) at a 0.05 significance level, the critical value (zα) is 1.645.
**6. Make a Decision**
* Since the calculated z-score (2.18) is greater than the critical value (1.645), we **reject the null hypothesis (H0)**.
**7. Conclusion**
* There is sufficient evidence at the 0.05 significance level to support the claim that a majority of people in the city support gun control.
**In Summary:**
* The sample data provides strong evidence to suggest that more than 50% of the city's population supports gun control.

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