Question 1179923: A political strategist claims that 56% of voters in Madison County support his candidate. In a poll of 300 randomly selected voters, 150 of them support the strategist's candidate. At is the political strategist's claim warranted?
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's how to conduct a hypothesis test to determine if the strategist's claim is warranted:
**1. State the Hypotheses:**
* **Null Hypothesis (H0):** The proportion of voters who support the candidate is 56% (p = 0.56).
* **Alternative Hypothesis (H1):** The proportion of voters who support the candidate is *not* 56% (p ≠ 0.56). This is a two-tailed test.
**2. Significance Level:** α = 0.05 (If not specified, we will assume this common value)
**3. Calculate the Sample Proportion (p̂):**
* p̂ = (Number of voters supporting the candidate) / (Total number of voters)
* p̂ = 150 / 300 = 0.50
**4. Calculate the Test Statistic (z-score):**
z = (p̂ - p) / √(p(1 - p) / n)
Where:
* p̂ = sample proportion (0.50)
* p = hypothesized population proportion (0.56)
* n = sample size (300)
z = (0.50 - 0.56) / √(0.56 * (1 - 0.56) / 300)
z = -0.06 / √(0.56 * 0.44 / 300)
z = -0.06 / √(0.2464 / 300)
z = -0.06 / √0.0008213
z = -0.06 / 0.02866
z ≈ -2.09
**5. Determine the P-value:**
Since this is a two-tailed test, we need to find the probability of getting a z-score as extreme as -2.09 or 2.09. Using a z-table or calculator:
* P(z < -2.09) ≈ 0.0183
* P(z > 2.09) ≈ 0.0183
* P-value = 2 * 0.0183 ≈ 0.0366
**6. Make a Decision:**
Compare the p-value to the significance level (α):
* p-value (0.0366) < α (0.05)
Since the p-value is *less than* the significance level, we *reject* the null hypothesis.
**7. Conclusion:**
There is sufficient evidence at the α = 0.05 level of significance to conclude that the proportion of voters in Madison County who support the candidate is *not* 56%. Therefore, the political strategist's claim is *not* warranted.
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