SOLUTION: 8. A politician claim that she will receive 60% of the votes in the upcoming election. Of a random sample of 200 voters, there were 100 who will surely vote for her. Test the polit

Question 1186607: 8. A politician claim that she will receive 60% of the votes in the upcoming election. Of a random sample of 200 voters, there were 100 who will surely vote for her. Test the politician’s assertion at the 0.05 level of significance.
answer the problem using the steps in hypothesis testing.
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Here's how to conduct a hypothesis test for the politician's claim:
**1. State the Hypotheses:**
* **Null Hypothesis (H₀):** The politician's claim is correct. p = 0.60 (where p is the proportion of voters who will vote for her)
* **Alternative Hypothesis (H₁ or Ha):** The politician's claim is incorrect. p ≠ 0.60 (This is a two-tailed test because we're not specifying if the true proportion is greater or less than 60%.)
**2. Determine the Level of Significance (alpha):**
* α = 0.05 (This is given in the problem.)
**3. Calculate the Test Statistic:**
* We'll use a z-test for proportions since the sample size is large (n > 30).
* **Sample Proportion (p̂):** p̂ = (Number of voters who will vote for her) / (Total number of voters sampled) = 100/200 = 0.50
* **Standard Error (SE):** SE = sqrt[(p₀ * (1 - p₀)) / n] = sqrt[(0.60 * (1 - 0.60)) / 200] = sqrt(0.24/200) ≈ 0.0346
* **Z-score:** z = (p̂ - p₀) / SE = (0.50 - 0.60) / 0.0346 ≈ -2.89
**4. Determine the Critical Value(s) or the P-value:**
* **Critical Values (Two-tailed test):** For α = 0.05, the critical z-values are ±1.96. (You can find these using a z-table or a calculator.)
* **P-value:** The p-value is the probability of observing a test statistic as extreme as, or more extreme than, the one calculated, assuming the null hypothesis is true. Since it's a two-tailed test, we need to find the area in both tails of the standard normal distribution that is beyond our z-score of -2.89. Using a z-table or calculator, the area in one tail is approximately 0.0019. Therefore, the p-value is 2 * 0.0019 = 0.0038.
**5. Make a Decision:**
* **Using Critical Values:** Since our calculated z-score (-2.89) falls outside the critical region (-1.96 to +1.96), we reject the null hypothesis.
* **Using P-value:** Since the p-value (0.0038) is less than the level of significance (0.05), we reject the null hypothesis.
**6. Conclusion:**
There is sufficient evidence at the 0.05 level of significance to reject the politician's claim. The sample data suggests that the true proportion of voters who will vote for her is not 60%.

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