Model Answer
0 min readIntroduction
Hypothesis testing is a crucial statistical method used to determine whether there is enough evidence to reject a null hypothesis in favor of an alternative hypothesis. In the context of public policy and decision-making, such as the construction of a chemical plant, understanding public opinion is paramount. This question presents a scenario where a poll is conducted to assess the difference in support for a proposed chemical plant between town residents and those in the surrounding area. The goal is to statistically determine if the observed difference in proportions is significant, or simply due to random chance, using a 5% level of significance.
Hypothesis Formulation
Let p1 be the proportion of town voters favoring the proposal and p2 be the proportion of surrounding area voters favoring the proposal.
- Null Hypothesis (H0): p1 = p2 (There is no significant difference in the proportion of voters favoring the proposal between town and surrounding area residents.)
- Alternative Hypothesis (H1): p1 > p2 (The proportion of town voters favoring the proposal is higher than that of surrounding area voters.)
Data Summary
We are given the following data:
- Town voters: n1 = 200, x1 = 120 (number favoring the proposal)
- Surrounding area voters: n2 = 500, x2 = 240 (number favoring the proposal)
Calculating Sample Proportions
The sample proportions are calculated as follows:
- p̂1 = x1 / n1 = 120 / 200 = 0.6
- p̂2 = x2 / n2 = 240 / 500 = 0.48
Calculating the Pooled Proportion
The pooled proportion (p̂) is calculated as a weighted average of the sample proportions:
p̂ = (x1 + x2) / (n1 + n2) = (120 + 240) / (200 + 500) = 360 / 700 = 0.5143 (approximately)
Calculating the Test Statistic (Z-score)
The z-score is calculated using the following formula:
z = (p̂1 - p̂2) / √[p̂(1-p̂)(1/n1 + 1/n2)]
z = (0.6 - 0.48) / √[0.5143(1-0.5143)(1/200 + 1/500)]
z = 0.12 / √[0.5143(0.4857)(0.005 + 0.002)]
z = 0.12 / √[0.2499(0.007)]
z = 0.12 / √0.0017493
z = 0.12 / 0.04183 = 2.87 (approximately)
Determining the Critical Value
Since we are using a 5% level of significance and conducting a one-tailed test (H1: p1 > p2), we need to find the critical z-value (zα) corresponding to α = 0.05. Using a standard z-table, the critical value for a one-tailed test with α = 0.05 is approximately 1.645.
Decision Rule
Reject H0 if z > zα. Otherwise, fail to reject H0.
Conclusion of the Hypothesis Test
Since our calculated z-score (2.87) is greater than the critical z-value (1.645), we reject the null hypothesis. This indicates that there is statistically significant evidence at the 5% level of significance to conclude that the proportion of town voters favoring the proposal is higher than that of surrounding area voters.
Conclusion
Based on the hypothesis test, we can agree that the proportion of town voters favoring the chemical plant proposal is significantly higher than that of surrounding area voters. This suggests that the concerns of surrounding area voters regarding the potential for the proposal to pass due to the larger proportion of town voters in favor may be valid. Further investigation into the reasons behind this difference in opinion could be beneficial for informed decision-making and community engagement.
Answer Length
This is a comprehensive model answer for learning purposes and may exceed the word limit. In the exam, always adhere to the prescribed word count.