Math Problem Statement

One month before an election, a poll of 620 randomly selected voters showed 59% planning to vote for a certain candidate. A week later it became known that he had been a juvenile delinquent, and a new poll showed only 56% of 1080 voters supporting him. Do these results indicate a decrease in voter support for his candidacy? a) Test an appropriate hypothesis and state your conclusion. b) If you concluded there was a difference, estimate that difference with a 95% confidence interval and interpret your interval in context.

Solution

Let's go through the problem step-by-step to answer the questions presented.

Problem Summary

  1. Initial Poll: 620 voters, 59% support (p1 = 0.59).
  2. Subsequent Poll: 1080 voters, 56% support (p2 = 0.56).

We are tasked with:

  1. Testing if there is a significant decrease in voter support for the candidate (hypothesis test).
  2. If there is a significant difference, estimating this difference with a 95% confidence interval.

Step-by-Step Solution

a) Hypothesis Test

To test if there is a decrease in support, we can set up the hypotheses as follows:

  • Null Hypothesis (H0): p1=p2p_1 = p_2
  • Alternative Hypothesis (H1): p1>p2p_1 > p_2 (indicating a decrease in support)

Since this is a test for proportions, we use the z-test for two proportions.

  1. Calculate the pooled proportion: p^=x1+x2n1+n2=(0.59×620)+(0.56×1080)620+1080\hat{p} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{(0.59 \times 620) + (0.56 \times 1080)}{620 + 1080}

  2. Calculate the test statistic (z): z=(p1p2)p^(1p^)(1n1+1n2)z = \frac{(p_1 - p_2)}{\sqrt{\hat{p}(1 - \hat{p}) \left( \frac{1}{n_1} + \frac{1}{n_2} \right)}}

Given that we need to use z=1.20z = 1.20 (as provided in the question), we can find the p-value using standard normal tables or a calculator.

For z=1.20z = 1.20, the p-value (for a one-tailed test) is approximately: P=0.1151P = 0.1151

So, P=0.1151P = 0.1151 when rounded to four decimal places.

b) 95% Confidence Interval for the Difference in Proportions

If we conclude that there was a significant difference, we can calculate a 95% confidence interval for the difference in proportions (p1p2)(p_1 - p_2) as follows:

  1. Formula for the confidence interval: (p1p2)±Zα/2p1(1p1)n1+p2(1p2)n2(p_1 - p_2) \pm Z_{\alpha/2} \sqrt{\frac{p_1(1 - p_1)}{n_1} + \frac{p_2(1 - p_2)}{n_2}}

    Where Zα/2=1.96Z_{\alpha/2} = 1.96 for a 95% confidence interval.

  2. Plug in the values to compute the interval.

This interval will provide an estimate of the range within which the true difference in proportions lies, with 95% confidence.


Summary of Answers

  • Test Statistic (z): 1.20
  • P-value (P): 0.1151

Would you like further details on the calculations or clarification on any step?

Additional Related Questions

  1. What does the p-value represent in the context of hypothesis testing?
  2. How would the interpretation change if the p-value was less than 0.05?
  3. Why is a confidence interval useful in interpreting the difference in voter support?
  4. How would you calculate the confidence interval if the sample sizes were different?
  5. What other factors could impact voter support that aren't measured by this data?

Tip

Always double-check the conditions for using a z-test with proportions, such as ensuring both n×pn \times p and n×(1p)n \times (1 - p) are greater than 10, to ensure the test’s validity.

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Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
Proportion Testing
Confidence Intervals
Z-Test for Two Proportions

Formulas

Pooled Proportion Formula: p̂ = (x1 + x2) / (n1 + n2)
Z-Test Formula: z = (p1 - p2) / sqrt(p̂ * (1 - p̂) * (1/n1 + 1/n2))
Confidence Interval Formula: (p1 - p2) ± Zα/2 * sqrt((p1 * (1 - p1) / n1) + (p2 * (1 - p2) / n2))

Theorems

Central Limit Theorem

Suitable Grade Level

Undergraduate Statistics