Math Problem Statement

All answers

You are conducting a study to see if the proportion of voters who prefer Candidate A is significantly different from 56% at a significance level of α = 0.05. According to your sample, 40 out of 71 potential voters prefer Candidate A.

Solution

Let's break this down step by step to solve the problem from your exam.

Question Details:

You are conducting a study to determine if the proportion of voters who prefer Candidate A differs from 56%, at a significance level of α=0.05\alpha = 0.05. The sample includes 71 voters, and 40 out of those 71 prefer Candidate A.

We will go through each part of the question and find the appropriate answers.

1. Study Proportion pp:

For this study, you are testing if the true population proportion pp is different from 0.56 (or 56%).

This is a two-tailed test since you want to check if it is "different," not specifically greater or less.

2. Null and Alternative Hypotheses:

  • H0:p=0.56H_0: p = 0.56 (the population proportion is 56%)
  • Ha:p≠0.56H_a: p \neq 0.56 (the population proportion is different from 56%)

3. Test Statistic:

The test statistic zz is calculated using the formula for a proportion hypothesis test:

z=p^−p0p0(1−p0)nz = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}

Where:

  • p^\hat{p} = Sample proportion = 4071≈0.563\frac{40}{71} \approx 0.563
  • p0=0.56p_0 = 0.56 (null hypothesis proportion)
  • n=71n = 71 (sample size)

Now, let's compute this:

= \frac{0.003}{\sqrt{\frac{0.2464}{71}}} = \frac{0.003}{\sqrt{0.00347}} = \frac{0.003}{0.059}$$ So, $$z \approx 0.051$$. ### 4. P-value: The p-value corresponds to the probability of observing a test statistic as extreme as the one calculated (in both directions, since it's two-tailed). Since the z-value is very small (near 0), the p-value will be close to 1. To get an approximate p-value for $$z = 0.051$$, we can look this up in a standard normal distribution table or use a calculator. The p-value is about 0.959. ### 5. Compare P-value to $$\alpha$$: Since the p-value (0.959) is much greater than $$\alpha = 0.05$$, we **fail to reject** the null hypothesis. ### 6. Conclusion: Based on the test, there is **not enough evidence** to suggest that the proportion of voters who prefer Candidate A is different from 56%. Thus, the final answer should be: - "The sample data suggest that the population proportion is not significantly different from 56% at $$\alpha = 0.05$$, so there is not sufficient evidence to conclude that the proportion of voters who prefer Candidate A is different from 56%." --- Do you want more details on any of the steps? Let me know if you have any questions! ### Related Questions: 1. How is a two-tailed test different from a one-tailed test? 2. What is the significance level, and how does it impact hypothesis testing? 3. How do we interpret the p-value in hypothesis testing? 4. Why do we use a z-test for proportions instead of other tests? 5. What is the relationship between the sample size and the test statistic? ### Tip: Always double-check your p-values and z-scores when performing hypothesis tests, as they directly influence your conclusions.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
Proportions
Z-test
P-value

Formulas

z = (p̂ - p₀) / sqrt((p₀(1 - p₀)) / n)
p̂ = 40/71 ≈ 0.563
p₀ = 0.56

Theorems

Z-test for proportion
Null and alternative hypothesis testing

Suitable Grade Level

College-level Statistics

Related Recommendation

Hypothesis Test for Proportion: Candidate A Study

Testing Rival's Claim About Election Candidate's Votes (40%)

Hypothesis Test for Voter Proportion at 56%

Hypothesis Test for Proportion: Politician Supported by Majority - Test Statistic Calculation

Hypothesis Test for Proportion of Voters Preferring Democratic Candidate at 0.01 Significance Level