Math Problem Statement

A news article that you read stated that 56% of voters prefer the Democratic candidate. You think that the actual percent is different. 159 of the 300 voters that you surveyed said that they prefer the Democratic candidate. What can be concluded at the 0.05 level of significance?

For this study, we should use Select an answer The null and alternative hypotheses would be:
Ho: ?

Select an answer

(please enter a decimal)
H1: ?

Select an answer

(Please enter a decimal) The test statistic ?

(please show your answer to 3 decimal places.) The p-value = (Please show your answer to 4 decimal places.) The p-value is ?

Based on this, we should Select an answer the null hypothesis. Thus, the final conclusion is that ... a) The data suggest the population proportion is not significantly different 56% at = 0.05, so there is sufficient evidence to conclude that the proportion of voters who prefer the Democratic candidate is equal to 56%. b) The data suggest the populaton proportion is significantly different 56% at = 0.05, so there is sufficient evidence to conclude that the proportion of voters who prefer the Democratic candidate is different 56% c) The data suggest the population proportion is not significantly different 56% at = 0.05, so there is not sufficient evidence to conclude that the proportion of voters who prefer the Democratic candidate is different 56%. Interpret the p-value in the context of the study. a) There is a 29.52% chance of a Type I error. b) If the sample proportion of voters who prefer the Democratic candidate is 53% and if another 300 voters are surveyed then there would be a 29.52% chance that we would conclude either fewer than 56% of all voters prefer the Democratic candidate or more than 56% of all voters prefer the Democratic candidate. c) If the population proportion of voters who prefer the Democratic candidate is 56% and if another 300 voters are surveyed then there would be a 29.52% chance that either fewer than 53% of the 300 voters surveyed prefer the Democratic candidate or more than 59% of the 300 voters surveyed prefer the Democratic candidate. d) There is a 29.52% chance that the percent of all voters who prefer the Democratic candidate differs from 56%. Interpret the level of significance in the context of the study. a) If the proportion of voters who prefer the Democratic candidate is different 56% and if another 300 voters are surveyed then there would be a 5% chance that we would end up falsely concluding that the proportion of voters who prefer the Democratic candidate is equal to 56%. b) There is a 5% chance that the earth is flat and we never actually sent a man to the moon. If the population proportion of voters who prefer the Democratic candidate is 56% and if another 300 voters are surveyed then there would be a 5% chance that we would end up falsely concluding that the proportion of voters who prefer the Democratic candidate is different 56% c) There is a 5% chance that the proportion of voters who prefer the Democratic candidate is different 56%.

Solution

Step-by-Step Solution:

Type of Test:

This is a two-tailed test for population proportion because we are testing whether the proportion is different from 0.56.

Step 1: Null and Alternative Hypotheses

The null hypothesis H0H_0 and alternative hypothesis H1H_1 are:

H0:p=0.56(The population proportion is equal to 0.56.)H_0: p = 0.56 \quad (\text{The population proportion is equal to 0.56.}) H1:p≠0.56(The population proportion is different from 0.56.)H_1: p \neq 0.56 \quad (\text{The population proportion is different from 0.56.})

Step 2: Sample Statistics

Given:

  • Sample size n=300n = 300
  • Sample proportion p^=159300=0.53\hat{p} = \frac{159}{300} = 0.53
  • Hypothesized proportion p0=0.56p_0 = 0.56

Step 3: Test Statistic

The test statistic for a proportion test is calculated as:

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

Substitute the values:

= \frac{-0.03}{\sqrt{\frac{0.56 \cdot 0.44}{300}}} = \frac{-0.03}{\sqrt{\frac{0.2464}{300}}} = \frac{-0.03}{\sqrt{0.00082133}} = \frac{-0.03}{0.02866}$$ $$z = -1.047$$ --- #### **Step 4: P-Value** The p-value for a two-tailed test is calculated as: $$p\text{-value} = 2 \cdot P(Z < -1.047)$$ From the standard normal table or a calculator: $$P(Z < -1.047) = 0.1476$$ So: $$p\text{-value} = 2 \cdot 0.1476 = 0.2952$$ --- #### **Step 5: Decision** - Level of significance $$\alpha = 0.05$$ - Compare the p-value to $$\alpha$$: - $$p\text{-value} = 0.2952 > 0.05$$ Thus, **we fail to reject the null hypothesis $$H_0$$**. --- #### **Step 6: Conclusion** The final conclusion is: - **c)** The data suggest the population proportion is not significantly different from 56% at $$\alpha = 0.05$$, so there is not sufficient evidence to conclude that the proportion of voters who prefer the Democratic candidate is different from 56%. --- ### Interpretations: #### **P-Value Interpretation:** - Correct answer: **c)** If the population proportion of voters who prefer the Democratic candidate is 56%, and if another 300 voters are surveyed, then there would be a 29.52% chance that either fewer than 53% of the 300 voters surveyed prefer the Democratic candidate or more than 59% of the 300 voters surveyed prefer the Democratic candidate. #### **Level of Significance Interpretation:** - Correct answer: **a)** If the proportion of voters who prefer the Democratic candidate is different from 56% and if another 300 voters are surveyed, then there would be a 5% chance that we would end up falsely concluding that the proportion of voters who prefer the Democratic candidate is equal to 56%. --- ### Recap of Final Answers: 1. **Type of Test:** Two-tailed 2. **Null Hypothesis $$H_0$$:** $$p = 0.56$$ 3. **Alternative Hypothesis $$H_1$$:** $$p \neq 0.56$$ 4. **Test Statistic:** $$z = -1.047$$ 5. **P-Value:** $$0.2952$$ 6. **Decision:** Fail to reject $$H_0$$ 7. **Conclusion:** **c)** Not sufficient evidence to conclude that the proportion is different from 56%. --- ### Questions to Deepen Understanding: 1. Why do we use a two-tailed test instead of a one-tailed test in this scenario? 2. How does the sample size affect the calculation of the test statistic? 3. What would happen if the significance level $$\alpha$$ were changed to 0.10? 4. Can you explain how the p-value reflects the probability of observing a test statistic at least as extreme as the one calculated? 5. What are the potential consequences of failing to reject the null hypothesis? ### Tip: Always ensure the assumptions for a hypothesis test (e.g., random sampling, normal approximation for proportions) are satisfied before interpreting results.

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Math Problem Analysis

Mathematical Concepts

Hypothesis Testing
Population Proportion
Significance Testing

Formulas

z = (p_hat - p_0) / sqrt((p_0 * (1 - p_0)) / n)
p-value for two-tailed test = 2 * P(Z < -|z|)

Theorems

Central Limit Theorem
Properties of Standard Normal Distribution

Suitable Grade Level

Undergraduate Statistics or AP Statistics

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