A number of initiatives on the topic of legalized gambling have appeared on state ballots. A political candidate has decided to support legalization of casino gambling if he is convinced that more than two-thirds of American adults approve of casino gambling. Suppose that 1,025 of the people in a random sample of 1,531 American adults said they approved of casino gambling. Is there convincing evidence that more than two-thirds approve? Test the relevant hypotheses using a significance level of 0.05.
State the appropriate null and alternative hypotheses using exact numerical values. (Let p be the population proportion of American adults who approve of casino gambling. Enter != for ≠ as needed.)


To test whether more than two-thirds of American adults approve of casino gambling, we need to set up the null and alternative hypotheses based on the population proportion \( p \).

1. **Null Hypothesis (\( H_0 \))**: \( p \leq \frac{2}{3} \) (or \( p \leq 0.6667 \))
2. **Alternative Hypothesis (\( H_a \))**: \( p > \frac{2}{3} \) (or \( p > 0.6667 \))

Next, we can perform a hypothesis test using the sample data provided.

### Step 1: Sample Proportion Calculation

The sample proportion (\( \hat{p} \)) can be calculated as follows:

\[
\hat{p} = \frac{x}{n} = \frac{1025}{1531} \approx 0.669
\]

### Step 2: Standard Error Calculation

The standard error (SE) for the sample proportion under the null hypothesis can be calculated using the formula:

\[
SE = \sqrt{\frac{p_0(1 - p_0)}{n}}
\]

Where \( p_0 = \frac{2}{3} \) (or approximately 0.6667) and \( n = 1531 \):

\[
SE = \sqrt{\frac{0.6667 \times (1 - 0.6667)}{1531}} \approx \sqrt{\frac{0.6667 \times 0.3333}{1531}} \approx \sqrt{\frac{0.2222}{1531}} \approx \sqrt{0.000145} \approx 0.0121
\]

### Step 3: Z-Statistic Calculation

We can now calculate the Z-statistic using:

\[
Z = \frac{\hat{p} - p_0}{SE} = \frac{0.669 - 0.6667}{0.0121} \approx \frac{0.0023}{0.0121} \approx 0.190
\]

### Step 4: P-Value Calculation

Using a Z-table or standard normal distribution, we find the p-value associated with \( Z = 0.190 \). This p-value represents the probability of observing a sample proportion as extreme as \( \hat{p} \) given that \( H_0 \) is true.

### Step 5: Decision Rule

With a significance level of \( \alpha = 0.05 \), we compare the p-value to \( \alpha \):

- If the p-value < 0.05, we reject \( H_0 \).
- If the p-value ≥ 0.05, we do not reject \( H_0 \).

Given that the calculated Z-value is not very extreme, the corresponding p-value is likely greater than 0.05.

### Conclusion

Based on these calculations, we would conclude whether there is sufficient evidence to support the alternative hypothesis that more than two-thirds of American adults approve of casino gambling.

Do you want more details on any part of the hypothesis testing process or further questions? Here are five related questions:

1. How do you interpret the p-value in the context of this test?
2. What assumptions are made when performing this hypothesis test?
3. How would the conclusion change if the sample size were larger?
4. What are potential implications of legalizing casino gambling based on public approval?
5. How can we improve the accuracy of our estimate of \( p \)? 

And here’s a tip: Always ensure your sample is representative of the population to avoid biased results in surveys.

A number of initiatives on the topic of legalized gambling have appeared on state ballots. A political candidate has decided to support legalization of casino gambling if he is convinced that more than two-thirds of American adults approve of casino gambling. Suppose that 1,025 of the people in a random sample of 1,531 American adults said they approved of casino gambling. Is there convincing evidence that more than two-thirds approve? Test the relevant hypotheses using a significance level of 0.05.

Explore the hypothesis testing process to determine if more than two-thirds of American adults approve of casino gambling based on a sample of 1,531, with 1,025 approving. This guide covers key statistical concepts and calculations.

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