To calculate the 95% confidence interval for the true population percentage of party members who want a change in leadership, we can use the formula for the confidence interval of a proportion:

$\text{Confidence Interval} = \hat{p} \pm Z \cdot \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$

where:

• $\hat{p}$ is the sample proportion,
• $Z$ is the Z-score corresponding to the 95% confidence level (approximately 1.96),
• $n$ is the sample size.

Step 1: Calculate the Sample Proportion ($\hat{p}$)

$\hat{p} = \frac{\text{number who want change}}{\text{total sample size}} = \frac{1200}{1600} = 0.75$

Step 2: Plugging Values into the Formula

Using $Z = 1.96$ for a 95% confidence level, $n = 1600$, and $\hat{p} = 0.75$:

$\text{Margin of Error} = 1.96 \cdot \sqrt{\frac{0.75 \times (1 - 0.75)}{1600}}$

1. Calculate $0.75 \times (1 - 0.75) = 0.75 \times 0.25 = 0.1875$,
2. Divide by 1600: $\frac{0.1875}{1600} = 0.0001171875$,
3. Take the square root: $\sqrt{0.0001171875} \approx 0.01083$,
4. Multiply by 1.96: $1.96 \times 0.01083 \approx 0.02123$.

Step 3: Determine the Confidence Interval

$\text{Confidence Interval} = 0.75 \pm 0.02123$

Converting to percentages, we get:

$\text{Confidence Interval} = 75\% \pm 2.12\%$

Final Answer

The 95% confidence interval for the true population percentage of party members who want a change in leadership is approximately:

$[72.88\%, 77.12\%]$

Would you like further details or have any questions?

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Here are some related questions:

1. How would the confidence interval change if the confidence level were 99%?
2. What if the sample size were doubled to 3200 members?
3. How does the choice of confidence level affect the width of the interval?
4. What is the interpretation of a 95% confidence interval in this context?
5. How would the results differ if the sample proportion were closer to 0.5?

Tip: Larger sample sizes generally produce narrower confidence intervals, reflecting higher precision in the estimate of the population parameter.