To calculate the required sample size for estimating the proportion of Labour party supporters within ±2% (or 0.02) at a 95% confidence level, we use the formula for sample size in proportion estimation:

$n = \frac{z^2 \cdot p \cdot (1-p)}{E^2}$

Where:

• $z$ is the z-value corresponding to the confidence level (for 95%, $z = 1.96$),
• $p$ is the proportion of Labour party supporters ($p = 0.12$),
• $E$ is the margin of error ($E = 0.02$).

Step-by-step calculation:

1. Substitute the values into the formula:

$n = \frac{(1.96)^2 \cdot 0.12 \cdot (1 - 0.12)}{(0.02)^2}$

2. Simplify the terms:

$n = \frac{(3.8416) \cdot (0.12) \cdot (0.88)}{0.0004}$

$n = \frac{3.8416 \cdot 0.1056}{0.0004}$

$n = \frac{0.4051}{0.0004}$

$n = 1012.7$

3. Round to the nearest whole number:

$n = 1013$

Final Answer:

The required sample size is 1013.

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Related Questions:

1. How does changing the confidence level affect the required sample size?
2. What happens to the sample size if the estimated proportion $p$ changes?
3. How do we calculate sample size when the population size is limited?
4. What is the impact of a smaller margin of error (e.g., ±1%) on sample size?
5. Why is $z = 1.96$ used for 95% confidence?

Tip:

Always check the assumptions of the formula—like the expected proportion and whether the population size affects the calculation.