You want to obtain a sample to estimate a population proportion. Based on previous evidence, you believe the population proportion is approximately p = 88%. You would like to be 99.5% confident that your estimate is within 1% of the true population proportion. How large of a sample size is required?

Answer:

Given:

The population proportion $(p)=0.88$

The margin of error $(e)=0.01$

The confidence interval level $=99.5$

Solution:

→ The critical value at 99.5% confidence interval = 2.807

→ The sample size $(n):$

Here we use a margin of error formula to find out the sample size

$$\therefore e=z_c\cdot \sqrt{\frac{p(1–p)}{n}}$$ $$\therefore 0.01=2.807\cdot \sqrt{\frac{0.88(1–0.88)}{n}}$$ $$\therefore n=\mathrm{8,320}$$

Final Answer:
The sample size $(n)=8320$