Question

A politician wants to estimate the proportion of constituents favoring a controversial piece of proposed legislation. suppose that a 99% confidence interval that extends at most 0.05 on each side of the sample proportion is required. how many sample observation are needed?

Answer 1

Answer:

We are given:

Confidence level = 99%. Therefore, the critical value at 0.01 significance level using the standard normal table is given below:

$z_{\frac{0.01}{2}}=2.576$

Margin of error is given in the question as:

$E=0.05$

Since the previous proportion is not given, therefore, we need to assume $\hat{p} = 0.5$

Therefore, the sample size is:

$n=\hat{p}(1-\hat{p}) \left( \frac{z_{\frac{0.01}{2}} }{E} \right)$

  $=0.5(1-0.5) \left( \frac{2.576}{0.05} \right)$

  $=0.25 \times 51.52^2$

  $=663.6 \approx 664$

Therefore, 664 sample observations are required.