Question

In Exercises 6.15 to 6.18 , what sample size is needed to give the desired margin of error in estimating a population proportion with the indicated level of confidence? A margin of error within \(\pm 5 \%\) with \(95 \%\) confidence.

Short answer

The required sample size for a confidence level of 95% with a margin of error of ±5% is approximately 385.

Step-by-step solution

Determine the z-score

Based on the criteria of 95% confidence level, the value of z-score (\(Z_{\alpha/2}\)) is approximately 1.96. This value can be obtained from Z-tables or other statistical tools which provide the Z-score corresponding to the given confidence level.

Set the Error Margin

The problem states an error margin of ±5 %. This translates to 0.05 in decimal form. Therefore, E = 0.05.

Assume Population Proportion

Since the question does not provide a known population proportion (p), we assume p to be 0.5. This gives the worst case scenario and is commonly used when the true population proportion is unknown.

Calculate the Sample Size

Substitute the values of z-score, margin of error, and assumed population proportion in the formula for sample size: \(n = (\frac{Z_{\alpha/2}*\sqrt{p(1-p)}}{E})^2 = (\frac{1.96*\sqrt{0.5(1-0.5)}}{0.05})^2\). After completing the calculations, round the final answer to the nearest whole number since sample size must be a whole number.