Question

A random sample of \(n=900\) observations from a binomial population produced \(x=655\) successes. Estimate the binomial proportion \(p\) and calculate the margin of error.

Short answer

Answer: The estimated binomial proportion is approximately 0.728, with a margin of error of ±0.027. The 95% confidence interval for the true binomial proportion in the population is approximately 0.701 to 0.755.

Step-by-step solution

Calculate the Sample Proportion

To estimate the binomial proportion \(p\), we'll first calculate the sample proportion \(\hat{p}\) by dividing the number of successes (\(x = 655\)) by the total number of observations (\(n = 900\)): \(\hat{p} = \frac{x}{n} = \frac{655}{900} \approx 0.728\)

Calculate the Standard Error of the Sample Proportion

Next, we calculate the standard error of the sample proportion using the formula: \(SE(\hat{p}) = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}\) Plugging in the given values \(\hat{p}\) and \(n\), we get: \(SE(\hat{p}) = \sqrt{\frac{0.728(1 - 0.728)}{900}} \approx 0.014\)

Calculate the Margin of Error

To calculate the margin of error, we need to determine the critical value for a 95% confidence level, which is commonly denoted as \(z_{\frac{\alpha}{2}}\). For a 95% confidence level, the critical value is \(1.96\) (from the standard normal distribution table). Next, we multiply the critical value by the standard error: \(ME = z_{\frac{\alpha}{2}} \cdot SE(\hat{p}) = 1.96 \cdot 0.014 \approx 0.027\)

Report the Estimate and Margin of Error

Now that we have calculated the sample proportion and margin of error, we can report our estimate of the binomial proportion \(p\) and the margin of error: The estimated binomial proportion \(p\) is \(\approx 0.728\), with a margin of error of \(\pm 0.027\). This means that we can be 95% confident that the true binomial proportion \(p\) in the population lies between \(0.728 - 0.027 \approx 0.701\) and \(0.728 + 0.027 \approx 0.755\).