Question

Using the sample information given in Exercises \(22-23,\) give the best point estimate for the binomial proportion \(p\) and calculate the margin of error. A random sample of \(n=900\) observations from a binomial population produced \(x=655\) successes.

Short answer

Answer: The best point estimate for the binomial proportion is approximately 0.728, and the margin of error for this estimate is approximately 0.0283.

Step-by-step solution

Calculate the sample proportion (p̂)

To find the sample proportion, we will divide the number of successes (x) by the total number of observations (n): p̂ = x/n = 655/900

Calculate the margin of error

Now, we will calculate the margin of error using the formula: Margin of error = Zα/2 * √((p̂ * (1-p̂))/n) From the given data, p̂ ≈ 0.728, n = 900 and for a 95% confidence level, the Z-score (Zα/2) = 1.96. Margin of error = 1.96 * √((0.728 * (1-0.728))/900) Calculating the margin of error, we get: Margin of error ≈ 0.0283

Present the best point estimate and margin of error

The best point estimate for the binomial proportion p is the sample proportion p̂ ≈ 0.728, and the margin of error for this estimate is approximately 0.0283.

Key concepts

Sample Proportion

To understand binomial proportion estimation, it's vital to start with the basics of the sample proportion. It represents a key aspect of inferential statistics, where the goal is to learn about a population by examining a subset of it—that is, a sample.

In the context of our problem, this involves finding the proportion of successes in a binomial distribution. This is reflective of certain 'yes' or 'no' outcomes, like the flipping of a coin, where you'd count the number of successful outcomes (e.g., heads) to the whole number of trials or observations.

In practice, the sample proportion is symbolized by \( \hat{p} \) and is calculated by dividing the number of observed successes \( x \) by the total number of observations \( n \): \[ \hat{p} = \frac{x}{n} \]In our exercise, with 655 successes out of 900 observations, the sample proportion is \( \hat{p} = \frac{655}{900} = 0.728 \) approximately. This calculation serves as the best point estimate for the parameter \( p \) of the population proportion.

Margin of Error Calculation

The margin of error is a statistic expressing the amount of random sampling error in a survey's results. It represents the radius of the confidence interval for a particular statistic. Higher margins of error indicate less confidence in the precision of the sample proportion as a point estimate.

The margin of error is calculated using a simple formula: \[ \text{Margin of Error} = Z_{\alpha/2} \times \sqrt{\frac{\hat{p} \times (1 - \hat{p})}{n}} \]where \( Z_{\alpha/2} \) refers to the critical value from the Z-distribution that corresponds to the desired confidence level, \( \hat{p} \) is the sample proportion, and \( n \) is the sample size.

For instance, with a 95% confidence level, the Z-score (also known as \( Z_{\alpha/2} \) value) is 1.96. If we apply this along with our sample proportion \( \hat{p} \) and the sample size \( n = 900 \), we find the margin of error to be approximately 0.0283, as calculated in the step-by-step solution.

Confidence Interval

A confidence interval is used to estimate the range within which a population parameter will fall a certain percentage of the time. It takes into account the sample proportion and the margin of error to provide an interval estimate, rather than a single value, which indicates where the true population proportion is likely to lie.

The confidence interval for a sample proportion is calculated as: \[ \text{Confidence Interval} = (\hat{p} - \text{Margin of Error}, \hat{p} + \text{Margin of Error}) \]For example, using our previous results, a 95% confidence interval for the true binomial proportion \( p \) based on our sample data would be: \[ (0.728 - 0.0283, 0.728 + 0.0283) \]This indicates that there is a 95% chance that the true population proportion lies within this range, showing us a complete picture of the estimation undoubtedly more informative than the point estimate alone.