Question

Independent random samples of \(n_{1}=500\) and \(n_{2}=500\) observations were selected from binomial populations 1 and \(2,\) and \(x_{1}=120\) and \(x_{2}=147\) successes were observed. a. What is the best point estimator for the difference \(\left(p_{1}-p_{2}\right)\) in the two binomial proportions? b. Calculate the approximate standard error for the statistic used in part a. c. What is the margin of error for this point estimate?

Short answer

Based on the given information, we have the following: Sample proportions: - Population 1: \(\hat{p}_1 = \frac{120}{500}\) - Population 2: \(\hat{p}_2 = \frac{147}{500}\) Point estimator for the difference in proportions: \(\hat{p}_1 - \hat{p}_2 = \left(\frac{120}{500}\right) - \left(\frac{147}{500}\right)\) Approximate standard error: \(SE = \sqrt{\frac{\left(\frac{120}{500}\right)\left(1 - \frac{120}{500}\right)}{500} + \frac{\left(\frac{147}{500}\right)\left(1 - \frac{147}{500}\right)}{500}}\) Margin of error (assuming a 95% confidence level): \(ME = 1.96 \times SE\) You may provide the final numerical values when carrying out the calculations, along with the interpretation of the results.

Step-by-step solution

Calculate the sample proportions \(\hat{p}_1\) and \(\hat{p}_2\)

To calculate the sample proportions, we will use the formula \(\hat{p} = \frac{x}{n}\), where \(x\) is the number of successes and \(n\) is the sample size. For population 1, we have: $$\hat{p}_1 = \frac{x_1}{n_1} = \frac{120}{500}$$ For population 2, we have: $$\hat{p}_2 = \frac{x_2}{n_2} = \frac{147}{500}$$

Find the point estimator for the difference in proportions \(p_1 - p_2\)

The best point estimator for the difference in proportions is the difference in sample proportions, i.e., $$\hat{p}_1 - \hat{p}_2 = \left(\frac{120}{500}\right) - \left(\frac{147}{500}\right)$$

Calculate the approximate standard error

The standard error for the difference in proportions can be found using the formula: $$SE = \sqrt{\frac{\hat{p}_1(1 - \hat{p}_1)}{n_1} + \frac{\hat{p}_2(1 - \hat{p}_2)}{n_2}}$$ Substitute the values we found in Step 1: $$SE = \sqrt{\frac{\left(\frac{120}{500}\right)\left(1 - \frac{120}{500}\right)}{500} + \frac{\left(\frac{147}{500}\right)\left(1 - \frac{147}{500}\right)}{500}}$$

Determine the margin of error

To find the margin of error, we will use the formula: $$ME = Z_{\alpha/2} \times SE$$ Where \(ME\) is the margin of error, \(Z_{\alpha/2}\) is the critical value corresponding to the desired confidence level (we will assume a 95% confidence level, if not specified), and \(SE\) is the standard error we calculated in Step 3. For a 95% confidence level, \(Z_{\alpha/2} = 1.96\). Thus, $$ME = 1.96 \times SE$$

Key concepts

Binomial Distribution

A binomial distribution is a probability distribution that summarizes the likelihood that a value will take one of two independent states. It's often used to model the number of successes in a fixed number of trials, such as flipping a coin or observing a specific outcome in a survey. Each trial is independent, and the probability of success remains the same for each trial. This type of distribution is particularly useful when you're working with scenarios that involve yes/no questions, success/failure outcomes, or any two-outcome experiments.

For example, in the problem you encountered, you have two binomial distributions. You are comparing outcomes from independent sampling processes but within similar contexts. By looking at the number of successes in your sample (like 120 successes out of 500 trials for population 1 and 147 successes out of 500 for population 2), we can start to understand the behaviors of these distributions.

Standard Error

The standard error (SE) is a statistic that provides an estimate of the variability or dispersion in a sample distribution. When dealing with proportions or means, it is especially useful as it indicates how much the sample proportion (\(\hat{p}\)) or sample mean (\(\bar{x}\)) is expected to fluctuate around the overall population proportion or mean, due to sampling variability alone.

To find the standard error for the difference in proportions in your exercise, the formula is utilized:\[SE = \sqrt{\frac{\hat{p}_1(1 - \hat{p}_1)}{n_1} + \frac{\hat{p}_2(1 - \hat{p}_2)}{n_2}}\]This formula gives you an idea of the standard deviation in the distribution of the difference between sample proportions from population samples. Lower SE values suggest less variability, indicating the proportion from the sample is a closer reflection of the true population proportion.

Point Estimator

A point estimator is a single-value estimate of a population parameter. It's a simple yet powerful way to summarize an aspect of your data based on your samples. For proportions, the best point estimator is the sample proportion itself. It provides a snapshot of the proportion of interest in your population based on the sample data, and it is very influential in statistical estimation and hypothesis testing scenarios.

In the given task, the difference in your sample proportions (\(\hat{p}_1 - \hat{p}_2\)) acts as the point estimator for the difference in the true population proportions (\(p_1 - p_2\)). It is calculated simply by taking the difference between the two observed sample proportions, and it offers a direct estimate of the gap between the two population characteristics you're studying.

Margin of Error

The margin of error (ME) reflects the range within which we can expect the true population parameter to fall, with a certain degree of confidence. It illustrates the precision of your point estimator and helps to quantify how confident one can be about the range in which the true value resides.

Calculation of ME involves multiplying the standard error by a critical value, often derived from a standard normal distribution or a t-distribution. In your problem, assuming a 95% confidence level, this critical value is typically 1.96 (denoted by \(Z_{\alpha/2}\)). Hence, the calculation is:\[ME = 1.96 \times SE\]The resulting margin of error gives a boundary around your point estimate. For example, if your ME turned out to be 0.05, then you'd expect the true difference in proportions to be your point estimate ± 0.05, with 95% confidence. This concept is key to building confidence intervals and communicating the certainty of statistics.