Question

Use the formula for the standard error of \(\hat{p}\) to explain why a. The standard error is greater when the value of the population proportion \(p\) is near 0.5 than when it is near \(1 .\) b. The standard error of \(\hat{p}\) is the same when the value of the population proportion is \(p=0.2\) as it is when \(p=0.8\)

Short answer

The standard error is greater when the value of the population proportion \(p\) is near 0.5 because the function \(\sqrt{\frac{p(1-p)}{n}}\) reaches its maximum when \(p = 0.5\). Moreover, the standard error of \(\hat{p}\) is the same when \(p = 0.2\) as when \(p = 0.8\) because the expressions \(\sqrt{\frac{0.2 (1-0.2)}{n}}\) and \(\sqrt{\frac{0.8 (1-0.8)}{n}}\) are identical.

Step-by-step solution

Consider case when the population proportion \(p\) is near 0.5

In this case, the student should plot the function \(\sqrt{\frac{p(1-p)}{n}}\) with \(p\) ranging from 0 to 1. The maximum value of the function would be at \(p = 0.5\), hence the standard error is greater when the value of \(p\) is near 0.5 than when it is near 1.

Evaluate the standard error for \(p = 0.2\) and \(p = 0.8\)

Using the formula for the standard error of a proportion \(\hat{p}\), the standard error for \(p = 0.2\) is \(\sqrt{\frac{0.2 (1-0.2)}{n}}\) and for \(p = 0.8\) is \(\sqrt{\frac{0.8 (1-0.8)}{n}}\). It can be easily verified that both expressions are equal, therefore the standard error of \(\hat{p}\) is the same when \(p = 0.2\) as it is when \(p = 0.8\).

Key concepts

Population Proportion

The population proportion, denoted as \( p \) in statistics, represents the percentage of a particular characteristic present in the entire population. Understanding \( p \) is essential as it serves as a benchmark for comparison when we conduct surveys or experiments.

For instance, if we have a population of voters and we want to know the proportion that favors a specific candidate, that's our \( p \) value. The population proportion becomes especially crucial when we are dealing with samples. When we draw a sample from the population, what we're really looking at is how well the sample proportion \( \hat{p} \) estimates the true population proportion \( p \).

In the context of the exercise you mentioned, recognizing how the value of \( p \) affects the standard error is fundamental. When \( p \) is near 0.5, each possible result of a single observation (like a voter choosing or not choosing the candidate) has nearly the same chance of occurring, which makes the outcome's variability - and thus the standard error - higher.

Sampling Distribution

The concept of a sampling distribution is pivotal in understanding the behavior of sample statistics. It refers to the probability distribution of a given statistic, like the sample proportion \( \hat{p} \) over many samples from a population.

Imagine you repeatedly took samples of a specific size from the population and calculated the sample proportion of each. If you were to plot these sample proportions, you'd get the sampling distribution of \( \hat{p} \) which will tend to be normally distributed due to the Central Limit Theorem. This theorem applies when you have a large sample size or when the population is normally distributed to begin with.

The standard error is directly related to this concept, because it measures the variability of the sample proportion around the population proportion. As shown in the exercise, when the population proportion is 0.5, the standard error is maximum, indicating maximum variability in \( \hat{p} \) across all samples.

Binomial Distribution

The binomial distribution describes the number of successes in a fixed number of independent trials, with only two possible outcomes, often labeled as 'success' and 'failure' for each trial. The parameters that define a binomial distribution are \( n \) (number of trials) and \( p \) (probability of success on a single trial).

In the case of standard error, we assume that the sampling process follows a binomial distribution since each sample item can either have the characteristic of interest or not (success or failure). So, when you take a sample and look at the proportion of success, that's a sample from a binomial distribution.

The formula for the standard error of the sample proportion, \( \sqrt{\frac{p(1-p)}{n}} \), stems from the variance of the binomial distribution. This relationship emphasizes why the standard error is largest when \( p \) is 0.5 - because the variance, which measures the spread of distribution, is maximized at that point.