Question

Consider random sampling from a dichotomous population with \(p=0.3,\) and let \(E\) be the event that \(\hat{P}\) is within ±0.05 of \(p .\) Use the normal approximation (without the continuity correction) to calculate \(\operatorname{Pr}\\{E\\}\) for a sample of size \(n=400\).

Short answer

The probability that \( \hat{P} \) is within ±0.05 of \(p = 0.3\) for a sample size of 400 is found by calculating the Z-scores for the lower and upper bounds and then finding the area between these Z-scores in the standard normal distribution.

Step-by-step solution

Identify the Sampling Distribution

The sample proportion, \( \hat{P} \), follows a normal distribution by the Central Limit Theorem since the sample size is large. The mean of the sampling distribution of \( \hat{P} \) is \( p \), and the standard deviation is \( \sqrt{\frac{p(1-p)}{n}} \).

Calculate the Standard Deviation of \( \hat{P} \)

Use the formula for the standard deviation of a sample proportion. Since we have \( p = 0.3 \) and \( n = 400 \), the standard deviation \( \sigma_{\hat{P}} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.3(1-0.3)}{400}} \).

Express the Event \(E\) in Terms of Standard Deviation

The event \(E\) is the probability that \( \hat{P} \) is within ±0.05 of \(p\). We define the event as \( p - 0.05 \leq \hat{P} \leq p + 0.05 \). To express this in terms of standard deviations, we find the z-scores for the boundaries around \(p\).

Calculate the Z-scores

The Z-scores are calculated as \( Z = \frac{\hat{P}-p}{\sigma_{\hat{P}}} \). For the lower bound, \( Z_{lower} = \frac{0.3 - 0.05 - 0.3}{\sigma_{\hat{P}}} \) and for the upper bound \( Z_{upper} = \frac{0.3 + 0.05 - 0.3}{\sigma_{\hat{P}}} \). Simplified, we need to calculate \( Z_{lower} = \frac{-0.05}{\sigma_{\hat{P}}} \) and \( Z_{upper} = \frac{0.05}{\sigma_{\hat{P}}} \).

Compute the Probability

Use standard normal distribution tables or technology to find the probability that a standard normal random variable falls between \( Z_{lower} \) and \( Z_{upper} \). This gives us \( \operatorname{Pr}\{E\} \) which is the probability that \( \hat{P} \) is within ±0.05 of \(p\).

Key concepts

Central Limit Theorem

The Central Limit Theorem (CLT) is a fundamental statistical principle that explains why the shape of sampling distributions tends to look like a normal distribution as the sample size grows, regardless of the shape of the population distribution. This theorem is crucial for making inferences about populations from samples. When we're taking samples from a population, especially when the size is 30 or larger, the distribution of sample means will approximate a normal distribution. This is why, even when the original population is not normally distributed, the CLT allows us to use normal approximation methods.

In our exercise, we are dealing with a large sample size of 400, which is why the CLT applies. It suggests that the sample proportion, \( \bar{P} \), of successes will follow a normal distribution around the population proportion, \( p = 0.3 \), with a standard deviation that shrinks as the sample size increases. Because of the CLT, we are justified in using the normal approximation to solve this problem.

Sampling Distribution

A sampling distribution is a probability distribution of a statistic obtained through a large number of samples drawn from a specific population. It is a key concept in statistics because it allows us to see the variability and distribution of a sample statistic, such as the sample mean or sample proportion. The shape and spread of the sampling distribution depend on three factors: the population distribution, the size of the samples, and the sample size. Importantly, if the sample size is large enough, the Central Limit Theorem ensures that the sampling distribution of the sample mean or proportion will be approximately normal.

In the context of our exercise, we focus on the sampling distribution of the sample proportion \( \bar{P} \). Since we are drawing a large sample (n=400) from the population, the CLT guarantees that the sampling distribution of the sample proportion will approximate a normal distribution.

Standard Deviation of Sample Proportion

The standard deviation of the sample proportion is a measure of the variability of the proportion in different samples from a population. More formally, for a sample proportion \( \bar{P} \), the standard deviation of the sampling distribution of \( \bar{P} \), often denoted as \( \sigma_{\bar{P}} \), represents how much the proportions in the samples are expected to differ from the true population proportion \( p \). This is calculated using the formula \( \sigma_{\bar{P}} = \sqrt{\frac{p(1-p)}{n}} \), where \( p \) is the population proportion of successes and \( n \) is the sample size.

The smaller the standard deviation, the closer the sample proportions are likely to be to the population proportion. In our exercise, with \( p = 0.3 \) and \( n = 400 \), we calculate a specific standard deviation which is then used to assess the variability of our sample proportion and to compute z-scores in the subsequent steps of the solution.

Z-score Calculation

A z-score represents the number of standard deviations a data point is from the mean of a distribution. In statistical analysis and in our context, the z-score calculation is a step that transforms the sample proportion into a standard normal distribution so that we can use standard statistics tables or software to find probabilities.

The formula for calculating a z-score in the context of sample proportions is given by \( Z = \frac{\hat{P} - p}{\sigma_{\hat{P}}} \). To determine the probability of an event such as 'the sample proportion being within a specific range', we compute the z-scores for the range limits.

Lower and Upper Z-scores

Using the standard deviation calculated previously, we find the z-scores for the bounds set by the event \( E \). If the z-scores for the range fall within the common 'z-table', we can use the table to find the probabilities of a sample proportion falling within that range. This part of the solution is where we combine the concept of standard deviation, z-scores, and the normal distribution to make practical inferences about our sample proportion in relation to the population proportion.