Question

A random sample of size \(n=80\) is selected from a binomial distribution with population proportion \(p=.25 .\) a. What will be the approximate shape of the sampling distribution of \(\hat{p} ?\) b. What will be the mean and standard deviation (or standard error) of the sampling distribution of \(\hat{p} ?\) c. Find the probability that the sample proportion \(\hat{p}\) is between 18 and .44 7.42 a. Is the normal approximation to the sampling distribution of \(\hat{p}\) appropriate when \(n=400\) and \(p=.8 ?\) b. Use the results of part a to find the probability that \(\hat{p}\) is greater than \(.83 .\) c. Use the results of part a to find the probability that \(\hat{p}\) lies between .76 and .84

Short answer

Answer: The approximate probability that a sample proportion lies between 0.18 and 0.44 is 92.58%.

Step-by-step solution

Central Limit Theorem for Binomial Distribution

In order to determine the approximate shape of the sampling distribution of \(\hat{p}\), we have to see if the Central Limit Theorem (CLT) applies to our case. The CLT states that the sampling distribution of the sample proportion will be approximate normal if both \(np \geq 10\) and \(n(1-p) \geq 10\). Let's check if these conditions are met: $$n = 80$$ $$p = 0.25$$ $$np = 80*0.25 =20 \geq 10$$ $$n(1-p) = 80*(1-0.25) = 60 \geq 10$$ Since both conditions are met, we can conclude that the shape of the sampling distribution of \(\hat{p}\) is approximately normal. #b. Mean and Standard Deviation of the Sampling Distribution#

Mean and Standard Error

The mean of the sampling distribution of the sample proportion is equal to the binomial distribution's population proportion, which is given by \(\mu_{\hat{p}} = p\). The standard deviation, also known as the standard error, is given by \(\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}\). Let's calculate them: $$\mu_{\hat{p}} = p = 0.25$$ $$\sigma_{\hat{p}} = \sqrt{\frac{0.25(1-0.25)}{80}} = \sqrt{\frac{0.25*0.75}{80}} = \sqrt{\frac{0.1875}{80}} \approx 0.0483$$ So, the mean of the sampling distribution is 0.25 and the standard deviation (standard error) is approximately 0.0483. #c. Probability of sample proportion between 0.18 and 0.44#

Z-score and Probability

To find the probability that the sample proportion \(\hat{p}\) is between 0.18 and 0.44, we need to convert these values to Z-scores and use the standard normal table (or software) to find the probabilities. The Z-score formula is given as follows: $$Z = \frac{\hat{p} - \mu_{\hat{p}}}{\sigma_{\hat{p}}}$$ Let's calculate the Z-scores for 0.18 and 0.44: $$Z_{0.18} = \frac{0.18 - 0.25}{0.0483} \approx -1.45$$ $$Z_{0.44} = \frac{0.44 - 0.25}{0.0483} \approx 3.93$$ Now, we can use the standard normal table (or software) to find the probability that the Z-score lies between -1.45 and 3.93: $$P(-1.45 \leq Z \leq 3.93) \approx P(Z \leq 3.93 ) - P(Z \leq -1.45) \approx 0.9999 - 0.0741 \approx 0.9258$$ So, there is a 92.58% chance that the sample proportion \(\hat{p}\) is between 0.18 and 0.44.

Key concepts

Sampling Distribution

When dealing with binomial distributions, we often use a concept known as "sampling distribution". It describes the probability distribution of a given statistic based on a random sample. For a sample proportion, denoted by \( \hat{p} \), the sampling distribution is crucial in understanding how the sample mean behaves in relation to the population mean.

In our exercise, we are working with a sample size of \( n = 80 \) and a population proportion \( p = 0.25 \). According to the Central Limit Theorem, when both conditions \( np \geq 10 \) and \( n(1-p) \geq 10 \) are satisfied, the shape of the sampling distribution of \( \hat{p} \) is approximately normal. This is vital because it allows statisticians to use normal distribution techniques when analyzing binomial distributions.

Thus, the main takeaway is that understanding the sampling distribution of \( \hat{p} \) helps in predicting the sample results based on the known population characteristics with a substantial level of confidence.

Normal Approximation

The "normal approximation" is a method that allows us to use the normal distribution as an estimate for a binomial distribution when certain conditions are met. Specifically, when the sample sizes are large, the binomial distribution's discrete probability outcomes can be approximated using a continuous normal distribution.

This approximation is particularly useful in cases where computing exact probabilities using the binomial formula is complex. In the current context, because both \( np \) and \( n(1-p) \) are greater than 10, the normal approximation is valid. This means we can calculate probabilities regarding \( \hat{p} \) using the normal probability rules.

The benefit of using a normal approximation is that it simplifies the process of finding probabilities, especially when dealing with a large number of trials. By approximating the binomial distribution, we can quickly ascertain likely outcomes and probabilities, making decision-making more efficient.

Standard Error of Proportion

The "standard error of proportion" is a measure of how much the sample proportion \( \hat{p} \) might fluctuate from the true population proportion \( p \). It provides insights into the stability and reliability of the sample statistics.

In the formula \( \sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} \), the standard error helps quantify the expected amount of variation or "error" inherent in the sampling process. In the scenario given, with \( p = 0.25 \) and \( n = 80 \), the standard error calculates to approximately 0.0483.

This means that in repeated sampling, the sample proportions would typically be within this range of variation. A small standard error indicates that the sample proportion is a reliable estimate of the population proportion. This is crucial for statisticians as it reinforces confidence in the findings drawn from sample data.