**Problem: Sampling Distribution of Delinquent Accounts**The internal auditing staff of a local manufacturing company performs a sample audit each quarter to estimate the proportion of accounts that are delinquent more than 90 days overdue. The historical records of the company show that over the past 8 years, 14 percent of the accounts are delinquent. For this quarter, the auditing staff randomly selected 250 customer accounts. What are the mean and the standard deviation of the sampling distribution of $\hat{p}$?*Explanation:*This problem deals with the estimation of proportions using a sample. It requires calculating the mean and standard deviation of the sampling distribution of a sample proportion, $\hat{p}$.1. **Mean of the Sampling Distribution of $\hat{p}$:** The mean of the sampling distribution of $\hat{p}$ is equal to the population proportion, $p$. \[ \mu_{\hat{p}} = p = 0.14 \]2. **Standard Deviation of the Sampling Distribution of $\hat{p}$:** The standard deviation of the sampling distribution, often called the standard error, is calculated using the formula: \[ \sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} \] where: - $p = 0.14$ (proportion of delinquent accounts), - $n = 250$ (sample size). Substitute the values into the formula: \[ \sigma_{\hat{p}} = \sqrt{\frac{0.14 \times 0.86}{250}} \] *Note:* The actual computation should be carried out to find the numerical value of $\sigma_{\hat{p}}$.This provides an understanding of the variability one can expect from the sample proportion $\hat{p}$, around the population proportion $p$, in repeated sampling.

Given: Population proportion = p = 0.14, Sample size = n = 250.

44. The internal auditing staff of a local manufacturing company performs a sample audit each quarter to estimate the proportion of accounts that are delinquent more than 90 days overdue. The historical records of the company show that over the past 8 years 14 percent of the accounts are delinquent. For this quarter, the auditing staff randomly selected 250 customer accounts. What are the mean and the standard deviation of the sampling distribution of

44. The internal auditing staff of a local manufacturing company performs a sample audit each quarter to estimate the proportion of accounts that are delinquent more than 90 days overdue. The historical records of the company show that over the past 8 years 14 percent of the accounts are delinquent. For this quarter, the auditing staff randomly selected 250 customer accounts. What are the mean and the standard deviation of the sampling distribution of

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Concept explainers

Estimation

A new framework of data analysis where the new statistical methods are used for interpreting the results and analyzing the data is known as estimation in statistics.

Question

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**Problem: Sampling Distribution of Delinquent Accounts**

The internal auditing staff of a local manufacturing company performs a sample audit each quarter to estimate the proportion of accounts that are delinquent more than 90 days overdue. The historical records of the company show that over the past 8 years, 14 percent of the accounts are delinquent. For this quarter, the auditing staff randomly selected 250 customer accounts. What are the mean and the standard deviation of the sampling distribution of $\hat{p}$?

*Explanation:*

This problem deals with the estimation of proportions using a sample. It requires calculating the mean and standard deviation of the sampling distribution of a sample proportion, $\hat{p}$.

1. **Mean of the Sampling Distribution of $\hat{p}$:**

The mean of the sampling distribution of $\hat{p}$ is equal to the population proportion, $p$.

\[
\mu_{\hat{p}} = p = 0.14
\]

2. **Standard Deviation of the Sampling Distribution of $\hat{p}$:**

The standard deviation of the sampling distribution, often called the standard error, is calculated using the formula:

\[
\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}
\]

where:
- $p = 0.14$ (proportion of delinquent accounts),
- $n = 250$ (sample size).

Substitute the values into the formula:

\[
\sigma_{\hat{p}} = \sqrt{\frac{0.14 \times 0.86}{250}}
\]

*Note:* The actual computation should be carried out to find the numerical value of $\sigma_{\hat{p}}$.

This provides an understanding of the variability one can expect from the sample proportion $\hat{p}$, around the population proportion $p$, in repeated sampling.

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