Q4(ch 9). The libraries around the country face a crisis, the patrons now have so many choices to borrow a book in person becoming outdated Budget cuts around many libraries are becoming slated. Only 57% of patrons borrow books anymore, Claims the American Library Association, suddenly! This statistic is wrong to its core thinks the director of the library in Plainsboro, Kentucky. She hires a stats wizard to conduct a study. He did his thing and what did he find! That in Plainsboro many a person and their buddy Borrowed books to expand their mind. Of the 250 people he sampled with diligence, about 185 borrowed actual books with no indifference. Using a 1% level of significance, he made a report To show that the library was not falling short He calculated the possible proportion of patrons, loyally Borrowing books in good ol' Plainsboro, Kentucky! a. Find the standard Error . b. Find the Margin Error (ME) c. What is the Confidence interval = mean +/-ME

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### Statistical Analysis of Borrowing Habits at Plainsboro Library

#### Study Overview

**Problem Context:**
The libraries across the country are facing a crisis with the increasing number of patrons opting for alternatives to borrowing books in person. Budget cuts are contributing to the decline of many libraries. According to the American Library Association, only 57% of patrons borrow books nowadays. However, the library director in Plainsboro, Kentucky, is skeptical of this statistic and hires a statistics expert to conduct a detailed study.

**Research Findings:**

The expert surveyed a sample of 250 people and found that 185 of them actually borrowed books. This study was conducted with a 1% level of significance.

#### Analyzing the Data

**a. Find the standard error:**

\[ \text{Standard Error (SE)} = \sqrt{\frac{p(1-p)}{n}} \]
Where:
- \( p \) is the sample proportion
- \( n \) is the sample size

**b. Find the Margin of Error (ME):**

\[ \text{Margin of Error (ME)} = z \times SE \]
Where \( z \) is the z-score corresponding to the desired confidence level

**c. Calculate the Confidence Interval:**

\[ \text{Confidence Interval} = \text{mean} \pm \text{Margin of Error (ME)} \]

#### Detailed Explanation:

1. **Sample Proportion (p):** 
   \[ p = \frac{185}{250} = 0.74 \]

2. **Standard Error (SE):**
   Using the sample proportion \( p \) and sample size \( n = 250 \),
   \[ SE = \sqrt{\frac{0.74 \times (1 - 0.74)}{250}} \]

3. **Z-Score for 1% Level of Significance:**
   The z-score for a 99% confidence level (1% significance level) typically approximates to 2.576.

4. **Margin of Error (ME):**
   \[ ME = 2.576 \times SE \]

5. **Confidence Interval:**
   The confidence interval can be calculated using the sample mean:
   \[ \text{95% Confidence Interval} = 0.74 \pm \text{ME} \]

This detailed analysis helps in understanding the library's borrowing habits and to determine whether the American Library
Transcribed Image Text:### Statistical Analysis of Borrowing Habits at Plainsboro Library #### Study Overview **Problem Context:** The libraries across the country are facing a crisis with the increasing number of patrons opting for alternatives to borrowing books in person. Budget cuts are contributing to the decline of many libraries. According to the American Library Association, only 57% of patrons borrow books nowadays. However, the library director in Plainsboro, Kentucky, is skeptical of this statistic and hires a statistics expert to conduct a detailed study. **Research Findings:** The expert surveyed a sample of 250 people and found that 185 of them actually borrowed books. This study was conducted with a 1% level of significance. #### Analyzing the Data **a. Find the standard error:** \[ \text{Standard Error (SE)} = \sqrt{\frac{p(1-p)}{n}} \] Where: - \( p \) is the sample proportion - \( n \) is the sample size **b. Find the Margin of Error (ME):** \[ \text{Margin of Error (ME)} = z \times SE \] Where \( z \) is the z-score corresponding to the desired confidence level **c. Calculate the Confidence Interval:** \[ \text{Confidence Interval} = \text{mean} \pm \text{Margin of Error (ME)} \] #### Detailed Explanation: 1. **Sample Proportion (p):** \[ p = \frac{185}{250} = 0.74 \] 2. **Standard Error (SE):** Using the sample proportion \( p \) and sample size \( n = 250 \), \[ SE = \sqrt{\frac{0.74 \times (1 - 0.74)}{250}} \] 3. **Z-Score for 1% Level of Significance:** The z-score for a 99% confidence level (1% significance level) typically approximates to 2.576. 4. **Margin of Error (ME):** \[ ME = 2.576 \times SE \] 5. **Confidence Interval:** The confidence interval can be calculated using the sample mean: \[ \text{95% Confidence Interval} = 0.74 \pm \text{ME} \] This detailed analysis helps in understanding the library's borrowing habits and to determine whether the American Library
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