Confidence Intervals for Population Proportion Summary

Confidence Intervals for Population Proportion

Introduction

• Deals with populations where items fall into two categories.
• Aims to construct a confidence interval for the proportion of individuals in a specified category.

Notation

• $P$: Population proportion of individuals in a specified category.
• $X$: Number of individuals in the sample in the specified category.
• $N$: Sample size.
• $\hat{P}$: Sample proportion of individuals in the specified category, calculated as $\frac{X}{N}$.

Constructing a Confidence Interval

• Point Estimate:
  • $\hat{P} = \frac{X}{N}$ is the point estimate for the population proportion $P$.
• Standard Error:
  • By the Central Limit Theorem for proportions, the standard error is $\sqrt{\frac{\hat{P}(1-\hat{P})}{N}}$.
• Margin of Error:
  • $E = Z<em>{\alpha/2} \times \sqrt{\frac{\hat{P}(1-\hat{P})}{N}}$, where $Z</em>{\alpha/2}$ is the critical value.
• Confidence Interval:
  • Point estimate plus or minus the margin of error: $\hat{P} \pm E$.

Assumptions for Approximately Normal Sampling Distribution

1. Simple Random Sample
2. Population Size: The population is at least 20 times the sample size.
3. Two Categories: Items in the population are divided into two categories.
4. Sample Size in Each Category: The sample must contain at least 10 individuals in each category.

Example: Music Education Survey

• Scenario: Survey of 800 parents, 632 believe music education has a positive effect.
• Goal: Construct a 95% confidence interval.
• Checking Assumptions:
  • Simple random sample.
  • Population of parents is more than 20 times the sample size.
  • Two categories: believe vs. don't believe.
  • 632 > 10 and 800 - 632 = 168 > 10.
• Calculations:
  • $N = 800$, $X = 632$.
  • $\hat{P} = \frac{632}{800} = 0.79$.
  • Critical value for 95% confidence: $Z_{\alpha/2} = 1.96$.
  • Margin of error: $E = 1.96 \times \sqrt{\frac{0.79(1-0.79)}{800}} = 0.028225$.
  • Confidence interval: $0.79 \pm 0.028225$, which is $(0.762, 0.818)$.
• Interpretation: We are 95% confident that the proportion of parents who believe music education has a positive effect is between 0.762 and 0.818.

Determining Sample Size

• Goal: To find the sample size necessary to obtain a confidence interval with a given width (margin of error).
• Problem: A large margin of error makes the confidence interval too wide to be useful.
• Solutions:
  • Reduce the level of confidence (undesirable).
  • Increase the sample size.
• Formula Derivation:
  • Starting with the margin of error formula: $E = Z_{\alpha/2} \times \sqrt{\frac{\hat{P}(1-\hat{P})}{N}}$.
  • Rearranging to solve for N: $N = \hat{P}(1-\hat{P}) \times (\frac{Z_{\alpha/2}}{E})^2$.
• Using the Formula:
  • This formula gives the minimum sample size needed for a desired margin of error.
  • Always round the result up to the nearest whole number.
• Estimating $\hat{P}$:
  • If an estimate is available from a prior study, use it.
  • If no estimate is available, use $\hat{P} = 0.5$. This maximizes the margin of error for any sample size.

Example: Defective Computer Chips

• Scenario: Preliminary study of 100 chips, 17 found defective. Want to construct a 95% confidence interval with a margin of error of 0.06 for a new sample.
• Knowns:
  • $Z_{\alpha/2} = 1.96$ (for 95% confidence).
  • $E = 0.06$.
  • $\hat{P} = \frac{17}{100} = 0.17$ (from the preliminary study).
• Calculation:
  • $N = 0.17(1-0.17) \times (\frac{1.96}{0.06})^2 = 150.5694$.
• Result: Round up to $N = 151$. 151 computer chips must be tested.

Example: Incumbent Mayor Election

• Scenario: Conduct a survey for residents supporting a major with 90% confidence interval and margin of error no larger than 0.04.
• Knowns:
  • $Z_{\alpha/2} = 1.645$ (for 90% confidence).
  • $E = 0.04$.
  • No prior estimate of P, so use $\hat{P} = 0.5$.
• Calculation:
  • $N = 0.5(1-0.5) \times (\frac{1.645}{0.04})^2 = 422.8164$.
• Result: Round up to $N = 423$. Need to poll 423 residents.

Example: Internet Service Interruption

• Scenario: 540 customers sampled, 75 experienced an interruption.
  • Part A: Find a point estimate for the population proportion.
  • Part B: Construct a 90% confidence interval.
  • Part C: Determine if a claim that no more than 10% experienced interruptions is contradicted by the interval.
• Checking Assumptions:
  • Simple random sample.
  • Population is at least 20 times sample size.
  • Two categories: interruption vs. no interruption.
  • 75 > 10 and 540 - 75 = 465 > 10.
• Procedure Steps:
  1. Find the point estimate $\hat{P}$.
  2. Find the critical value $Z_{\alpha/2}$.
  3. Compute the standard error and margin of error.
  4. Construct the confidence interval: $\hat{P} \pm E$.
  5. Interpret the result.
• Calculations:
  • $N = 540$, $X = 75$.
  • $\hat{P} = \frac{75}{540} = 0.1389$.
  • $Z_{\alpha/2} = 1.645$ (for 90% confidence).
  • Standard error: $\sqrt{\frac{0.1389(1-0.1389)}{540}} = 0.0149$.
  • Margin of error: $1.645 \times 0.0149 = 0.0245$.
  • Confidence interval: $0.1389 \pm 0.0245 = (0.1144, 0.1634)$.
• Interpretation:
  • We are 90% confident that the proportion of customers who experienced an interruption is between 11.4% and 16.3%.
  • The quality control manager's claim of no more than 10% is contradicted by the confidence interval.

Example: Adults Believing Economic Conditions are Improving

• Part A: Calculating Sample Size with Prior Estimate
• Part B: Calculating Sample Size without Prior Estimate
• Formulas:
  • With Prior Estimate: $N = \hat{P}(1-\hat{P}) \times (\frac{Z_{\alpha/2}}{M})^2$
  • Without Prior Estimate: $N = 0.25 \times (\frac{Z_{\alpha/2}}{M})^2$
• Given:
  • 95% confidence interval ($Z_{\alpha/2} = 1.96$)
  • Margin of Error ($M = 0.03$)
• Part A (With Estimate $\hat{P} = 0.33$):
  • $N = 0.33(1-0.33) \times (\frac{1.96}{0.03})^2 = 943.8$
  • Rounded Up: $N = 944$
• Part B (Without Estimate, Using 0.25):
  • $N = 0.25 \times (\frac{1.96}{0.03})^2 = 1067.1$
  • Rounded Up: $N = 1068$