Day 9

MATH 313: Survey Design and Sampling

Bastola

Overview of Population Proportion Estimation

Framework: Observations in a sample are indicators (\(Y_i\)) representing the presence (1) or absence (0) of a characteristic.
Mathematical Basis: \(Y_i\) follows a Binomial distribution, useful in estimating proportions where outcomes are binary.
Calculation: The sample proportion \(\hat{p}\) is computed as \(\hat{p} = \bar{y} = \frac{1}{n} \sum Y_i\), where \(n\) is the sample size.

Political Campaign Analysis

Objective: Estimate the proportion of voters supporting a new policy.
Survey: Randomly sample 1,500 voters; \(Y_i=1\) if a voter supports, and 0 otherwise.
Statistical Relevance: Results direct strategic decisions and policy development with precision due to the binary nature of survey responses.

Educational Research

Study Goal: Assess the integration rate of digital tools in classrooms.
Data Collection: A survey of 350 schools, where \(Y_i=1\) indicates digital tool adoption.
Implications: Estimating \(\hat{p}\) assists in evaluating the effectiveness of tech policies and funding allocations in education.

Environmental Conservation Efforts

Focus: Measure public support for new environmental conservation laws.
Methodology: Conduct a survey where each response \(Y_i=1\) if the individual supports conservation efforts.
Statistical Analysis: Utilizing the binomial model to estimate \(p\), the proportion of the population favoring conservation, guiding legislative and community actions.

Bernoulli Distribution: Estimation

Binary Data: \(Y_i = 1\) for responses in the category; \(0\) otherwise.
Total Responses: \(\sum Y_i\) counts responses in the category.
Proportion Estimate:
- \(\hat{p} = \bar{Y} = \frac{1}{n} \sum Y_i\), the sample proportion.
Bernoulli Distribution: \(Y_i\) follows with success \(p\).
Variance: \(\sigma^2 = pq\) where \(q = 1 - p\).

Estimation of Population Proportion

Estimator for Proportion \(p\):
- \(\hat{p}=\bar{y}=\frac{1}{n}\sum_{i=1}^n y_i\)
Variance of Estimator \(\hat{p}\):
- \(\hat{V}(\hat{p})=\left(1-\frac{n}{N}\right)\cdot\frac{\hat{p}(1-\hat{p})}{n-1}\)
- Using \(n-1\) provides an unbiased estimator.
Error Bound on Estimation:
- \(B_{\hat{p}}=t_{1-\frac{\alpha}{2}, n-1}\cdot\sqrt{\hat{V}(\hat{p})}\)
- Represents the \((1-\alpha)\) confidence interval for \(\hat{p}\).

Example 1: A simple random sample of \(n=200\) college seniors was selected to estimate the proportion of \(N=1200\) seniors going on to graduate school. Suppose 37 of them indicated that he or she plans to attend graduate school. Using the sample data:

Estimate \(p\), the proportion of seniors planing to attend graduate school.

Compute the \(95 \%\) bounds on the error of estimation.

Compute the \(95 \%\) confidence interval for the estimation.

Estimate the number of seniors who plan to attend graduate college.

Sample Size for Estimating \(p\)

Sample Size with Known \(\hat{p}\):
- \(n=\frac{N \hat{p} \hat{q}}{(N-1)\left(\frac{B_{\hat{p}}}{Z_{1-\frac{\alpha}{2}}}\right)^2+\hat{p} \hat{q}}\)
- Uses \(\hat{p}\) from similar past surveys to minimize error bound \(B_{\hat{p}}\).
Sample Size with Unknown \(\hat{p}\):
- \(n=\frac{N(0.5)(0.5)}{(N-1)\left(\frac{B_p}{Z_{1-\frac{\alpha}{2}}}\right)^2+(0.5)(0.5)}\)
- Applies a conservative estimate, assuming \(\hat{p}=0.5\), to ensure a robust sample size calculation.

Example 2: Student government leaders at a college want to conduct a survey to determine the proportion of students who favor a proposed honor code. Because interviewing \(N=2000\) students in a reasonable length of time is almost impossible, determine the sample size (numbers of students to be interviewed) needed to estimate \(p\) with a \(95 \%\) bound on the error of estimation \(B=0.05\) for the following cases:

No prior information is available to estimate \(p\)

A similar survey performed in another college provides \(\hat{p}=0.61\).

Example 3:Suppose you are a public health researcher aiming to estimate the proportion of residents in a large city who are willing to participate in a new health program designed to improve community health outcomes. From previous smaller-scale studies conducted in similar demographic settings, you estimate that around 45% of the population is likely to participate (\(\hat{p} = 0.45\)). The city has a population of 100,000 residents (\(N = 100,000\)), and you want to design a survey that will provide a reliable estimate of the population proportion with a high level of confidence.

Calculate the minimum sample size required to estimate the population proportion with a margin of error (\(B_{\hat{p}}\)) of 5% at a 95% confidence level.

Once the sample size is determined, simulate survey data assuming the estimated proportion is accurate, then compute the 95% confidence interval for the population proportion based on the sample data.

Click for answer

# Load necessary library
library(stats)
set.seed(123)
# Given values
p_hat <- 0.45
q_hat <- 1 - p_hat
N <- 100000
B_p <- 0.05
z <- qnorm(0.975)  # Z-value for 95% confidence

# Calculating sample size
n <- N * p_hat * q_hat / ((N - 1) * (B_p / z)^2 + p_hat * q_hat)
n <- ceiling(n)

# Set seed for reproducibility
set.seed(42)

# Simulate survey data
survey_responses <- rbinom(n, size = 1, prob = p_hat)
sample_p_hat <- mean(survey_responses)
sample_q_hat <- 1 - sample_p_hat
t <- qt(0.975, df = n-1)
# Calculate the standard error and confidence interval
standard_error <- sqrt((1 - n/N)*sample_p_hat * sample_q_hat / (n - 1))
confidence_interval <- c(sample_p_hat - t * standard_error, sample_p_hat + t * standard_error)

# Output results
list(sample_size = n, confidence_interval = confidence_interval)

$sample_size
[1] 379

$confidence_interval
[1] 0.3827059 0.4827295

Based on the simulated survey data, the 95% confidence interval for the proportion of residents willing to participate in the health program is approximately (0.391, 0.491).