Lecture 25: Confidence Intervals – Proportions, Means, and the t-Distribution

Confidence Intervals

Introduction to Confidence Intervals

Confidence intervals are a fundamental concept in inferential statistics, providing a range of plausible values for an unknown population parameter based on sample data. They quantify the uncertainty associated with sample estimates and are widely used for both proportions and means.

• Confidence Interval (CI): An interval estimate, calculated from the sample, that is likely to contain the true value of a population parameter with a specified probability (confidence level).

• Confidence Level: The probability (e.g., 95%) that the interval contains the true parameter value in repeated sampling.

Confidence Intervals for Proportions

Constructing a Confidence Interval for a Population Proportion

Suppose a simple random sample of size n is taken from a population, or the data are from a randomized experiment. The (1-α)100% confidence interval for the population proportion p is given by:

• Lower bound: $\hat{p} - z_{\alpha/2} \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

• Upper bound: $\hat{p} + z_{\alpha/2} \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

Conditions: To use this interval, ensure that $n\hat{p}(1-\hat{p}) \geq 10$ and $n \leq 0.05N$ (sample size is less than 5% of the population).

Example: Proportion Confidence Interval

Suppose a random sample of 100 people shows that 25 are left-handed. Find a 95% confidence interval for the true proportion of left-handers.

• Point estimate: $\hat{p} = \dfrac{25}{100} = 0.25$

• Margin of error: $E = 1.96 \times \sqrt{\dfrac{0.25 \times 0.75}{100}} = 1.96 \times 0.0433 = 0.085$

• CI: $0.25 \pm 0.085 = (0.165, 0.335)$

Margin of Error for Proportions

The margin of error quantifies the maximum expected difference between the true population parameter and a sample estimate:

• $E = z_{\alpha/2} \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

Example: Survey on Teen Texting While Driving

In a survey of 800 teenagers, 272 reported texting while driving. Find a 95% confidence interval for the proportion.

• $\hat{p} = \dfrac{272}{800} = 0.34$

• Check: $n\hat{p}(1-\hat{p}) = 800 \times 0.34 \times 0.66 = 179.52 \geq 10$

• Margin of error: $E = 1.96 \times \sqrt{\dfrac{0.34 \times 0.66}{800}} = 0.033$

• CI: $0.34 \pm 0.033 = (0.307, 0.373)$

Interpretation: We are 95% confident that the true proportion of 16- to 17-year-olds who text while driving is between 0.307 and 0.373.

Effect of Confidence Level on Margin of Error

• Increasing the confidence level (e.g., from 95% to 99%) increases the margin of error, resulting in a wider confidence interval.

• Example: For the same data, a 99% CI uses $z_{0.005} = 2.575$ and yields $E = 0.043$.

Factors Affecting Margin of Error

• Level of confidence: Higher confidence increases margin of error.

• Sample size: Larger sample size decreases margin of error.

• Population standard deviation: Greater variability increases margin of error.

Sample Size Determination for Proportions

Determining Required Sample Size

To achieve a desired margin of error E at a given confidence level, the minimum sample size n is:

• If $\hat{p}$ is known: $n = \hat{p}(1-\hat{p}) \left( \dfrac{z_{\alpha/2}}{E} \right)^2$

• If $\hat{p}$ is unknown: $n = 0.25 \left( \dfrac{z_{\alpha/2}}{E} \right)^2$ (using $\hat{p} = 0.5$ for maximum variability)

Always round up to the next integer.

Example: Carpooling Survey

• Desired margin of error: 2% ($E = 0.02$), 90% confidence ($z_{0.05} = 1.645$)

• (a) Using prior estimate $\hat{p} = 0.10$: $n = 0.10 \times 0.90 \times \left( \dfrac{1.645}{0.02} \right)^2 = 608.9 \rightarrow 609$

• (b) No prior estimate: $n = 0.25 \times \left( \dfrac{1.645}{0.02} \right)^2 = 1691.3 \rightarrow 1692$

Conclusion: Not having a prior estimate of $p$ can more than double the required sample size.

Point and Interval Estimation for the Mean

Point Estimation for the Mean

• Point estimate: The value of a statistic used to estimate a parameter. For the mean, the sample mean $\bar{X}$ is the point estimate for the population mean $\mu$.

• Formula: $\bar{X} = \dfrac{1}{n} \sum_{i=1}^n X_i$

Interval Estimation for the Mean

The (1-α)100% confidence interval for the population mean $\mu$ depends on:

• Sample size: Large ($n \geq 30$) or small ($n < 30$)

• Population standard deviation $\sigma$: Known or unknown

• Sampling distribution of the sample mean

Summary Table: Confidence Intervals for the Mean

	Large Sample ($n \geq 30$)	Small Sample ($n < 30$)
$\sigma$ known	$\bar{X} \pm z_{\alpha/2} \dfrac{\sigma}{\sqrt{n}}$	$\bar{X} \pm t_{\alpha/2, n-1} \dfrac{s}{\sqrt{n}}$
$\sigma$ unknown	$\bar{X} \pm z_{\alpha/2} \dfrac{s}{\sqrt{n}}$	$\bar{X} \pm t_{\alpha/2, n-1} \dfrac{s}{\sqrt{n}}$

Confidence Interval for $\mu$, $\sigma$ Known

• Assumptions: Simple random sample, population standard deviation $\sigma$ known, population normal or $n \geq 30$

• Formula: $\bar{X} \pm z_{\alpha/2} \dfrac{\sigma}{\sqrt{n}}$

Example

• Sample: $n = 100$, $\bar{X} = 72.5$, $\sigma = 12$, 95% CI ($z_{0.025} = 1.96$)

• CI: $72.5 \pm 1.96 \times \dfrac{12}{10} = 72.5 \pm 2.35 = (70.15, 74.85)$

Interpretation: We are 95% confident that the interval (70.15, 74.85) contains the true population mean.

Confidence Interval for $\mu$, $\sigma$ Unknown

• For large $n$ ($n \geq 30$): Replace $\sigma$ with sample standard deviation $s$.

• Formula: $\bar{X} \pm z_{\alpha/2} \dfrac{s}{\sqrt{n}}$

• For small $n$ ($n < 30$): Use Student's t-distribution.

• Formula: $\bar{X} \pm t_{\alpha/2, n-1} \dfrac{s}{\sqrt{n}}$

Student's t-Distribution

Definition and Properties

When the population standard deviation is unknown and the sample size is small, the sampling distribution of the sample mean follows the Student's t-distribution with $n-1$ degrees of freedom.

• The t-distribution is symmetric and centered at 0.

• It has fatter tails than the standard normal distribution, reflecting greater uncertainty.

• As sample size increases, the t-distribution approaches the standard normal distribution.

Properties of the t-Distribution

1. Different for each degree of freedom (df = n-1).

2. Centered at 0 and symmetric.

3. Total area under the curve is 1; area to the right of 0 equals area to the left.

4. As df increases, the curve approaches the normal distribution.

5. More area in the tails than the normal distribution (reflects extra variability from estimating $\sigma$ with $s$).

Critical Values and Confidence Intervals Using t-Distribution

• Critical value $t_{\alpha/2, n-1}$ is used instead of $z_{\alpha/2}$.

• Formula: $\bar{X} \pm t_{\alpha/2, n-1} \dfrac{s}{\sqrt{n}}$

Example: Confidence Interval with t-Distribution

• Sample: $n = 20$, $\bar{X} = 12.5$, $s = 5$

• 95% CI ($t_{0.025, 19} = 2.093$): $12.5 \pm 2.093 \times \dfrac{5}{\sqrt{20}} = 12.5 \pm 2.34 = (10.16, 14.84)$

• 98% CI ($t_{0.01, 19} = 2.539$): $12.5 \pm 2.539 \times \dfrac{5}{\sqrt{20}} = 12.5 \pm 2.84 = (9.66, 15.34)$

Normality Condition and Robustness

When to Use the t-Distribution

• For $n < 30$, check that the data are approximately normal and have no outliers (use normal probability plot and boxplot).

• For $n < 15$, use t-distribution only if data are symmetric and have no outliers.

• For $15 < n < 30$, t-distribution can be used if data are not extremely skewed and have no outliers.

• For $n \geq 30$, t-distribution can be used even for skewed distributions (Central Limit Theorem applies).

Summary Table: Confidence Intervals for the Mean

	Large Sample ($n \geq 30$)	Small Sample ($n < 30$)
$\sigma$ known	$\bar{X} \pm z_{\alpha/2} \dfrac{\sigma}{\sqrt{n}}$	$\bar{X} \pm t_{\alpha/2, n-1} \dfrac{s}{\sqrt{n}}$
$\sigma$ unknown	$\bar{X} \pm z_{\alpha/2} \dfrac{s}{\sqrt{n}}$	$\bar{X} \pm t_{\alpha/2, n-1} \dfrac{s}{\sqrt{n}}$

Additional Information

• Student's t-distribution was discovered by William Sealy Gosset, who published under the pseudonym "Student" while working at Guinness brewery.

• The t-distribution is robust to minor departures from normality, especially as sample size increases.