Statistical Inference

📖 Learn

Statistical inference is the process of using data from a sample to draw conclusions about a larger population. Rather than measuring every individual in a population, we use sample statistics to estimate population parameters and assess the reliability of those estimates.

Key Terminology

Term	Definition	Symbol
Population	The entire group you want to study	N (size)
Sample	A subset of the population you actually measure	n (size)
Parameter	A numerical value describing the population	μ (mean), σ (std dev), p (proportion)
Statistic	A numerical value calculated from sample data	x̄ (mean), s (std dev), p̂ (proportion)

Sampling Methods

Random Sampling

For valid statistical inference, samples must be collected using proper random sampling methods:

Simple Random Sample (SRS): Every individual has an equal chance of being selected
Stratified Random Sample: Population divided into groups (strata), then SRS from each
Cluster Sample: Randomly select groups (clusters), then sample all in selected clusters
Systematic Sample: Select every kth individual from a list

Bias in Sampling

Selection bias: Some individuals are more likely to be chosen

Response bias: Answers are influenced by question wording or social desirability

Nonresponse bias: People who don't respond differ from those who do

Biased samples lead to unreliable conclusions about the population.

Sampling Distributions

Sampling Distribution

A sampling distribution shows the distribution of a statistic (like x̄ or p̂) across all possible samples of a given size from a population. It tells us how much we expect the statistic to vary from sample to sample.

Central Limit Theorem (CLT)

Central Limit Theorem

For sufficiently large sample sizes (typically n ≥ 30), the sampling distribution of the sample mean x̄ is approximately normal, regardless of the shape of the population distribution.

Mean of x̄: μx̄ = μ

Standard Error: SE = σ/√n

Confidence Intervals

Confidence Interval

A confidence interval is a range of values that is likely to contain the true population parameter with a specified level of confidence.

CI = point estimate ± margin of error

CI for mean: x̄ ± z* · (σ/√n)

CI for proportion: p̂ ± z* · √(p̂(1-p̂)/n)

Common Critical Values (z*)

Confidence Level	z*	Interpretation
90%	1.645	90% of intervals capture the true parameter
95%	1.96	95% of intervals capture the true parameter
99%	2.576	99% of intervals capture the true parameter

Interpreting Confidence Intervals

Correct Interpretation

"We are 95% confident that the true population mean lies between [lower bound] and [upper bound]."

This means: If we took many samples and built intervals the same way, about 95% of them would contain the true parameter.

Common mistake: Saying "There is a 95% probability that μ is in this interval." (The parameter is fixed, not random.)

Margin of Error

MOE = z* · (σ/√n) or MOE = z* · √(p̂(1-p̂)/n)

The margin of error depends on:

Confidence level: Higher confidence → larger MOE
Sample size: Larger n → smaller MOE
Variability: More spread → larger MOE

Sample Size Determination

Required Sample Size

To achieve a desired margin of error E:

For means: n = (z* · σ / E)²

For proportions: n = p̂(1-p̂) · (z* / E)²

If p̂ is unknown, use p̂ = 0.5 for the most conservative (largest) sample size.

💡 Examples

Example 1: Constructing a Confidence Interval for a Mean

Problem: A sample of 64 students has a mean test score of x̄ = 72 with a known population standard deviation σ = 8. Construct a 95% confidence interval for the population mean.

Step 1: Identify the values:

x̄ = 72, σ = 8, n = 64, z* = 1.96 (for 95%)

Step 2: Calculate the standard error:

SE = σ/√n = 8/√64 = 8/8 = 1

Step 3: Calculate the margin of error:

MOE = z* · SE = 1.96 · 1 = 1.96

Step 4: Construct the interval:

CI = x̄ ± MOE = 72 ± 1.96

CI = (70.04, 73.96)

Answer: We are 95% confident that the true population mean test score is between 70.04 and 73.96.

Example 2: Confidence Interval for a Proportion

Problem: In a survey of 400 voters, 220 support a ballot measure. Construct a 90% confidence interval for the true proportion of supporters.

Step 1: Calculate the sample proportion:

p̂ = 220/400 = 0.55

Step 2: Check conditions (np̂ ≥ 10 and n(1-p̂) ≥ 10):

400(0.55) = 220 ≥ 10 ✓

400(0.45) = 180 ≥ 10 ✓

Step 3: Calculate standard error:

SE = √(p̂(1-p̂)/n) = √(0.55 · 0.45/400) = √(0.000619) = 0.0249

Step 4: Calculate MOE (z* = 1.645 for 90%):

MOE = 1.645 · 0.0249 = 0.041

Step 5: Construct interval:

CI = 0.55 ± 0.041 = (0.509, 0.591)

Answer: We are 90% confident that between 50.9% and 59.1% of all voters support the measure.

Example 3: Interpreting a Confidence Interval

Problem: A 95% confidence interval for the mean weight of apples from an orchard is (180g, 200g). Which statements are correct?

a) 95% of all apples weigh between 180g and 200g.

b) There is a 95% probability that μ is between 180g and 200g.

c) We are 95% confident that the true mean weight is between 180g and 200g.

a) INCORRECT: The CI is for the mean, not individual values. Individual apple weights have much more variation.

b) INCORRECT: The parameter μ is fixed (though unknown). It's either in the interval or not. Probability doesn't apply.

c) CORRECT: This properly expresses confidence in the method. If we repeated this process many times, about 95% of the intervals would contain μ.

Example 4: Determining Sample Size

Problem: A researcher wants to estimate the mean commute time with a margin of error of 2 minutes at 95% confidence. If σ = 10 minutes, how large a sample is needed?

Formula: n = (z* · σ / E)²

where z* = 1.96, σ = 10, E = 2

Calculate:

n = (1.96 · 10 / 2)² = (9.8)² = 96.04

Answer: n = 97 (always round up for sample size)

The researcher needs at least 97 participants.

Example 5: Effect of Sample Size on MOE

Problem: A 95% CI based on n = 100 has MOE = 4. What would the MOE be if n = 400?

Key insight: MOE is proportional to 1/√n

When n quadruples (100 → 400), √n doubles (10 → 20)

So MOE is halved: 4 → 2

Verification:

MOE₁/MOE₂ = √(n₂/n₁) = √(400/100) = 2

MOE₂ = MOE₁/2 = 4/2 = 2

Answer: The new MOE would be 2.

✏️ Practice

Apply your understanding of statistical inference.

Problem 1: A random sample of 49 light bulbs has a mean life of 1,200 hours. If σ = 140 hours, construct a 95% confidence interval for the mean life of all bulbs.

Problem 2: In a poll of 600 adults, 342 favor a new policy. Construct a 99% confidence interval for the true proportion.

Problem 3: A 90% confidence interval is (45, 55). (a) What is the point estimate? (b) What is the margin of error?

Problem 4: If we increase the sample size from 100 to 900, by what factor does the margin of error change?

Problem 5: A researcher wants a 95% CI for a proportion with MOE ≤ 0.03. What sample size is needed? (Use p̂ = 0.5)

Problem 6: Explain why convenience samples (like surveying friends) cannot be used for valid statistical inference.

Problem 7: A 95% CI for mean salary is ($48,000, $56,000). Can we conclude the population mean is $50,000?

Problem 8: Two researchers study the same population. Researcher A uses n = 50 and gets CI (10, 20). Researcher B uses n = 200. Will B's interval be wider, narrower, or the same? Why?

Problem 9: Calculate the standard error for x̄ if σ = 15 and n = 225.

Problem 10: A survey asks "Don't you agree this harmful policy should be stopped?" What type of bias might this cause?

Click to reveal answers

SE = 140/√49 = 20; CI = 1200 ± 1.96(20) = (1160.8, 1239.2) hours
p̂ = 0.57; SE = √(0.57·0.43/600) = 0.0202; CI = 0.57 ± 2.576(0.0202) = (0.518, 0.622)
(a) Point estimate = (45+55)/2 = 50 (b) MOE = (55-45)/2 = 5
MOE decreases by factor of √(900/100) = 3; new MOE is 1/3 of original
n = 0.5(0.5)(1.96/0.03)² = 0.25(4268.4) = 1067.1, so n = 1068
Convenience samples are biased because they don't give everyone in the population an equal chance of being selected. Results can't be generalized to the population.
We cannot conclude the mean IS $50,000. We can only say $50,000 is a plausible value since it's within the interval. Many values are plausible.
B's interval will be narrower because larger sample sizes produce smaller margins of error.
SE = σ/√n = 15/√225 = 15/15 = 1
Response bias due to leading question wording. The question pushes respondents toward agreeing.

✅ Check Your Understanding

Question 1: What is the difference between a parameter and a statistic?

Show answer

A parameter is a numerical value that describes a characteristic of the entire population (e.g., population mean μ). It is usually unknown and fixed.

A statistic is a numerical value calculated from sample data (e.g., sample mean x̄). It is known but varies from sample to sample.

We use statistics to estimate parameters.

Question 2: Why is random sampling important for statistical inference?

Show answer

Random sampling is essential because:

It eliminates selection bias by giving everyone an equal chance of being chosen
It allows us to use probability theory to quantify uncertainty
It makes our sample representative of the population
It justifies generalizing from sample to population

Without random sampling, we cannot make valid inferences about the population.

Question 3: How does increasing the confidence level affect the width of a confidence interval?

Show answer

Increasing the confidence level makes the interval wider. To be more confident that we capture the true parameter, we need a larger range of values.

For example, a 99% CI is wider than a 95% CI for the same data because we need more "room" to be more certain we've captured the parameter.

The trade-off: Higher confidence = wider (less precise) interval.

Question 4: What does the Central Limit Theorem tell us and why is it important?

Show answer

The Central Limit Theorem states that for large enough sample sizes, the sampling distribution of the sample mean is approximately normal, regardless of the population's shape.

This is crucial because:

It allows us to use normal distribution methods even when the population isn't normal
It provides the mathematical foundation for confidence intervals and hypothesis tests
It explains why the normal distribution appears so often in statistics
It tells us exactly how much sample means vary (via the standard error formula)

🚀 Next Steps

Review any concepts that felt challenging
Move on to the next lesson when ready
Return to practice problems periodically for review