Hypothesis Tests for Means
Master z-tests and t-tests for testing claims about population means
Lesson Objectives
By the end of this lesson, you will be able to:
- Determine when to use a z-test versus a t-test
- Calculate test statistics for means (z and t)
- Find critical values and make decisions using the critical value approach
- Calculate p-values and make decisions using the p-value approach
- Interpret test results in context
1. Z-Test vs. T-Test: When to Use Which?
When testing hypotheses about population means, we use either a z-test or a t-test:
| Test | When to Use | Requirements |
|---|---|---|
| Z-Test | Population standard deviation (σ) is KNOWN |
• σ is known • n ≥ 30 OR population is normal |
| T-Test | Population standard deviation (σ) is UNKNOWN |
• σ is unknown (use sample s) • n ≥ 30 OR population is normal |
Decision Flowchart
Testing a claim about μ (population mean)?
↓
Do we know the population standard deviation (σ)?
- YES, σ is known → Use Z-TEST
- NO, σ is unknown → Use T-TEST (use sample standard deviation s)
2. Z-Test for Population Mean
When to Use
Use the z-test when:
- You're testing a claim about a population mean (μ)
- The population standard deviation (σ) is known
- Either n ≥ 30 OR the population is normally distributed
Test Statistic for Z-Test
Where:
x̄ = sample mean
μ₀ = hypothesized population mean (from H₀)
σ = population standard deviation
n = sample size
Critical Value Approach
Steps:
- State H₀ and Hₐ, choose α
- Calculate test statistic z
- Find critical value(s) from z-table based on α and test type
- Compare test statistic to critical value(s)
- Make decision and interpret
| Test Type | Critical Value(s) | Rejection Rule |
|---|---|---|
| Two-tailed (Hₐ: μ ≠ μ₀) |
α = 0.05: ±1.96 α = 0.01: ±2.576 |
Reject H₀ if |z| > critical value |
| Right-tailed (Hₐ: μ > μ₀) |
α = 0.05: 1.645 α = 0.01: 2.326 |
Reject H₀ if z > critical value |
| Left-tailed (Hₐ: μ < μ₀) |
α = 0.05: -1.645 α = 0.01: -2.326 |
Reject H₀ if z < critical value |
Example 1: Z-Test (Two-Tailed)
Problem: A manufacturer claims light bulbs have a mean lifespan of 1000 hours. A consumer group tests 50 bulbs and finds x̄ = 980 hours. The population standard deviation is known to be σ = 60 hours. Test at α = 0.05.
Step 1: State hypotheses
- H₀: μ = 1000 (manufacturer's claim is correct)
- Hₐ: μ ≠ 1000 (claim is incorrect) — two-tailed test
- α = 0.05
Step 2: Calculate test statistic
z = (980 - 1000) / (60 / √50)
z = -20 / 8.49
z = -2.36
Step 3: Find critical value
Two-tailed test, α = 0.05 → critical values = ±1.96
Step 4: Make decision
|z| = 2.36 > 1.96, so z is in the critical region
Decision: Reject H₀
Step 5: Conclusion
There is sufficient evidence at the 0.05 significance level to reject the manufacturer's claim that the mean lifespan is 1000 hours. The bulbs appear to have a shorter lifespan than claimed.
P-Value Approach
Definition: P-Value
The p-value is the probability of obtaining a test statistic as extreme as (or more extreme than) the one calculated, assuming the null hypothesis is true.
Decision Rule:
- If p-value ≤ α → Reject H₀
- If p-value > α → Fail to reject H₀
- p < 0.01: Very strong evidence against H₀
- 0.01 ≤ p < 0.05: Strong evidence against H₀
- 0.05 ≤ p < 0.10: Weak evidence against H₀
- p ≥ 0.10: Little or no evidence against H₀
Example 2: P-Value Approach (Right-Tailed)
Problem: A gym claims their program increases vertical jump by more than 5 inches. Test 40 participants: x̄ = 6.2 inches, σ = 2.5 inches. Use α = 0.05.
Step 1: State hypotheses
- H₀: μ = 5 (no more than 5-inch increase)
- Hₐ: μ > 5 (more than 5-inch increase) — right-tailed
- α = 0.05
Step 2: Calculate test statistic
z = 1.2 / 0.395
z = 3.04
Step 3: Find p-value
For right-tailed test: p-value = P(Z > 3.04)
From z-table: P(Z < 3.04) = 0.9988
p-value = 1 - 0.9988 = 0.0012
Step 4: Make decision
p-value (0.0012) < α (0.05)
Decision: Reject H₀
Step 5: Conclusion
There is very strong evidence (p = 0.0012) that the gym's program increases vertical jump by more than 5 inches.
3. T-Test for Population Mean
When to Use
Use the t-test when:
- You're testing a claim about a population mean (μ)
- The population standard deviation (σ) is unknown
- Either n ≥ 30 OR the population is normally distributed
Test Statistic for T-Test
Where:
x̄ = sample mean
μ₀ = hypothesized population mean (from H₀)
s = sample standard deviation
n = sample size
Degrees of freedom: df = n - 1
- We use sample standard deviation s instead of σ
- We use the t-distribution instead of standard normal distribution
- Critical values depend on degrees of freedom (df = n - 1)
- The t-distribution has heavier tails than normal distribution (more conservative)
Finding Critical Values for T-Test
Critical values come from the t-distribution table using:
- Degrees of freedom: df = n - 1
- Significance level: α
- Test type: one-tailed or two-tailed
Example 3: T-Test (Left-Tailed)
Problem: A weight-loss program claims participants lose 10 pounds on average. A study of 25 participants shows x̄ = 8.2 pounds, s = 3.5 pounds. Test if weight loss is less than claimed (α = 0.05).
Step 1: State hypotheses
- H₀: μ = 10 (weight loss is 10 pounds as claimed)
- Hₐ: μ < 10 (weight loss is less than 10 pounds) — left-tailed
- α = 0.05
Step 2: Calculate test statistic
t = (8.2 - 10) / (3.5 / √25)
t = -1.8 / 0.7
t = -2.57
Step 3: Find critical value
df = n - 1 = 25 - 1 = 24
Left-tailed test, α = 0.05, df = 24 → critical value = -1.711 (from t-table)
Step 4: Make decision
t = -2.57 < -1.711, so t is in the critical region
Decision: Reject H₀
Step 5: Conclusion
There is sufficient evidence at the 0.05 significance level that the average weight loss is less than the claimed 10 pounds. The program appears to be less effective than advertised.
Example 4: T-Test (Two-Tailed) with P-Value
Problem: A coffee shop claims the average wait time is 5 minutes. A customer times 30 visits: x̄ = 5.8 minutes, s = 1.5 minutes. Test at α = 0.01.
Step 1: State hypotheses
- H₀: μ = 5 (wait time is 5 minutes)
- Hₐ: μ ≠ 5 (wait time differs from 5 minutes) — two-tailed
- α = 0.01
Step 2: Calculate test statistic
t = 0.8 / 0.274
t = 2.92
Step 3: Find critical value and p-value
df = 30 - 1 = 29
Two-tailed test, α = 0.01, df = 29 → critical values = ±2.756
For t = 2.92, df = 29: p-value ≈ 0.007 (from t-table or calculator)
Step 4: Make decision
Critical value approach: |t| = 2.92 > 2.756 → Reject H₀
P-value approach: p-value (0.007) < α (0.01) → Reject H₀
Step 5: Conclusion
There is strong evidence (p = 0.007) that the average wait time is different from the claimed 5 minutes. The actual average appears to be higher.
Check Your Understanding
Question 1: A researcher wants to test if the mean salary in a profession is $50,000. Should they use a z-test or t-test? Why?
Answer: Use a t-test.
Reasoning: In real-world scenarios, we almost never know the true population standard deviation (σ) for salaries. We would only have the sample standard deviation (s) from our study, so we must use the t-test.
Question 2: Calculate the test statistic: n = 36, x̄ = 102, μ₀ = 100, s = 12. Is this a z-test or t-test?
This is a t-test because we're given the sample standard deviation (s), not the population standard deviation (σ).
Calculation:
t = (102 - 100) / (12 / √36)
t = 2 / 2
t = 1.0
df = n - 1 = 36 - 1 = 35
Question 3: For a two-tailed test with α = 0.05, the critical values are ±1.96 (z-test). If your calculated z = 1.85, what is your decision?
Decision: Fail to reject H₀
Reasoning: For a two-tailed test, we reject H₀ if |z| > 1.96.
|1.85| = 1.85, which is NOT greater than 1.96, so the test statistic is not in the critical region.
We do not have sufficient evidence to reject the null hypothesis at the 0.05 significance level.
Summary
- Use z-test when σ is known; use t-test when σ is unknown (almost always)
- Z-test statistic: z = (x̄ - μ₀) / (σ / √n)
- T-test statistic: t = (x̄ - μ₀) / (s / √n) with df = n - 1
- Critical value approach: Compare test statistic to critical value(s)
- P-value approach: Compare p-value to α
- Both approaches always give the same conclusion
- Always interpret results in the context of the original question