13. Paired Means

Math 115 - Examples for Topic 13

Author

Yurk

These worked examples demonstrate inference for paired means using the t-distribution (model-based approach).

Example 1: Hypothesis Test for Paired Means

Scenario: A fitness app company wants to test whether their new “7-Minute Morning Workout” actually improves users’ resting heart rate. They recruit 40 volunteers and measure each person’s resting heart rate (in beats per minute) before starting the program and again after 8 weeks of daily use.

Sample results (difference = after − before):

n \(\bar{x}_d\) \(s_d\)
40 -4.2 bpm 8.5 bpm

A negative mean difference indicates heart rate decreased (improved) on average.

Research question: Is there evidence that the workout program changes resting heart rate?

Task: Conduct a hypothesis test at the α = 0.05 level using the t-distribution.

Step 1: Identify the parameter and why data are paired

Why paired? Each person is measured twice (before and after). The same individuals appear in both measurements, creating a natural pairing.

Let \(\mu_d\) = the true mean change in resting heart rate (after − before) for all people who would use this program.

Sample mean difference: \(\bar{x}_d = -4.2\) bpm

Step 2: State the hypotheses

Since we want to test whether heart rate changes (in either direction), this is a two-sided test:

  • \(H_0: \mu_d = 0\) (no change in resting heart rate)
  • \(H_A: \mu_d \neq 0\) (resting heart rate changes)

Step 3: Check conditions for using t-distribution

Independence:

  • Random sample of volunteers ✓
  • Each person’s change is independent of others ✓

Normality:

  • n = 40 ≥ 30 ✓
  • With n ≥ 30, moderate skewness in differences is acceptable

Assuming at most moderate skewness, conditions are met for using the t-distribution.

Step 4: Calculate the test statistic

Degrees of freedom: \(df = n - 1 = 40 - 1 = 39\)

Standard error:

\[SE = \frac{s_d}{\sqrt{n}} = \frac{8.5}{\sqrt{40}} = 1.344\]

T-statistic:

\[T = \frac{\bar{x}_d - 0}{SE} = \frac{-4.2 - 0}{1.344} = -3.13\]

Step 5: Calculate the p-value using Jamovi

Randomize module → Model-Based Inference

To calculate the p-value:

  1. Under Distribution, choose t
  2. Enter degrees of freedom: 39
  3. Check “Calculate area (p-value)”
  4. Enter the T-statistic: -3.13
  5. Select tail direction: Both (for two-sided test)

P-value = 0.0033

Step 6: Interpret the p-value

Interpretation: If the workout program had no effect on resting heart rate, there is approximately a 0.33% chance of observing a mean change of 4.2 bpm or more extreme in a random sample of 40 people.

Step 7: Make a decision

Using the significance level α = 0.05:

  • P-value = 0.0033
  • Compare: Is 0.0033 < 0.05? YES

Decision: We reject \(H_0\) at the α = 0.05 level.

Step 8: State the conclusion in context

The data provide convincing evidence that the 7-Minute Morning Workout program changes resting heart rate (T = -3.13, df = 39, p-value = 0.0033). On average, participants’ resting heart rate decreased by 4.2 bpm after 8 weeks of the program, suggesting the workout may improve cardiovascular health.

Example 2: Confidence Interval for Paired Means

Scenario: A cognitive psychologist studies whether listening to music affects typing speed. She recruits 55 office workers and measures each person’s typing speed (in words per minute) under two conditions: with background music playing and in silence. The order of conditions is randomized for each participant.

Sample results (difference = music − silence):

n \(\bar{x}_d\) \(s_d\)
55 3.8 wpm 12.4 wpm

A positive mean difference indicates faster typing with music.

Research question: What is the mean difference in typing speed between the music and silence conditions?

Task: Construct a 95% confidence interval for \(\mu_d\) using the t-distribution.

Step 1: Identify the parameter and why data are paired

Why paired? Each person is measured under both conditions (music and silence). The same individuals provide both measurements, creating a natural pairing.

Let \(\mu_d\) = the true mean difference in typing speed (music − silence) for all office workers.

Point estimate: \(\bar{x}_d = 3.8\) wpm

Step 2: Check conditions for using t-distribution

Independence:

  • Random sample of office workers ✓
  • Each person’s difference is independent of others ✓

Normality:

  • n = 55 ≥ 30 ✓

Assuming at most moderate skewness, conditions are met for using the t-distribution.

Step 3: Calculate the standard error

Degrees of freedom: \(df = n - 1 = 55 - 1 = 54\)

Standard error:

\[SE = \frac{s_d}{\sqrt{n}} = \frac{12.4}{\sqrt{55}} = 1.67\]

Step 4: Find the critical value using Jamovi

Randomize module → Model-Based Inference

To find \(t^*\):

  1. Under Distribution, choose t
  2. Enter degrees of freedom: 54
  3. Check “Calculate CI multiplier”
  4. Enter confidence level: 95

Critical value: \(t^*_{54} = 2.005\)

Step 5: Calculate the confidence interval

Margin of error:

\[ME = t^*_{df} \times SE = 2.005 \times 1.67 = 3.35\]

Confidence interval:

\[\text{CI} = \bar{x}_d \pm ME = 3.8 \pm 3.35\]

\[\text{95\% CI} = (0.45, 7.15)\]

Step 6: Interpret the confidence interval

Interpretation: We are 95% confident that the true mean difference in typing speed (music − silence) is between 0.45 and 7.15 words per minute.

In practical terms:

  • Both bounds are positive, indicating people type faster with music than in silence
  • We’re confident the increase is somewhere between 0.45 and 7.15 wpm

Note: Since the confidence interval does not include 0, this is consistent with rejecting \(H_0: \mu_d = 0\) at the α = 0.05 level. Because the order of conditions was randomized for each participant, we can conclude that listening to music causes an increase in typing speed (on average) for these office workers.

Example 3: Paired or Independent?

These questions ask you to identify whether each scenario involves paired data or independent groups.

Part A: Sleep Study

Scenario: Researchers want to know if a new mattress improves sleep quality. They recruit 100 volunteers and randomly assign 50 to sleep on the new mattress for one month, while the other 50 continue sleeping on their current mattress. At the end of the month, all participants rate their sleep quality on a scale of 1-10.

Question: Are these data paired or independent? Explain your reasoning.

These data are independent.

Different people are in each group. The 50 people who tried the new mattress are not the same as the 50 people who kept their old mattress. There is no natural pairing between an observation in one group and a specific observation in the other group.

To make this paired: Have the same 50 people sleep on their current mattress for one month, then switch to the new mattress for another month. Each person would provide two measurements.

Part B: Gas Mileage Comparison

Scenario: A consumer group wants to compare gas mileage between regular and premium gasoline. They select 30 cars and test each car twice: once with regular gas and once with premium gas (in random order). For each car, they record the miles per gallon achieved with each type of fuel.

Question: Are these data paired or independent? Explain your reasoning.

These data are paired.

Each car is tested under both conditions (regular and premium). The same 30 cars provide measurements for both fuel types, creating a natural pairing. The difference in MPG for each car can be calculated and analyzed as a single sample.

Why pairing matters here: Different cars have very different fuel efficiency. By using the same car for both fuel types, we control for car-to-car variability and isolate the effect of fuel type.

Part C: Exam Anxiety Treatment

Scenario: A university counseling center offers a workshop to reduce exam anxiety. To evaluate effectiveness, they measure anxiety levels (on a standardized scale) for 45 students immediately before and immediately after the 3-hour workshop.

Question: Are these data paired or independent? Explain your reasoning.

These data are paired.

The same 45 students are measured twice: before the workshop and after the workshop. Each student has a “before” score and an “after” score, creating a natural pairing. The change in anxiety for each student can be calculated and analyzed.

Why pairing matters here: Students vary widely in their baseline anxiety levels. By measuring the same students before and after, we can see how much each individual’s anxiety changed, rather than comparing different groups of students.

Part D: Coffee Shop Comparison

Scenario: A food critic wants to compare espresso quality between two popular coffee chains. She visits 25 locations of Chain A and 25 different locations of Chain B, ordering an espresso at each location and rating its quality on a scale of 1-100.

Question: Are these data paired or independent? Explain your reasoning.

These data are independent.

The 25 locations of Chain A are completely different from the 25 locations of Chain B. There is no natural connection between a specific Chain A location and a specific Chain B location. Different locations are being compared, not the same location under two conditions.

To make this paired: The critic could visit locations that are near each other (e.g., a Chain A and Chain B that are in the same shopping center) and pair them by location. Or she could have the same barista make espresso using both chains’ equipment and beans.

Summary: How to Identify Paired Data

Data are PAIRED when:

  • The same subjects are measured twice (before/after, two conditions)
  • The same items are measured under two different conditions
  • There is a deliberate matching between observations in the two groups

Data are INDEPENDENT when:

  • Different subjects are in each group
  • There is no natural or deliberate connection between observations across groups
  • Observations in one group tell you nothing about which observation they should be compared to in the other group

Key question to ask: “Can I match each observation in one group with exactly one specific observation in the other group based on the study design?”

  • If YES → Paired
  • If NO → Independent
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