Statistical Analysis — Mean, SD, % Change, T-Tests for Significance

1. INTRODUCTION TO STATISTICS IN SPORT SCIENCE

1.1 Why Statistics Matter in Sport Science

Statistics are essential tools for making sense of data in sport science. They allow us to:

1.2 Types of Statistics

1.3 Levels of Measurement

Understanding data types is crucial for selecting appropriate statistics:

Most sport science data is ratio level (times, distances, weights, physiological measures), allowing use of parametric statistics.

2. MEASURES OF CENTRAL TENDENCY

2.1 What Is Central Tendency?

Central tendency describes the "typical" or "average" value in a dataset — the center around which data clusters.

2.2 The Mean (Average)

Definition

The mean is the sum of all values divided by the number of values. It represents the arithmetic average of the data.

Formula

Mean (x̄) = Σx ÷ n

Where:
x̄ = sample mean (pronounced "x-bar")
Σx = sum of all values
n = number of values

Calculation Example

Sprint times (seconds): 5.12, 5.18, 5.15, 5.22, 5.13

Step 1: Sum all values
Σx = 5.12 + 5.18 + 5.15 + 5.22 + 5.13 = 25.80

Step 2: Count number of values
n = 5

Step 3: Calculate mean
Mean = 25.80 ÷ 5 = 5.16 seconds

Symbols

Properties of the Mean

Advantages of the Mean

Disadvantages of the Mean

Example of Outlier Effect

Without outlier: 5.12, 5.18, 5.15, 5.22, 5.13

• Mean = 5.16 seconds

With outlier: 5.12, 5.18, 5.15, 5.22, 6.50 (one very slow time)

• Mean = 5.43 seconds

The single outlier (6.50) pulled the mean up by 0.27 seconds, misrepresenting typical performance.

2.3 The Median

Definition

The median is the middle value when data is arranged in order. It divides the distribution in half — 50% of values fall above, 50% below.

Calculation

Example

Odd number of values: 5.12, 5.13, 5.15, 5.18, 5.22

• Median = 5.15 (middle value)

Even number of values: 5.12, 5.13, 5.15, 5.18, 5.22, 5.25

• Median = (5.15 + 5.18) ÷ 2 = 5.165

Advantages of the Median

When to Use Median Over Mean

2.4 The Mode

Definition

The mode is the most frequently occurring value in a dataset.

Characteristics

Example

Sprint times: 5.12, 5.15, 5.18, 5.15, 5.22, 5.15

• Mode = 5.15 (occurs 3 times)

2.5 Comparison of Central Tendency Measures

3. MEASURES OF VARIABILITY (DISPERSION)

3.1 Why Measure Variability?

Central tendency alone doesn't tell the whole story. We also need to know how spread out the data is around the center.

Example: Two groups both have mean = 5.20 seconds, but:

• Group A: 5.18, 5.19, 5.20, 5.21, 5.22 (very consistent)

• Group B: 4.80, 5.00, 5.20, 5.40, 5.60 (highly variable)

The mean is the same, but the groups are very different.

3.2 Range

Definition

The range is the difference between the highest and lowest values.

Formula

Range = Maximum value - Minimum value

Example

Sprint times: 5.12, 5.18, 5.15, 5.22, 5.13

• Range = 5.22 - 5.12 = 0.10 seconds

Advantages and Limitations

3.3 Variance

Definition

Variance is the average of the squared deviations from the mean. It quantifies the spread of data around the mean.

Formula (Sample Variance)

s² = Σ(x - x̄)² ÷ (n - 1)

Where:
s² = sample variance
x = each individual value
x̄ = sample mean
n = sample size
(n - 1) = degrees of freedom

Why (n - 1)?

Calculation Example

Sprint times: 5.12, 5.18, 5.15, 5.22, 5.13 Mean = 5.16

Variance = 0.0066 ÷ (5 - 1) = 0.0066 ÷ 4 = 0.00165

Limitation of Variance

The variance is in squared units (e.g., seconds²), making interpretation difficult. This is why we typically use standard deviation.

3.4 Standard Deviation (SD)

Definition

Standard deviation is the square root of the variance. It measures the typical deviation of values from the mean, expressed in the original units of measurement.

Formula

s = √[Σ(x - x̄)² ÷ (n - 1)]

Or simply:

s = √s² (square root of variance)

Symbols

Calculation Example (Continued)

From previous example:

• Variance = 0.00165

Standard Deviation = √0.00165 = 0.041 seconds

Reporting: Mean ± SD = 5.16 ± 0.04 seconds

Interpretation of Standard Deviation

The Empirical Rule (68-95-99.7 Rule)

For normally distributed data:

                    68%
               ←─────────────→
           95%
      ←─────────────────────────→
        99.7%
   ←───────────────────────────────→
   
   ─────────────────────────────────────
        -3SD  -2SD  -1SD  Mean  +1SD  +2SD  +3SD

Example Application

If mean sprint time = 5.16s and SD = 0.04s:

• ~68% of times fall between 5.12s and 5.20s (5.16 ± 0.04)

• ~95% of times fall between 5.08s and 5.24s (5.16 ± 0.08)

• ~99.7% of times fall between 5.04s and 5.28s (5.16 ± 0.12)

3.5 Coefficient of Variation (CV%)

Definition

The coefficient of variation expresses the standard deviation as a percentage of the mean. It allows comparison of variability between variables with different units or scales.

Formula

CV% = (SD ÷ Mean) × 100

Calculation Example

Mean = 5.16 seconds, SD = 0.04 seconds

CV% = (0.04 ÷ 5.16) × 100 = 0.78%

Interpretation: The typical variation is 0.78% of the mean value.

Uses of CV%

CV% Interpretation for Reliability

Advantage of CV%

Example: Comparing variability of two different tests:

Raw SD values (0.04 vs 3.5) are not comparable because units differ. But CV% shows sprint time (0.78%) is more consistent than vertical jump (6.4%).

4. PRESENTING DESCRIPTIVE STATISTICS

4.1 Standard Format

The most common way to present descriptive statistics:

Mean ± SD

Examples:

• Sprint time: 5.16 ± 0.04 seconds

• VO2max: 52.3 ± 4.8 mL/kg/min

• 1RM bench press: 85.5 ± 12.3 kg

4.2 Complete Descriptive Summary

4.3 Reporting in Tables

Example Table:

5. PERCENTAGE CHANGE

5.1 Definition

Percentage change quantifies the relative change between two measurements, typically pre- and post-intervention. It expresses change as a proportion of the original value.

5.2 Formula

% Change = [(Post - Pre) ÷ Pre] × 100

Or equivalently:

% Change = [(New Value - Original Value) ÷ Original Value] × 100

5.3 Calculation Examples

Example 1: Improvement in VO2max

• Pre-training VO2max: 48.0 mL/kg/min

• Post-training VO2max: 52.5 mL/kg/min

% Change = [(52.5 - 48.0) ÷ 48.0] × 100
% Change = [4.5 ÷ 48.0] × 100
% Change = 0.09375 × 100
% Change = +9.4%

Interpretation: VO2max improved by 9.4%

Example 2: Improvement in Sprint Time (Decrease = Improvement)

• Pre-training 40m sprint: 5.50 seconds

• Post-training 40m sprint: 5.35 seconds

% Change = [(5.35 - 5.50) ÷ 5.50] × 100
% Change = [-0.15 ÷ 5.50] × 100
% Change = -0.0273 × 100
% Change = -2.7%

Interpretation: Sprint time decreased (improved) by 2.7%

Note: For measures where LOWER is better (times, errors), negative % change indicates improvement.

Example 3: Decrease in Body Fat

• Pre: 18% body fat

• Post: 15% body fat

% Change = [(15 - 18) ÷ 18] × 100
% Change = [-3 ÷ 18] × 100
% Change = -16.7%

Interpretation: Body fat decreased by 16.7%

5.4 Interpreting Percentage Change

Direction and meaning depend on variable:

5.5 Meaningful Change

Critical consideration: Is the percentage change REAL (greater than measurement error)?

Example:

• If typical error (CV%) for VO2max testing = 3%

• Observed change = +5%

• This change (5%) exceeds typical error (3%), suggesting REAL improvement

5.6 Group Mean Percentage Change

When reporting group results, there are two approaches:

Approach 1: Mean of Individual % Changes

Calculate % change for each individual, then average:

Approach 2: % Change in Group Means

Calculate group means, then % change:

• Pre mean = 48.3 mL/kg/min

• Post mean = 52.2 mL/kg/min

• % Change = +8.1%

Note: These two approaches may give slightly different results. Be consistent and report which method was used.

5.7 Absolute Change vs Percentage Change

Both are useful:

• Absolute change: Actual units of improvement

• Percentage change: Relative to starting point; allows comparison across different baselines

5.8 Baseline Dependency

Important consideration: Same absolute change = different percentage change depending on baseline.

Example:

Same absolute improvement (+4 mL/kg/min), but different percentage change due to different starting points.

6. INFERENTIAL STATISTICS: T-TESTS

6.1 Purpose of Inferential Statistics

While descriptive statistics summarize data, inferential statistics allow us to:

6.2 Key Concepts

Hypothesis Testing

The Logic of Hypothesis Testing

1. Assume H₀ is true (no real difference)
         ↓
2. Calculate probability of observed results if H₀ is true (p-value)
         ↓
3. If p-value is very small (< α), reject H₀
         ↓
4. Conclude there IS a real difference (accept H₁)

p-value Interpretation

Type I and Type II Errors

6.3 What Is a T-Test?

A t-test is a statistical test used to determine whether the difference between two means is statistically significant. It produces a t-statistic, which is compared to critical values to determine significance.

When to Use T-Tests

6.4 Assumptions of T-Tests

6.5 Independent Samples T-Test

Purpose

Compare the means of two separate, unrelated groups to determine if they differ significantly.

When to Use

• Treatment group vs control group

• Males vs females

• Sport A vs Sport B

• Any comparison of two independent groups

Example Research Question

"Does a new training program improve sprint performance compared to standard training?"

• Group 1 (Experimental): 15 athletes using new program

• Group 2 (Control): 15 athletes using standard program

• Dependent variable: Post-intervention 40m sprint time

Hypotheses

Formula (Conceptual)

t = (Mean₁ - Mean₂) ÷ Standard Error of Difference

Where Standard Error depends on:
- Standard deviations of both groups
- Sample sizes of both groups

The full formula is:

t = (x̄₁ - x̄₂) ÷ √[(s₁²/n₁) + (s₂²/n₂)]

Example Calculation

Step 1: Calculate Standard Error
SE = √[(0.12²/15) + (0.14²/15)]
SE = √[(0.0144/15) + (0.0196/15)]
SE = √[0.00096 + 0.00131]
SE = √0.00227
SE = 0.048

Step 2: Calculate t-statistic
t = (5.15 - 5.28) ÷ 0.048
t = -0.13 ÷ 0.048
t = -2.71

Interpreting Results

1. Calculate degrees of freedom: df = n₁ + n₂ - 2 = 15 + 15 - 2 = 28

2. Look up critical t-value or calculate p-value

3. If |t| > critical value OR p < 0.05, result is statistically significant

For t = -2.71 with df = 28, p ≈ 0.011

Conclusion: p < 0.05, so we reject H₀. The experimental group is significantly faster than the control group.

Reporting Independent T-Test Results

Example format: "The experimental group (M = 5.15s, SD = 0.12) was significantly faster than the control group (M = 5.28s, SD = 0.14), t(28) = -2.71, p = 0.011."

6.6 Paired (Dependent) Samples T-Test

Purpose

Compare the means of the same group at two different time points (e.g., pre-test and post-test) to determine if there is a significant change.

When to Use

• Pre-test vs post-test comparisons

• Before vs after intervention

• Same subjects tested under two conditions

• Matched pairs designs

Example Research Question

"Does 8 weeks of plyometric training significantly improve vertical jump height?"

• Measurement 1: Pre-training vertical jump

• Measurement 2: Post-training vertical jump (same athletes)

Why "Paired"?

Each pre-test value is paired with its corresponding post-test value from the same individual. This pairing reduces error by accounting for individual differences.

Hypotheses

Formula (Conceptual)

t = Mean Difference ÷ Standard Error of Mean Difference

t = d̄ ÷ (SD_d ÷ √n)

Where:
d̄ = mean of the differences (post - pre for each individual)
SD_d = standard deviation of the differences
n = number of pairs

Example Calculation

Step 1: Calculate mean difference
d̄ = (4+2+5+3+5+3+3+5) ÷ 8 = 30 ÷ 8 = 3.75 cm

Step 2: Calculate SD of differences
SD_d = 1.16 cm (calculated from individual differences)

Step 3: Calculate Standard Error
SE = SD_d ÷ √n = 1.16 ÷ √8 = 1.16 ÷ 2.83 = 0.41

Step 4: Calculate t-statistic
t = d̄ ÷ SE = 3.75 ÷ 0.41 = 9.15

Interpreting Results

1. Degrees of freedom: df = n - 1 = 8 - 1 = 7

2. For t = 9.15 with df = 7, p < 0.001

Conclusion: p < 0.001, so we reject H₀. There is a statistically significant improvement in vertical jump following the training program.

Reporting Paired T-Test Results

Example format: "Vertical jump height significantly increased from pre-training (M = 51.5 cm, SD = 2.5) to post-training (M = 55.25 cm, SD = 3.4), t(7) = 9.15, p < 0.001. The mean improvement was 3.75 cm (95% CI: 2.78 to 4.72)."

6.7 One-Sample T-Test

Purpose

Compare a sample mean to a known or hypothesized population value to determine if the sample differs significantly from that value.

When to Use

• Compare sample to established norm

• Compare sample to target/threshold value

• Test if sample differs from a specific number

Example Research Question

"Does our team's average VO2max differ from the national average of 55 mL/kg/min?"

Formula

t = (x̄ - μ) ÷ (s ÷ √n)

Where:
x̄ = sample mean
μ = population/hypothesized value
s = sample standard deviation
n = sample size

Example

• Sample mean = 58.5 mL/kg/min

• Hypothesized value = 55 mL/kg/min

• Sample SD = 6.2 mL/kg/min

• n = 20

t = (58.5 - 55) ÷ (6.2 ÷ √20)
t = 3.5 ÷ (6.2 ÷ 4.47)
t = 3.5 ÷ 1.39
t = 2.52

For df = 19 and t = 2.52, p ≈ 0.021

Conclusion: Our team's VO2max (58.5) is significantly higher than the national average (55), p = 0.021.

6.8 Summary of T-Test Types

7. STATISTICAL SIGNIFICANCE VS PRACTICAL SIGNIFICANCE

7.1 The Distinction

7.2 Why the Distinction Matters

A result can be:

7.3 The Problem with Large Samples

With large sample sizes, even trivially small differences can be statistically significant.

Example:

• Group A mean = 5.15s (n = 500)

• Group B mean = 5.17s (n = 500)

• Difference = 0.02 seconds

With n = 1000 total, this tiny difference (0.02s) might yield p < 0.01. But is 0.02 seconds practically meaningful in a 40m sprint? Probably not.

7.4 Effect Size

Effect size quantifies the magnitude of the difference independent of sample size.

Cohen's d

The most common effect size measure for comparing two means.

Formula:

Cohen's d = (Mean₁ - Mean₂) ÷ Pooled SD

Where Pooled SD = √[(SD₁² + SD₂²) ÷ 2]

Interpretation (Cohen's conventions):

Example Calculation

• Experimental group: Mean = 5.15s, SD = 0.12s

• Control group: Mean = 5.28s, SD = 0.14s

Step 1: Calculate pooled SD
Pooled SD = √[(0.12² + 0.14²) ÷ 2]
Pooled SD = √[(0.0144 + 0.0196) ÷ 2]
Pooled SD = √[0.017]
Pooled SD = 0.13

Step 2: Calculate Cohen's d
Cohen's d = (5.15 - 5.28) ÷ 0.13
Cohen's d = -0.13 ÷ 0.13
Cohen's d = -1.0

Interpretation: |d| = 1.0, which is a large effect. The experimental group performed substantially better (faster).

7.5 Reporting Both Statistical and Practical Significance

Good practice: Report p-value AND effect size.

Example: "The experimental group (M = 5.15s, SD = 0.12) was significantly faster than the control group (M = 5.28s, SD = 0.14), t(28) = -2.71, p = 0.011, Cohen's d = 1.0. This represents a large effect."

8. CONFIDENCE INTERVALS (Continued)

8.3 Formula for 95% CI (for means)

95% CI = Mean ± (t-critical × Standard Error)

Where:
Standard Error (SE) = SD ÷ √n
t-critical = value from t-distribution for 95% confidence and given df

For large samples (n > 30), t-critical ≈ 1.96

8.4 Calculation Example

Data:

• Mean vertical jump = 55 cm

• SD = 5 cm

• n = 25

Step 1: Calculate Standard Error
SE = SD ÷ √n = 5 ÷ √25 = 5 ÷ 5 = 1.0 cm

Step 2: Find t-critical (df = 24, 95% confidence)
t-critical ≈ 2.064

Step 3: Calculate 95% CI
95% CI = 55 ± (2.064 × 1.0)
95% CI = 55 ± 2.06
95% CI = 52.94 to 57.06 cm

Reporting: "Mean vertical jump was 55 cm (95% CI: 52.9 to 57.1 cm)"

8.5 Confidence Interval for the Difference Between Means

When comparing two groups, we can calculate a CI for the difference between their means.

Key interpretation: If the 95% CI for the difference does NOT include zero, the difference is statistically significant at p < 0.05.

Example

• Mean difference = 5.15 - 5.28 = -0.13s

• SE of difference = 0.048 (calculated previously)

• 95% CI = -0.13 ± (2.048 × 0.048) = -0.13 ± 0.098

95% CI = -0.228 to -0.032 seconds

Interpretation: The CI does NOT include zero (-0.228 to -0.032), so the difference is statistically significant. The experimental group is significantly faster by 0.03 to 0.23 seconds.

8.6 Advantages of Confidence Intervals

8.7 Interpreting CI Width

9. CORRELATION

9.1 Definition

Correlation measures the strength and direction of the linear relationship between two variables. It indicates how well one variable predicts the other.

9.2 Pearson Correlation Coefficient (r)

Properties

Interpretation of r Values

Same interpretation for negative values (e.g., r = -0.85 is a strong negative correlation)

Visual Representation

POSITIVE CORRELATION (r = +0.85)     NEGATIVE CORRELATION (r = -0.80)
        Y                                     Y
        │       ●●                            │  ●●
        │     ●●                              │    ●●
        │   ●●                                │      ●●
        │ ●●                                  │        ●●
        │●                                    │          ●●
        └─────────── X                        └─────────── X

NO CORRELATION (r = 0.05)
        Y
        │    ●    ●
        │  ●   ●    ●
        │    ●   ●
        │ ●    ●   ●
        │   ●    ●
        └─────────── X

9.3 Coefficient of Determination (r²)

Definition

r² (r-squared) represents the proportion of variance in one variable explained by the other variable.

Formula

r² = r × r

Interpretation

Example

If correlation between VO2max and 5km run time is r = -0.85:

• r² = 0.72

• Interpretation: 72% of the variance in 5km time can be explained by VO2max. The remaining 28% is due to other factors (pacing, running economy, psychology, etc.)

9.4 Correlation in Sport Science Applications

9.5 Important Cautions

10. STANDARD ERROR OF MEASUREMENT (SEM)

10.1 Definition

Standard Error of Measurement (SEM) quantifies the typical error in individual measurements — the amount by which a measured score might differ from the true score due to measurement error.

10.2 Formula

SEM = SD × √(1 - ICC)

Or alternatively:

SEM = SD × √(1 - r)

Where:
SD = standard deviation of the measure
ICC = intraclass correlation coefficient (reliability)
r = test-retest reliability coefficient

10.3 Example Calculation

Data:

• SD of vertical jump scores = 5 cm

• Test-retest reliability (ICC) = 0.95

SEM = 5 × √(1 - 0.95)
SEM = 5 × √0.05
SEM = 5 × 0.224
SEM = 1.12 cm

Interpretation: Individual measurements typically vary by about ±1.1 cm due to measurement error.

10.4 Using SEM

True Score Estimation

If an athlete scores 55 cm on vertical jump (SEM = 1.1 cm):

• 68% confidence: True score is 55 ± 1.1 = 53.9 to 56.1 cm

• 95% confidence: True score is 55 ± (1.96 × 1.1) = 52.8 to 57.2 cm

Minimum Detectable Change (MDC)

The smallest change that can be confidently detected as real (not error):

MDC₉₅ = SEM × 1.96 × √2 = SEM × 2.77

Example: SEM = 1.1 cm

• MDC₉₅ = 1.1 × 2.77 = 3.05 cm

Interpretation: A change of at least 3.05 cm is needed to be 95% confident the change is real.

11. PRESENTING STATISTICAL RESULTS

11.1 Reporting Conventions

Descriptive Statistics

Mean ± SD (or Mean (SD))

Example: "Sprint time was 5.16 ± 0.12 seconds"

T-Test Results

t(df) = t-value, p = p-value

Example: "t(28) = 2.71, p = 0.011"

With Effect Size

t(df) = t-value, p = p-value, Cohen's d = effect size

Example: "t(28) = 2.71, p = 0.011, d = 1.0"

With Confidence Interval

Mean difference = X (95% CI: lower to upper)

Example: "The experimental group was 0.13s faster (95% CI: 0.03 to 0.23s)"

Correlation

r = correlation coefficient, p = p-value

Example: "r = 0.85, p < 0.001"

Or with r²:
"r = 0.85 (r² = 0.72), p < 0.001"

11.2 Example Results Section

Descriptive statistics: "Table 1 presents descriptive statistics for all variables. The experimental group (n = 15) had a mean 40m sprint time of 5.15 ± 0.12 seconds, while the control group (n = 15) averaged 5.28 ± 0.14 seconds."

Inferential statistics: "An independent samples t-test revealed that the experimental group was significantly faster than the control group, t(28) = -2.71, p = 0.011. The mean difference of 0.13 seconds (95% CI: 0.03 to 0.23) represents a large effect (Cohen's d = 1.0)."

Pre-post comparison: "A paired samples t-test showed vertical jump height significantly increased from pre-training (M = 51.5 cm, SD = 2.5) to post-training (M = 55.25 cm, SD = 3.4), t(7) = 9.15, p < 0.001. The mean improvement was 3.75 cm (7.3%), representing a large effect (d = 1.27)."

11.3 Tables for Presenting Data

Descriptive Statistics Table

Pre-Post Comparison Table

12. COMMON STATISTICAL MISTAKES

12.1 Mistakes to Avoid

12.2 Key Principle: Statistical vs Practical Significance

Always ask two questions:

1. Is the difference REAL? (statistical significance — p-value)

2. Is the difference IMPORTANT? (practical significance — effect size)

13. SUMMARY TABLE: KEY STATISTICAL FORMULAS

14. EXAM APPLICATION TIPS

14.1 Common Essay Questions

1. "Calculate and interpret the mean and standard deviation for a given dataset" (6-8 marks)

2. "Explain the difference between statistical significance and practical significance" (6 marks)

3. "Describe when you would use an independent t-test versus a paired t-test" (6 marks)

4. "Calculate percentage change and interpret whether the change is meaningful" (8 marks)

5. "Interpret the results of a t-test and explain what the p-value and effect size tell us" (10 marks)

6. "Explain how confidence intervals can be used to determine statistical significance" (6 marks)

14.2 Key Definitions to Know

14.3 Calculation Practice

Given data: Pre-training 40m sprint times for 6 athletes: 5.45, 5.52, 5.38, 5.61, 5.49, 5.55 seconds

Calculate:

1. Mean:

Mean = (5.45 + 5.52 + 5.38 + 5.61 + 5.49 + 5.55) ÷ 6
Mean = 33.00 ÷ 6 = 5.50 seconds

2. Standard Deviation:

Variance = 0.0320 ÷ (6-1) = 0.0064
SD = √0.0064 = 0.08 seconds

3. CV%:

CV% = (0.08 ÷ 5.50) × 100 = 1.45%

4. If post-training mean = 5.35 seconds, percentage change:

% Change = [(5.35 - 5.50) ÷ 5.50] × 100
% Change = -2.7% (improvement)

14.4 Interpretation Examples

Example 1: T-test result

"Results: t(18) = 3.45, p = 0.003, d = 1.2"

Interpretation: "The t-test was statistically significant (p = 0.003 < 0.05), indicating the difference between groups is unlikely due to chance. The effect size (d = 1.2) is very large, indicating a substantial practical difference between groups."

Example 2: Correlation result

"Results: r = 0.78, p < 0.001"

Interpretation: "There is a strong positive correlation between the two variables (r = 0.78). This relationship is statistically significant (p < 0.001). The coefficient of determination (r² = 0.61) indicates that 61% of the variance in one variable can be explained by the other."

Example 3: Percentage change with error consideration

"Pre: 52.0 mL/kg/min, Post: 55.5 mL/kg/min, typical error CV = 3%"

Interpretation:

% Change = [(55.5 - 52.0) ÷ 52.0] × 100 = +6.7%
Typical error at pre = 3% × 52.0 = 1.56 mL/kg/min
Change (3.5) > 2 × typical error (3.1)

"The improvement of 6.7% (3.5 mL/kg/min) exceeds twice the typical error, suggesting this represents a real physiological improvement rather than measurement variability."

15. KEY RESEARCHERS TO REFERENCE

16. SUMMARY: STATISTICAL ANALYSIS PRINCIPLES

17. COMPLETE MEASUREMENT AND TESTING SUMMARY

You have now completed comprehensive notes on all measurement and testing topics: