Why is the denominator of the corrected sample variance n-1 instead of n?

Conditions

• The data represents a sample drawn from a larger population.
• The goal is to calculate the corrected sample variance.
• The sample size $n$ is greater than 1.

Reasoning, step by step

1. Identify that the data is a sample, not the entire population.
2. Recall that the sample mean $\bar{x}$ is estimated from the sample data itself.
3. Understand that this estimation imposes one constraint on the $n$ deviations, leaving $n-1$ degrees of freedom.
4. Apply the corrected sample variance formula with denominator $n-1$ to ensure unbiasedness.

Example

The screen displays "S² = $1/(6-1)$ [(71-74)² + ... + (79-74)²]". The narration distinguishes sample and population, then calculates the pulse data through the sample mean, corrected sample variance and sample standard deviation.

Common misconceptions

• Believing that sample variance should also be divided by the sample size $n$.
• Confusing the corrected sample variance with the population variance, which divides by the population size.

Watch the explanation

Connected concepts

Explore next

• How is the sample variance calculated for the given pulse data?
• What is the difference between sample and population variance?
• Why does the class size of 40 not affect the sample variance calculation?