If we only sample a few people, how can we estimate everyone’s average?

95% CI = sample average ± $t_{0.975,n-1}\times \frac{\text{sample SD}}{\sqrt{\text{sample size}}}$

1. Start with the sampling distribution of the mean.

   For a normal population, standardizing the sample mean with the unknown population standard deviation gives

   Z = (x̄ − μ) / (σ / √n)  ∼  N(0, 1).
2. Account for estimating the unknown spread.

   The sample variance satisfies

   U = (n − 1)S² / σ²  ∼  χ²_n−1,

   and, for normal data, Z and U are independent. The n − 1 appears because estimating x̄ uses one degree of freedom.

3. Combine those two results.

   Dividing the standard normal variable by the square root of an independent chi-square variable divided by its degrees of freedom produces a Student t variable:

   T = Z / √(U / (n − 1))= (x̄ − μ) / (S / √n)  ∼  t_n−1.
4. Keep the middle 95% of that t distribution.

   Let α = 0.05 and let t* = t_{1−α/2,n−1} = t_0.975,n−1. By the definition of this critical value,

   P(−t* ≤ T ≤ t*) = 1 − α = 0.95.

   The 0.975 quantile leaves 2.5% in the upper tail; symmetry leaves another 2.5% in the lower tail.

5. Substitute T and isolate μ.
   P(−t* ≤ (x̄ − μ)/(S/√n) ≤ t*) = 0.95P(−t*S/√n ≤ x̄ − μ ≤ t*S/√n) = 0.95P(x̄ − t*S/√n ≤ μ ≤ x̄ + t*S/√n) = 0.95.

When is this exact? The derivation is exact for independent observations from a normal population. For non-normal populations, the same interval is commonly used as a large-sample approximation when the sampling design is valid and the mean’s sampling distribution is close to normal.