The t-Distribution and Small Sample Inference

When the population variance $\sigma^2$σ2 is unknown and the sample size is small, we cannot use the normal distribution directly for inference about the mean. Instead, we use the t-distribution, which accounts for the additional uncertainty from estimating $\sigma$σ with the sample standard deviation $s$s.

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Why the t-Distribution?

When $\sigma$σ is known:

$Z = \frac{\bar{X} - \mu}{\sigma / \sqrt{n}} \sim N(0, 1)$Z=σ/n​Xˉ−μ​∼N(0,1)

When $\sigma$σ is unknown and replaced by the sample standard deviation $s$s:

$T = \frac{\bar{X} - \mu}{s / \sqrt{n}} \sim t_{n-1}$T=s/n​Xˉ−μ​∼tn−1​

The statistic $T$T follows a t-distribution with $n - 1$n−1 degrees of freedom, not a standard normal. This is because $s$s is a random variable that introduces extra variability.

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Properties of the t-Distribution

Property	Detail
Shape	Symmetric, bell-shaped (similar to the normal)
Centre	Mean = 0
Tails	Heavier tails than the normal (more probability in the tails)
Parameter	Degrees of freedom $\nu = n - 1$ν=n−1
As $\nu \to \infty$ν→∞	The t-distribution approaches $N(0, 1)$N(0,1)

The heavier tails reflect the extra uncertainty from estimating $\sigma$σ. With small samples, extreme values of the test statistic are more likely than under the normal distribution.

$\nu$ν	$t_{0.025}$t0.025​ (two-tailed 5%)	$z_{0.025}$z0.025​
5	2.571	1.960
10	2.228	1.960
30	2.042	1.960
120	1.980	1.960

As $\nu$ν increases, the critical values approach the normal critical values.

Exam Tip: Use the t-distribution when $\sigma$σ is unknown and replaced by $s$s, and the sample comes from a normal (or approximately normal) population. If $\sigma$σ is known, use the normal distribution regardless of sample size.

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One-Sample t-Test

Hypotheses:

$H_0: \mu = \mu_0$H0​:μ=μ0​ vs $H_1: \mu \neq \mu_0$H1​:μ=μ0​ (two-tailed) or $H_1: \mu > \mu_0$H1​:μ>μ0​ / $H_1: \mu < \mu_0$H1​:μ<μ0​ (one-tailed).

Test statistic:

$T = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}$T=s/n​xˉ−μ0​​

Compare $|T|$∣T∣ with the critical value from $t_{n-1}$tn−1​ tables at the required significance level.

Worked Example

A factory claims that its light bulbs last 1000 hours on average. A sample of 12 bulbs gives $\bar{x} = 985$xˉ=985 and $s = 30$s=30. Test at the 5% level whether the mean lifetime is less than 1000.

$H_0: \mu = 1000$H0​:μ=1000, $H_1: \mu < 1000$H1​:μ<1000 (one-tailed).

$T = \frac{985 - 1000}{30/\sqrt{12}} = \frac{-15}{8.660} = -1.732$T=30/12​985−1000​=8.660−15​=−1.732

From tables: $t_{11, 0.05} = 1.796$t11,0.05​=1.796 (one-tailed).