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Confidence Intervals and Tests using the t-Distribution

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Confidence Intervals & Tests using the t-Distribution

Big idea: when you don't actually know how spread out a population is (its true variance), and you've only got a small sample to go on, you can't trust the normal distribution anymore — you need the t-distribution instead, because it's built to be a bit more "forgiving" about that uncertainty.

Why the t-distribution? Reading t-tables Confidence intervals Hypothesis test (one mean) Paired t-test Two means — CI Two means — test What to memorise Checklist Exam tips

Summary

The whole chapter in seven bite-sized ideas

Why not just use z? With a small sample and unknown variance, the sample variance isn't a reliable enough stand-in for the true variance — so we swap the normal distribution for the "wider" t-distribution.

Reading the tables t-tables are read using degrees of freedom (rows) and tail probability (columns). Always sketch the curve first.

CI for one mean x̄ ± t(n-1)(α/2) × s/√n gives a range of plausible values for the true population mean.

Hypothesis test for one mean Same 5-step process as z-tests, just swap in the t-statistic and t-critical-values.

Paired t-test For before/after data — test the mean of the differences, not the two samples separately.

Two independent means Pool the two sample variances into one shared estimate, then build a CI or run a test on the difference of the means.

1. Why do we need the t-distribution?

The problem it solves

Start with something you already know: if X is normally distributed with mean μ and (crucially) a known variance σ², then the sample mean X̄ is also normal: X̄ ~ N(μ, σ²/n). That lets you standardise using (X̄ − μ)/(σ/√n), which follows the standard normal N(0,1) — the familiar z-score.

But here's the catch: in real life, you almost never actually know σ². You only have your sample, from which you can calculate the sample variance, s². Two situations then arise:

Large n: s² is a good enough approximation for σ², so you can basically carry on as before and treat it like a z-problem.
Small n: s² is not reliably close to σ² — a small sample can easily give you a wonky estimate of spread just by bad luck. This extra uncertainty means (X̄ − μ)/(S/√n) is more spread out and has fatter tails than N(0,1). We give it a new name, t, and a new distribution to match: the t-distribution.

Core definition

If X₁, X₂, …, Xₙ is a random sample from N(μ, σ²) with unknown variance, then
t = (X̄ − μ) / (S/√n) follows a t-distribution with v = n − 1 degrees of freedom,
where S² = (1/n)(Σx² − nx̄²) is an unbiased estimator of σ².

In plain English: "how many standard errors away from μ is my sample mean" — but measured on a ruler that gets a bit more generous (wider) the smaller your sample is, to account for the fact that s is a shakier estimate of σ when n is small.

Think of it like this Imagine judging a diving competition. If you've watched a diver do 100 dives, you're pretty confident about their "typical" performance — you can be strict with your judgement (that's the z / normal world). If you've only seen them dive twice, you genuinely don't know how consistent they are — so you have to leave more room for error (that's the t-distribution: wider, fatter tails, more caution baked in).

What happens as n gets bigger?

As the degrees of freedom v = n − 1 → ∞, the t-distribution's shape converges to N(0,1). This makes sense: with a huge sample, your estimate s² becomes an excellent approximation of σ², so the "extra caution" the t-distribution builds in becomes unnecessary. This is exactly why the t-distribution is only used for small samples — for large samples the two distributions are practically identical.

Quick Check

Q: Why can't we just always use the sample variance s² and treat the test statistic as N(0,1), regardless of sample size?

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2. Reading t-distribution tables

The single most important practical skill in this chapter

t-tables give you the value t such that P(T > t) = p, for a given number of degrees of freedom v (the row) and tail probability p (the column). Every single question in this chapter — confidence intervals, hypothesis tests, paired tests — boils down to correctly reading one of these values. Always sketch the bell curve and shade the region you want before touching the table. It takes 10 seconds and prevents almost every silly mistake.

Worked Example

X has a t-distribution with 12 degrees of freedom. Find t such that:

(i) P(X > t) = 0.01	Read straight off the table: row v = 12, column 0.01 → `t₁₂(0.01) = 2.681`
(ii) P(X < t) = 0.9	The table only gives "greater than" probabilities, so convert first: if P(X < t) = 0.9, then P(X > t) = 1 − 0.9 = 0.1. Now read row v = 12, column 0.1 → `t₁₂(0.1) = 1.356`
(iii) P(\|X\| > t) = 0.1	This is a two-tailed statement: P(X < −t) + P(X > t) = 0.1. By symmetry each tail has area 0.05, so look up row v=12, column 0.05 → `t₁₂(0.05) = 1.782`. The two boundary values are +1.782 and −1.782.

Worked Example — probability the "wrong way round"

Y has a t₆-distribution. Find P(Y < 1.440).

Looking along the v = 6 row, you'll spot 1.440 sitting in the 0.10 column. That tells you P(Y > 1.440) = 0.1. But we were asked for P(Y < 1.440), so use "total area under the curve = 1":

P(Y < 1.440) = 1 − P(Y > 1.440) = 1 − 0.1 = 0.9

Watch out The table values are always positive and represent P(X > t). If a question asks about a negative t-value or "less than", you must flip it yourself using symmetry (for negative values) or 1 − p (for "less than" statements). The table will never do this conversion for you.

Practice Question

Q: The random variable T has a t₉-distribution. Find the value of t such that P(T < −t) = 0.025.

▶ Show Answer

3. Confidence interval for the mean (unknown variance)

Small sample, normal population, σ² unknown

When your sample size n is small and you don't know σ², you replace σ with your sample standard deviation s, and — critically — you must also replace the z critical value with a t critical value (using n − 1 degrees of freedom), because the extra uncertainty from estimating σ needs to be accounted for.

Formula

100(1 − α)% confidence limits for μ: x̄ ± t_n−1(α/2) × s/√n

100(1 − α)% confidence interval: ( x̄ − t_n−1(α/2)×s/√n , x̄ + t_n−1(α/2)×s/√n )

In plain English: take your sample mean, then step out an equal distance in both directions. How far you step depends on (a) how confident you want to be [more confidence = bigger step, bigger t-value] and (b) how much natural variability there is in your data relative to your sample size [s/√n].

Worked Example

8 biscuits are weighed (grams): 15, 14.5, 16.5, 14.7, 13.9, 15.4, 13.6, 16.7. Find a 95% CI for the true mean weight.

Find x̄ and s	Calculator gives x̄ = 15.0375, s = 1.1211
Identify n, α, degrees of freedom	n = 8, α = 0.05, so we need t₇(0.025) = 2.365 (half of α because a CI is two-tailed)
Substitute into the formula	15.0375 ± 2.365 × (1.211/√8)
Result	(14.025, 16.050)

Key habit For a confidence interval, always halve α before looking up your t-value — a 95% CI uses the 0.025 column, not 0.05, because there's an equal chunk of "leftover" probability in each tail.

Practice Question

Q: A random sample of 10 items from a normal population gives x̄ = 24.6 and s = 3.2. Find a 90% confidence interval for the population mean.

▶ Show Answer

4. Hypothesis test for the mean (unknown variance)

Same logic as a z-test, new distribution

This follows the exact same 5-step process you've used for every hypothesis test so far — the only thing that changes is which distribution (and therefore which table) you use for the critical value.

The 5 steps

1. State H₀ and H₁
2. State the significance level α and degrees of freedom v = n − 1
3. Find the critical region (using the t-table)
4. Calculate x̄, s² and the test statistic t = (x̄ − μ)/(s/√n)
5. Compare t to the critical region and write a conclusion in context

Worked Example

A census once found the mean height of adult women was 164cm. A new sample of 6 women gives mean 167.5cm, with s = 3cm. Test at the 5% level whether women have gotten taller.

Hypotheses	H₀: μ = 164, H₁: μ > 164 (one-tailed, since we're testing for an increase)
Significance & df	α = 0.05, v = 6 − 1 = 5
Critical value	t₅(0.05) = 2.015, so critical region is t ≥ 2.015
Test statistic	t = (167.5 − 164)/(3/√6) = 2.858
Decision	2.858 > 2.015, so t is in the critical region → reject H₀
Conclusion	There is sufficient evidence to suggest that the mean height of women has increased.

Common mistake Read the wording carefully! "Has the mean changed" needs a two-tailed test (H₁: μ ≠ …), but "has the mean increased/decreased" needs a one-tailed test. Getting this wrong changes which column of the table you use — for a two-tailed test at the 5% level, you actually look up the 0.025 column, not the 0.05 column, because the 5% risk is split across both tails.

Practice Question

Q: A machine is supposed to fill bottles with 500ml of liquid. A sample of 10 bottles has mean 496ml and s = 5ml. Test at the 5% significance level whether the machine is filling bottles with the wrong amount (a two-tailed test).

▶ Show Answer

5. The paired t-test

For "before and after" data

Sometimes two sets of results aren't independent — they come in natural pairs, like the same person's score before and after a treatment. In this case, don't try to compare the two samples directly. Instead, work with the differences, D, between each pair, and test whether the mean difference is zero.

Formula

(D̄ − μ_D) / (S/√n) ~ t_n−1

Here D̄ is the mean of the differences between paired observations. We usually test H₀: μ_D = 0, meaning "no real difference between the two conditions on average". This turns a two-sample problem into a single-sample t-test on the differences.

Worked Example

11 students take a maths test, attend a seminar on problem solving, then take a second test of equal difficulty. Test at the 5% level whether the seminar improved scores.

Hypotheses	H₀: μ_d = 0, H₁: μ_d > 0 (one-tailed — testing for improvement)
Significance & df	α = 0.05, v = 11 − 1 = 10
Critical value	t₁₀(0.05) = 1.812, so critical region is t ≥ 1.812
Find d̄ and s²	d̄ = Σd/n = 16/11 = 1.4545 s² = (Σd² − nd̄²)/(n−1) = (146 − 11×1.4545²)/10 = 14.4545
Test statistic	t = (1.4545 − 0)/(√14.4545/√11) = 1.2688
Decision & conclusion	1.2688 < 1.812, so t is not in the critical region → do not reject H₀. There is insufficient evidence that the seminar improved test performance.

Why this works By focusing on the differences, you cancel out any "baseline" variation between individuals (e.g. some students are just naturally better at maths than others). All that's left is the effect you actually care about — did scores go up or down after the seminar? This is what makes paired tests so much more powerful than treating the two samples as independent.

Practice Question

Q: Why must we assume the differences D are normally distributed for the paired t-test to be valid, rather than assuming the two original samples are normal?

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6. Confidence interval for the difference between two means

Independent samples, unknown but equal variances

Now imagine two completely separate (independent) samples, from two normal populations, and you want to estimate the difference between their true means, μ_x − μ_y. If we're willing to assume both populations share the same unknown variance σ², we combine ("pool") the two sample variances into one better, shared estimate — using more data always gives a more reliable estimate.

Step 1 — Pooled variance

s_p² = [(n_x − 1)s_x² + (n_y − 1)s_y²] / (n_x + n_y − 2)

A weighted average of the two sample variances, weighted by their degrees of freedom (bigger sample → more say in the pooled estimate).

Step 2 — Confidence interval

(x̄ − ȳ) ± t_c × s_p × √(1/n_x + 1/n_y)

where t_c comes from a t-distribution with v = n_x + n_y − 2 degrees of freedom (remember: halve α, since a CI is two-tailed).

Worked Example

Seedlings grown in two composts. Brand A: n=10, x̄=10.32, s²=0.484. Brand B: n=15, ȳ=12.473, s²=0.4735. Find a 90% CI for the difference in mean height (assume equal, unknown variances).

Pooled variance	s_p² = (9×0.484 + 14×0.4735)/(10+15−2) = 0.4776 → s_p = 0.6911
Degrees of freedom & t-value	v = 23, halve α: t₂₃(0.05) = 1.714
Substitute	(12.473 − 10.32) ± 1.714 × 0.6911 × √(1/15 + 1/10) = 2.153 ± 0.4836
Result	(1.6694, 2.6366)

Don't forget The degrees of freedom for two-sample problems is n_x + n_y − 2 (you "lose" one degree of freedom for each sample mean you had to estimate), not n_x + n_y − 1.

Practice Question

Q: Two independent samples give: n₁ = 12, x̄₁ = 45.2, s₁² = 6.1 and n₂ = 14, x̄₂ = 41.8, s₂² = 7.3. Find the pooled variance and the degrees of freedom you would use to find a confidence interval for the difference in means.

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7. Hypothesis test for the difference between two means

Putting it all together

This combines everything above: build the test statistic using the pooled variance, then compare it to a t-critical-value with n_x + n_y − 2 degrees of freedom, following the usual 5-step hypothesis testing structure. Watch closely for questions that test a specific difference (e.g. "boys are more than 5cm taller than girls") rather than simply "is there a difference" — the hypothesised difference then appears directly inside the test statistic.

Formula

t = [(x̄ − ȳ) − (μ_x − μ_y)] / [s_p × √(1/n_x + 1/n_y)]

(μ_x − μ_y) here is whatever H₀ claims the true difference is — often 0, but not always (see example below).

Worked Example

8 boys: x̄=152, s²=43. 8 girls: ȳ=134, s²=51. Test at 10% whether boys are, on average, more than 5cm taller than girls.

Hypotheses	H₀: μ_x = μ_y + 5, H₁: μ_x > μ_y + 5
Significance & df	α = 0.1 (one-tailed), v = 8+8−2 = 14
Critical value	t₁₄(0.1) = 1.345, critical region t ≥ 1.345
Pooled variance	s_p² = (7×43 + 7×51)/14 = 47 → s_p = 6.8556
Test statistic	t = [(152 − 134) − 5] / [√47 × √(1/8 + 1/8)] = 3.792
Decision & conclusion	3.792 > 1.345 → reject H₀. Sufficient evidence that boys' mean height exceeds girls' mean height by more than 5cm.

Exam-winning habit Whenever H₀ isn't a simple "no difference" claim, subtract that claimed difference explicitly in your test statistic — don't just plug in (x̄ − ȳ) on its own. Missing this is one of the most common lost marks in this topic.

Practice Question

Q: State, in your own words, the two key assumptions needed before you can use a pooled-variance t-test to compare two independent means.

▶ Show Answer

What to Memorise

Your one-stop formula reference

Situation	Formula
Test statistic (one mean)	t = (x̄ − μ)/(s/√n), v = n − 1
Unbiased variance estimator	S² = (1/n)(Σx² − nx̄²)
CI for one mean	x̄ ± t_n−1(α/2) × s/√n
Paired t-test statistic	t = (d̄ − μ_D)/(s/√n), v = n − 1 (n = number of pairs)
Pooled variance (two samples)	s_p² = [(n_x−1)s_x² + (n_y−1)s_y²]/(n_x+n_y−2)
CI for difference of two means	(x̄−ȳ) ± t_c·s_p√(1/n_x+1/n_y), v = n_x+n_y−2
Test statistic (difference of two means)	t = [(x̄−ȳ) − (μ_x−μ_y)] / [s_p√(1/n_x+1/n_y)]
As v → ∞	t-distribution → N(0,1)

Concepts Checklist

Tick each one off only once you could explain it out loud, no notes

I can explain why the t-distribution is needed instead of the normal distribution for small samples with unknown variance
I can read a t-table to find t given P(X > t)
I can convert "less than" and two-tailed probability statements into the correct table lookup
I can construct a confidence interval for a single population mean with unknown variance
I can carry out a full 5-step hypothesis test for a single mean using the t-distribution
I know when to use a one-tailed vs two-tailed test based on the wording of the question
I can set up and carry out a paired t-test on "before and after" data
I can calculate a pooled estimate of variance from two independent samples
I can build a confidence interval for the difference between two independent means
I can run a hypothesis test for the difference between two means, including when H₀ specifies a non-zero difference
I always sketch a diagram before reading values off the t-table

Exam Tips

What examiners are actually looking for — and where marks get lost

Forgetting to halve α for confidence intervals A confidence interval is always two-tailed by nature — a 95% CI needs the 0.025 column of the t-table, not 0.05. This is one of the most frequently lost marks in the whole chapter.

Using the wrong degrees of freedom One sample: v = n − 1. Paired test: v = (number of pairs) − 1. Two independent samples: v = n_x + n_y − 2. Mixing these up is an easy way to read the wrong critical value.

Misreading "increased/decreased" vs "changed" "Changed" = two-tailed test (H₁: μ ≠ …). "Increased" or "decreased" = one-tailed test (H₁: μ > … or μ < …). Always match your alternative hypothesis to the exact wording of the question.

Dropping the hypothesised difference in two-mean tests If H₀ claims μ_x − μ_y = 5 (not 0), you must subtract that 5 inside your test statistic. Forgetting it and just using (x̄ − ȳ) is a very common error.

Not writing the conclusion in context Examiners want the final line to refer back to the original scenario ("...evidence that women have gotten taller"), not just "reject H₀". A conclusion without context loses marks even if all the working is correct.

Always sketch first Before reading any value off a t-table, draw a quick bell curve and shade the region the question is asking about. This single habit prevents almost every sign error and "which column" mistake.

State assumptions explicitly For two-sample problems, examiners expect you to state that both populations are assumed normally distributed with equal (though unknown) variance — this is worth marks on its own in many mark schemes.

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