MathsFurther Statistics 2 › Confidence intervals and tests with the t-distribution

Confidence intervals and tests with the t-distribution

Estimating the variance from the same small sample costs something, and the t-distribution is the price. Wider tails, wider intervals, and three standard situations that use it.

Year FMEDEXCEL 9FM0 FS2

Builds on Comparing two normal means and Estimators, standard error and confidence intervals.

IN THIS TOPIC

  • Carry out a one-sample t test and build the matching confidence interval.
  • Recognise paired data and reduce it to a single sample of differences.
  • Pool two sample variances and run a two-sample t test.

WHAT YOU PROBABLY THINK

With a sample of 12 you should use z = 1.96 for a 95% interval, since the sample mean is normal.

Paying for the estimated variance

When σ is unknown and estimated by s from the same small sample, the standardised mean no longer follows the normal distribution:

sample mean - μSn is t on n - 1 degrees of freedom

The t distribution is symmetric and centred at zero like the normal, but with heavier tails, so the value cutting off 2.5% sits further out: 2.201 on 11 degrees of freedom against 1.96 for the normal. Using 1.96 anyway, as the opening claim suggests, makes every interval too narrow and every test too willing to reject. As n grows the two converge, and beyond about 30 the difference stops mattering much.

The t distribution against the standard normal: the same centre, heavier tails, and a wider 95% intervalnormal: 1.96t with 11 df: 2.201less known, so more allowance: t exceeds z at every df
FIG. 1The t distribution laid over the standard normal: the same centre, heavier tails, and a critical value further from zero.

WORKED EXAMPLE

A one-sample t test and interval

A sample of 12 has mean 25.4 and s = 3.2. Test at 5% whether the population mean exceeds 23, and give a 95% interval for it.

Standard error = 3.2/√12 = 0.924. t = (25.4 − 23)/0.924 = 2.60 on 11 degrees of freedom.

The one-tailed 5% critical value is 1.796, and 2.60 exceeds it, so reject: there is evidence the mean exceeds 23.

For the interval use 2.201: 25.4 ± 2.201 × 0.924 gives (23.37, 27.43), which excludes 23 as the test implies.

Paired data, and two independent samples

When each reading in one sample has a natural partner in the other, before and after on the same subject, subtract within pairs and test the single sample of differences against zero. That removes the variation between subjects, which is usually the largest source of noise, and is why a paired design detects smaller effects than an unpaired one with the same number of readings.

Seven paired differences reduced to one sample: mean 2, and t = 2.45 on 6 degrees of freedommean 2pair the data, then test the differences against 0t = 2 / (2.160/√7) = 2.45
FIG. 2Seven paired differences with their mean of 2, reduced to a single sample and tested against zero.

For two genuinely independent samples with a common unknown variance, pool the two estimates in proportion to their degrees of freedom:

s2 = (n1 - 1)s12 + (n2 - 1)s22n1 + n2 - 2

The test statistic then divides the difference of the means by s√(1/n1 + 1/n2), on n1 + n2 − 2 degrees of freedom. The assumption of equal variances is what an F test checks, so the two topics belong together.

YOUR TURN

Pooling two samples

Sample 1: n = 8, mean 42, s = 5. Sample 2: n = 10, mean 38, s = 4. Test at 5% in two tails whether the means differ, given a critical value of 2.120.

Show the working

Pooled variance = (7 × 25 + 9 × 16)/16 = 319/16 = 19.94, so s = 4.465.

Standard error = 4.465 × √(1/8 + 1/10) = 4.465 × 0.4743 = 2.118.

t = 4/2.118 = 1.89 on 16 degrees of freedom.

Since 1.89 < 2.120, do not reject: there is insufficient evidence at the 5% level that the population means differ.

THE EXAM BIT

  • Count the degrees of freedom before opening the tables: n − 1, or n₁ + n₂ − 2 when pooling.
  • Look for the word 'paired' or a natural pairing in the data; the unpaired test on paired data throws information away.
  • Pool by weighting each sample variance by its own degrees of freedom, never by averaging the two.
  • State the assumptions: normal populations, and equal variances for the pooled test.

CHECK YOURSELF

A sample of 9 has mean 51 and s = 6. Find the test statistic for H₀: μ = 48 and state the degrees of freedom.

Show a hint

Standard error first, then divide.

Show the answer

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With σ estimated from the sample, use t on n − 1 degrees of freedom: heavier tails than the normal, so wider intervals.

Paired data becomes one sample of differences; independent samples with a common variance use the pooled s² on n₁ + n₂ − 2 degrees of freedom.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

CHECK YOUR PROGRESS

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  • Carry out a one-sample t test and build the matching confidence interval.
  • Recognise paired data and reduce it to a single sample of differences.
  • Pool two sample variances and run a two-sample t test.

Open the full revision checklist to see every objective in the course in one place.

No animated video for this topic yet; these notes stand alone.