SG1 - One-sample t-tests

SG1 - One-sample t-tests

Test a claim about the mean of a normal population when its variance is unknown. You will estimate the variance correctly, use the appropriate t distribution and match the rejection tail to the contextual alternative hypothesis.

A normal mean with unknown variance

Nine measurements have mean 52 when a process is claimed to average 50. Is that enough evidence of an increase? The difference of 2 matters only in relation to the uncertainty in the sample mean.

Recall the notation X∼N(μ,σ2)X\sim\mathrm{N}(\mu,\sigma^2): μ\mu is the population mean, σ2\sigma^2 the population variance and σ\sigma the population standard deviation. Lower-case xˉ\bar x, s2s^2 and ss describe a realised sample; upper-case Xˉ\bar X and SS vary from one random sample to another.

When σ\sigma is known, the mean has standard error σ/n\sigma/\sqrt n, and the standardised mean is normal. When σ\sigma is unknown, we estimate it using ss. The denominator now varies too: replacing σ\sigma by an estimate creates additional uncertainty. The resulting reference distribution is Student's t distribution, with heavier tails than the standard normal distribution.

The one-sample t statistic

For a random sample of n≥2n\ge2 independent observations from a normal population with unknown positive variance, under H0:μ=μ0H_0:\mu=\mu_0,

T=Xˉ−μ0S/n∼tn−1T=\frac{\bar X-\mu_0}{S/\sqrt n}\sim t_{n-1}

The observed statistic uses xˉ\bar x and the positive sample standard deviation ss; s/ns/\sqrt n is the estimated standard error of the mean.

The t pivot, including its normal-population condition and n−1n-1 degrees of freedom, is in the formula booklet.

The normality assumption concerns the population of individual measurements, not a claim that a small set of observed numbers “looks symmetrical”. For a small sample, symmetry alone is insufficient. Independence and representative random sampling are also needed. Repeated measurements on the same person are not automatically independent observations for this procedure.

Notation recall: Z∼N(0,1)Z\sim\mathrm{N}(0,1) denotes a standard normal variable; ϕ\phi is its probability density function and Φ(z)=P(Z≤z)\Phi(z)=\mathrm{P}(Z\le z) its cumulative distribution function. These are normal-distribution symbols. Use a t cumulative distribution, rather than Φ\Phi, to calculate a finite-degree-of-freedom t-test p-value.

Key idea

Unknown population variance is estimated from the sample. For independent normal observations, that leads to a t reference distribution, not an exact standard normal one.

Estimating variance and standard error

A calculator may display both a standard deviation calculated with divisor nn and one with divisor n−1n-1. The one-sample t statistic needs the latter. Use s2s^2 here for the unbiased estimate of population variance. “Unbiased” means its average over repeated random samples equals σ2\sigma^2; it does not mean that every sample estimate equals the unknown true variance. Taking the square root does not make ss itself an unbiased estimator of σ\sigma.

Unbiased variance estimate

For a random sample of n≥2n\ge2 independent observations from one population with finite variance,

s2=∑(xi−xˉ)2n−1=∑xi2−(∑xi)2/nn−1,SE⁡^(Xˉ)=sns^2=\frac{\sum(x_i-\bar x)^2}{n-1} =\frac{\sum x_i^2-(\sum x_i)^2/n}{n-1},\qquad \widehat{\operatorname{SE}}(\bar X)=\frac{s}{\sqrt n}

The first expression for the unbiased variance estimator is in the formula booklet. The summary-statistic rearrangement and separate estimated-standard-error formula are not printed as standalone formulas; they follow from the supplied variance expression and t pivot.

Why n−1n-1? Deviations from the fitted sample mean sum to zero, so fitting that mean has used one independent piece of information. The next part makes this constraint concrete.

Worked example 1: variance from deviations

Question. Nine process measurements have xˉ=52\bar x=52 and ∑(xi−xˉ)2=72\sum(x_i-\bar x)^2=72. Find s2s^2, ss and the estimated standard error of the mean.

Plan. Divide the sum of squared deviations by 8, take its square root and then divide by 3.

s2=729−1=9,s=3,sn=39=1s^2=\frac{72}{9-1}=9,\qquad s=3,\qquad\frac{s}{\sqrt n}=\frac3{\sqrt9}=1

Answer. The estimated variance is 9 squared units, standard deviation is 3 units and standard error is 1 unit.

Check. Reconstruct the deviation sum: (n−1)s2=8×9=72(n-1)s^2=8\times9=72. Also, averaging nine observations gives a smaller spread than individual measurements: 1<31<3.

Keep the same sample but change the way its information is supplied. This is the common summary-statistic form of the same calculation.

Worked example 2: variance from summary totals

Question. The same nine measurements are summarised by ∑xi=468\sum x_i=468 and ∑xi2=24408\sum x_i^2=24408. Recover the mean, variance and standard error.

Plan. Find xˉ\bar x, then subtract the squared-mean contribution from the sum of squares before dividing by n−1n-1.

xˉ=468/9=52∑(xi−xˉ)2=24408−4682/9=72s2=72/8=9,s/9=1\begin{aligned} \bar x&=468/9=52\\ \sum(x_i-\bar x)^2&=24408-468^2/9=72\\ s^2&=72/8=9,\qquad s/\sqrt9=1 \end{aligned}

Answer. xˉ=52\bar x=52, s2=9s^2=9 and estimated standard error 1, agreeing with Worked example 1.

Check. The independent decomposition ∑xi2=(n−1)s2+nxˉ2\sum x_i^2=(n-1)s^2+n\bar x^2 gives 8(9)+9(522)=244088(9)+9(52^2)=24408, the supplied total.

The identity behind the rearrangement is ∑(xi−xˉ)2=∑xi2−2xˉ∑xi+nxˉ2=∑xi2−nxˉ2\sum(x_i-\bar x)^2=\sum x_i^2-2\bar x\sum x_i+n\bar x^2=\sum x_i^2-n\bar x^2. Retain full precision until the final statistic or interval.

Watch out

Wrong: use 72/972/9 for the unbiased variance, or put s2/ns^2/\sqrt n in the t denominator. Correct: s2=72/8s^2=72/8, then the denominator is s/ns/\sqrt n.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account