The One-Sample T-Test

A claim and twelve bottles

A bottling machine is claimed to fill bottles with a mean of 500 ml. Twelve bottles are taken from it and measured. Their mean is x̄ = 496 ml, and their standard deviation is s = 6 ml.

The hypotheses are about $\mu$μ, the mean fill of every bottle the machine makes: $H_0$H₀: $\mu =500$μ = 500 and $H_1$H₁: $\mu \ne 500$μ ≠ 500. The question is whether the machine is off in either direction, so the test is two-tailed, at the 5 percent level.

A z-test would need $\sigma$σ, the standard deviation of every bottle the machine makes. Nobody has given it. The only measure of spread is s, worked out from the same twelve bottles.

An estimated spread

s is the sample standard deviation with divisor n − 1: the squared deviations from x̄ are added and divided by 11, not by 12, and s is the square root of that.

Twelve bottles give only a rough estimate of $\sigma$σ. Some samples underestimate it, and then the gap from ${\mu }_0$μ₀ looks larger than it is. So the statistic made with s varies more than z does, and it follows a t distribution instead of the standard normal.

A t distribution is centered on 0 and symmetric, like the standard normal, but lower in the middle and heavier in the tails. Which t distribution depends on the degrees of freedom.

The gold curve is the t distribution with 11 degrees of freedom. The dashed curve is the standard normal. The t curve is lower at 0 and stays higher out in both tails.

The statistic and its degrees of freedom

The statistic has the same shape as z, with s in place of $\sigma$σ: $t=\frac{\bar{x}-{\mu }_0}{\frac{s}{\sqrt{n}}}$t = (x̄ − μ₀)/(s/√n). The top is the gap between the sample mean and the claimed mean. The bottom is the estimated standard error.

It has n − 1 degrees of freedom. The twelve deviations from x̄ always add up to zero, so once eleven of them are known, the twelfth is fixed. Working out s uses x̄, and that costs one degree of freedom: 12 bottles give 11.

The twelve bottles

The estimated standard error is $\frac{s}{\sqrt{n}}=\frac{6}{\sqrt{12}}=\frac{6}{3.464}=1.732$s/√n = 6/√12 = 6/3.464 = 1.732 ml.

The gap is 496 − 500 = −4 ml, so $t=-\frac{4}{1.732}=-2.309$t = −4/1.732 = −2.309, which the lesson rounds to −2.31. The sample mean is 2.31 estimated standard errors below the claimed mean.

The t distribution with 11 degrees of freedom. The shaded tail begins at −2.20 and holds 2.5 percent of the area; the same 2.5 percent lies above 2.20, at the other end. The bottles’ t = −2.31 lies just inside the shaded tail.

The verdict

With 11 degrees of freedom, $P\left(T\ge 2.201\right)=0.025$P(T ≥ 2.201) = 0.025. For a two-tailed test at 5 percent, 2.5 percent goes in each tail, so the critical values are −2.201 and 2.201: reject $H_0$H₀ if |t| > 2.201.

|t| = 2.31 > 2.20, so reject $H_0$H₀. Stated about the machine: there is evidence at the 5 percent level that its mean fill is not 500 ml.

The p-value is $2\times P\left(T\ge 2.309\right)=0.0413$2 × P(T ≥ 2.309) = 0.0413, below 0.05. At the 1 percent level the critical value is 3.106, and 2.31 falls short of it, so the same twelve bottles would not be enough evidence at 1 percent.

One tail

If the only question had been whether the machine underfills, $H_1$H₁: $\mu <500$μ < 500, all 5 percent would go in the lower tail. With 11 degrees of freedom, $P\left(T\le -1.796\right)=0.05$P(T ≤ −1.796) = 0.05, so $H_0$H₀ is rejected if t < −1.796.

t = −2.31 < −1.796, so $H_0$H₀ is rejected here too, with p-value $P\left(T\le -2.309\right)=0.0207$P(T ≤ −2.309) = 0.0207, half the two-tailed one. The tails are chosen from the question before the data are seen, never after.

t or z

When $\sigma$σ is given, the z-test applies and the two-tailed 5 percent critical value is 1.96. When the sample must estimate the spread with s, the t-test applies, on n − 1 degrees of freedom.

The smaller the sample, the more the two differ. The two-tailed 5 percent critical value of t is 2.201 for 11 degrees of freedom, 2.042 for 30 and 1.984 for 100. As the sample grows, s becomes a close estimate of $\sigma$σ and t gets closer to z.

Like the z-test, the t-test assumes the population is roughly normal. With twelve bottles there are too few for the central limit theorem to make up for a population that is far from normal.

The usual mistakes

Dividing by s instead of $\frac{s}{\sqrt{n}}$s/√n. $-\frac{4}{6}=-0.67$−4/6 = −0.67 compares the gap with the spread of single bottles; the test is about the mean of twelve.

Using n degrees of freedom. Twelve bottles give 11, because s is worked out from x̄.

Using 1.96. That critical value belongs to the normal distribution; with 11 degrees of freedom it is 2.201.

Saying $H_0$H₀ is proved when |t| falls short. Not rejecting means only that the sample did not give enough evidence against the claimed mean.

Nine batteries

In the application below, a consumer group tests 9 batteries against a claimed mean life of 20 hours. The variance is estimated from the nine lives with divisor 8, the test is one-tailed, and the statistic is compared with t on 8 degrees of freedom at 5 percent and at 1 percent.