T-Test

Testing whether a mean differs from a reference value or whether two group means differ, using the t-distribution when sample sizes are small.

A t-test asks whether the observed mean is unusually far from the null

Result

p = 0.0808

Fail to reject H0 at alpha = 0.05

df = 14

t=2.10

The p-value is the tail area at least as extreme as the observed t-statistic, assuming H0 is true.

Definition

The t-test is a statistical test used to compare means when the population variance is unknown and the sample is small.

It uses the t-statistic:

t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}

where $\bar{x}$ is the sample mean, $\mu_0$ is the hypothesized mean, $s$ is the sample standard deviation, and $n$ is the sample size.

The t-statistic follows a t-distribution with $n - 1$ degrees of freedom under $H_0$ .

One-sample t-test

A factory claims machines produce bolts with diameter 10 mm. You measure 9 bolts and get $\bar{x} = 10.3$ mm, $s = 0.4$ mm.

$t = \frac{10.3 - 10}{0.4/\sqrt{9}} = \frac{0.3}{0.133} = 2.25$

With 8 degrees of freedom, the two-tailed p-value $\approx 0.054$ . At $\alpha = 0.05$ , just fail to reject $H_0$ .

Try it

A diet claims to reduce weight by 5 kg. You weigh 16 people before and after. The mean reduction is 4.2 kg with $s = 3.2$ kg. Test $H_0: \mu = 5$ vs $H_1: \mu < 5$ at $\alpha = 0.05$ . ( $t_{0.05, 15} = 1.753$ )

Solution

$t = \frac{4.2 - 5}{3.2/\sqrt{16}} = \frac{-0.8}{0.8} = -1.0$

Since $|-1.0| < 1.753$ , fail to reject $H_0$ . Insufficient evidence the diet falls short.

Related concepts

Needs first

Hypothesis Testing Normal Distribution Standard Deviation

ANOVA Mann-Whitney Test

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