Try each one on paper first, then check a single step. That is worth far more than reading a finished solution.
Example 1 — a one-sample t-test
A sample of 16 has mean 105 and standard deviation 12. Test at the 5% level whether the population mean exceeds 100.
- 1H0:μ=100; H1:μ>100"Exceeds" gives a direction, so the test is one-tailed.
- 2σ is unknown and n<30, so use t with ν=15.One degree of freedom is lost to the sample mean.
- 3Critical value: t0.05,15=1.753Larger than the normal's 1.645 — the price of estimating σ.
- 4SE =1612=3
- 5t=3105−100=1.67
- 61.67<1.753: do not reject H0. Insufficient evidence at the 5% level.A z-test would have rejected — the distribution genuinely matters.
Example 2 — a confidence interval using t
For the same sample (n=16, xˉ=105, s=12), find a 95% confidence interval for μ.
- 1ν=15, and for a 95% interval the two-tailed value is t=2.131.Compare 1.96 for the normal.
- 2SE =3
- 3Margin of error =2.131×3=6.39
- 4105±6.39=(98.6, 111.4)
- 5The interval contains 100.Consistent with the test's failure to reject H0 — the two methods always agree.
Example 3 — choosing a paired test
Ten students take a test before and after a revision course. Which test applies, and what are the degrees of freedom?
- 1Each student appears twice, so the two sets of marks are not independent.A two-sample test would be invalid.
- 2Work with the differences d=after−before.Ten differences, one per student.
- 3Apply a one-sample t-test to those differences.H0:μd=0 — no average improvement.
- 4t=sd/ndˉ with n=10.
- 5ν=n−1=9Pairing removes the variation between students, which is what makes it powerful.