Tyre wear on five cars
Topic: Hypothesis tests · more questions on this topic
‹ Paired or two-sample? A one-sided sign test ›
Each of five cars carries one tyre of type A and one of type B. Tread thickness after 10000 km is
| Car | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Type A | 10.6 | 9.8 | 12.3 | 9.7 | 8.8 |
| Type B | 10.2 | 9.4 | 11.8 | 9.1 | 8.3 |
Test whether the mean tread differs between the two types, at level 0.05. Use t_{4;0.975}=2.776.
The differences A minus B are 0.4, 0.4, 0.5, 0.6 and 0.5, with mean \overline d=0.48 and standard deviation s_d=0.0837. Model them as a normal random sample with mean \mu_D, and test H_0:\mu_D=0 against \mu_D\neq0:
t=\frac{0.48}{0.0837/\sqrt5}=\frac{0.48}{0.0374}=12.8.
Since 12.8>t_{4;0.975}=2.776, reject H_0; the two-sided p-value is 0.0002. Type A keeps more tread. Ignoring the pairing and comparing the two columns as independent samples would be wrong: the variation between cars, from 8.3 to 12.3, is far larger than the difference between tyres.