Tips and coupons
Topic: Analysis of variance · more questions on this topic
‹ A Tukey interval for two varieties Two treatments in blocks ›
Hardness readings for four tips on four metal coupons are
| Tip 1 | Tip 2 | Tip 3 | Tip 4 | |
|---|---|---|---|---|
| Coupon 1 | 9.3 | 9.4 | 9.2 | 9.7 |
| Coupon 2 | 9.4 | 9.3 | 9.4 | 9.6 |
| Coupon 3 | 9.6 | 9.8 | 9.5 | 10.0 |
| Coupon 4 | 10.0 | 9.9 | 9.7 | 10.2 |
Treat coupons as blocks and tips as treatments in a randomised block design, with one observation per cell. Compute the two F statistics and state the conclusion of each test at level 0.05. The 0.95 quantile of F_{3,9} is about 3.86.
The grand mean of the 16 readings is 9.625. The tip means, the column means, are 9.575, 9.600, 9.450 and 9.875, so
\mathrm{SS}_{\mathrm{tip}}=4\sum_j(\overline y_{\cdot j}-9.625)^2=0.385, \qquad \mathrm{MS}_{\mathrm{tip}}=0.385/3=0.1283.
The coupon means, the row means, are 9.400, 9.425, 9.725 and 9.950, and \mathrm{SS}_{\mathrm{coupon}}=0.825, so \mathrm{MS}_{\mathrm{coupon}}=0.275. The total sum of squares is 1.290, so the error sum of squares is 1.290-0.385-0.825=0.080, on (4-1)(4-1)=9 degrees of freedom, and \mathrm{MS}_{\mathrm{error}}=0.008889.
F_{\mathrm{tip}}=\frac{0.1283}{0.008889}\approx 14.44, \qquad F_{\mathrm{coupon}}=\frac{0.275}{0.008889}\approx 30.94.
Both exceed 3.86. At level 0.05 the test rejects equal tip means, and the coupons also differ, which shows that blocking on the coupon removed a large source of variation from the error.