Question

Data from a one-way layout are given below. Four treatments correspond to area sizes 0.016, 0.030, 0.044 and 0.058. You may regard them simply as treatments A, B, C and D.

(a) Construct the ANOVA table (follow lecture example to get all relevant quantities computed. Use a single line R-command only for verification).

(b) Use F-test for no difference in their means. State your statistical conclusion in a full sentence (e.g. the treatments are found statistically significantly different at 5% level).

(c) Carry out multiple comparisons for treatment differences (construct 95% simultaneous confidence intervals by the Tukey’s method).

data:

area velocity

0.016 294.9

0.016 294.1

0.016 301.7

0.016 307.9

0.016 285.5

0.016 298.6

0.016 303.1

0.016 305.3

0.016 264.9

0.016 262.9

0.016 256

0.016 255.3

0.016 256.3

0.016 258.2

0.016 243.6

0.016 250.1

0.03 295

0.03 301.1

0.03 293.1

0.03 300.6

0.03 285

0.03 289.1

0.03 277.8

0.03 266.4

0.03 248.1

0.03 255.7

0.03 245.7

0.03 251

0.03 254.9

0.03 254.5

0.03 246.3

0.03 246.9

0.044 270.5

0.044 263.2

0.044 278.6

0.044 267.9

0.044 269.6

0.044 269.1

0.044 262.2

0.044 263.2

0.044 224.2

0.044 227.9

0.044 217.7

0.044 219.6

0.044 228.5

0.044 230.9

0.044 227.6

0.044 228.6

0.058 258.6

0.058 255.9

0.058 257.1

0.058 263.6

0.058 262.6

0.058 260.3

0.058 305.3

0.058 304.9

0.058 216

0.058 216

0.058 210.6

0.058 207.4

0.058 214.6

0.058 214.3

0.058 222.1

0.058 222.2

Answer 1

One-way ANOVA calculations

STEP 1 Compute CM, the correction for the mean.

= (TOTAL of all observations)^2/Ntotal

= 16590.4^2/64

= 4300646.44

STEP 2 Compute the total SS.

The total SS = SS(Total) = sum of squares of all observations −CM.

= (294.9^2+...+222.2^2) - 4300646.44

= 4352411.68-4300646.44 = 51765.24

STEP 3 Compute SST, the treatment sum of squares.

First we compute the total (sum) for each treatment.

T1 = 294.9+...250.1 = 4438.4

T2 = 295+...+246.9 = 4311.2

T3 = 270.5+...+228.6 = 3949.3

T4 = 258.6+...+222.2 = 3891.5

= (4438.4^2/16+4311.2^2/16+3949.3^2/16+3891.5^2/16) - 4300646.44

= 4314161.421 - 4300646.44

= 13514.9812

STEP 4 Compute SSE, the error sum of squares.

Here we utilize the property that the treatment sum of squares plus the error sum of squares equals the total sum of squares. Hence,

SSE = SS(TOTAL) - SST = 51765.24-13514.9812 = 38250.26

STEP 5 Compute MST, MSE, and their ratio, F.

MST = SST/(k-1) = 13514.9812/3 = 4504.99373

MSE = SSE/(N-k) = 38250.26/(64-4) = 38250.26/60 = 637.5043

Finally, compute F as

F = MST/MSE = 4504.99373/637.5043 = 7.0666

b)

The test statistic is the F value of 7.0666. Using an α of 0.05,
F_0.05;3,60 = 2.76

Since the test statistic is much larger than the critical value, we reject the null hypothesis of equal population means and conclude that there is a (statistically) significant difference among the population means.