Question

Consider a study involving turbulent water jets. Shape factor was determined for five different nozzle designs at six levels of jet efflux velocity. Researchers were particularly interested in potential differences in shape factor between nozzle designs, with jet efflux velocity considered a nuisance factor. The data for this experiment are provided below. The first column provides the nozzle design number, the second column provides the jet efflux velocity (in m/s), and the third column provides the shape factor measurement. Note that the run order is not provided. 1. Use SAS to perform ANOVA for a randomized complete block design to test for differences in the population mean shape factor among the five different designs. Use α = 0.05. Be sure to state the null and alternative hypotheses, identify the test statistic, identify the p-value for the test, determine whether to reject or fail to reject the null hypothesis, and state a conclusion within the context of the problem. 2. By hand, calculate the critical value for comparing pairs of mean using both the LSD and Tukey methods. In other words, determine Tα and LSD using the results of this experiment. HINT: Use the MSE from the SAS output to perform your calculations. 3. Determine which pairs of nozzle designs have significantly different treatment means using both the LSD and Tukey methods. NOTE: You may use the lsmeans or means statement in PROC GLM to make this determination. You do not need to perform calculations by hand. 4. Which method (LSD or Tukey’s) would you recommend using to determine significant differences among pairs of treatment means for this experiment? Clearly explain your choice. 5. Determine if the model assumptions are reasonably met. Be sure to state the assumptions of the test and identify which plot/test can be used to evaluate each assumption.

Design	JEV	SF
1	11.73	0.78
2	11.73	0.85
3	11.73	0.93
4	11.73	1.14
5	11.73	0.97
1	14.37	0.8
2	14.37	0.85
3	14.37	0.92
4	14.37	0.97
5	14.37	0.86
1	16.59	0.81
2	16.59	0.92
3	16.59	0.95
4	16.59	0.98
5	16.59	0.78
1	20.43	0.75
2	20.43	0.86
3	20.43	0.89
4	20.43	0.88
5	20.43	0.76
1	23.46	0.77
2	23.46	0.81
3	23.46	0.89
4	23.46	0.86
5	23.46	0.76
1	28.74	0.78
2	28.74	0.83
3	28.74	0.83
4	28.74	0.83
5	28.74	0.75

Answer 1

Ans:

Here is the ANOVA summary with  α = 0.05.

ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Sample	484.1863	5	96.83727	29145.9	2.52E-82	2.408514
Columns	5057.078	1	5057.078	1522070	1.1E-109	4.042652
Interaction	498.9345	5	99.78691	30033.68	1.22E-82	2.408514
Within	0.15948	48	0.003323
Total	6040.359	59

Anova: Two-Factor With Replication
SUMMARY	JEV	SF	Total
1
Count	5	5	10
Sum	58.65	4.67	63.32
Average	11.73	0.934	6.332
Variance	0	0.01863	32.38428
1
Count	5	5	10
Sum	71.85	4.4	76.25
Average	14.37	0.88	7.625
Variance	0	0.00435	50.55196
1
Count	5	5	10
Sum	82.95	4.44	87.39
Average	16.59	0.888	8.739
Variance	0	0.00777	68.49034
1
Count	5	5	10
Sum	102.15	4.14	106.29
Average	20.43	0.828	10.629
Variance	0	0.00457	106.7349
1
Count	5	5	10
Sum	117.3	4.09	121.39
Average	23.46	0.818	12.139
Variance	0	0.00317	142.407
1
Count	5	5	10
Sum	143.7	4.02	147.72
Average	28.74	0.804	14.772
Variance	0	0.00138	216.784
Total
Count	30	30
Sum	576.6	25.76
Average	19.22	0.858667
Variance	33.89855	0.007667