Question

Engin. 8.27 Wludyka and Nelson (1997) describe the following study. In the manufacture of soft contact lenses, a monomer is injected into a plastic frame, the monomer is subjected to ultraviolet light and heated the time, temperate, and light intensity are varied), the frame is removed, and the lens is hydrated. It is thought that temperature can be manipulated to target the power (strength of the lens), so comparing the variability in power is of interest. The data are coded deviations from the target power using monomers from five different suppliers given below. Sample Supplier 1 2 3 4 5 6 7 8 9 191. 9 178. 2 156.6 125.8 218.6 1 89.1 1 74.1 158.4 132.4 208.4 190.9 170.3 157.7 132.2 187.1 183.8 171.6 154.1 133.0 199.5 185.5 171.7 152.3 133.2 202.0 190.9 174.7 161.5 125.9 211.1 192.8 176.0 158.1 132.9 197.6 188.4 176.6 150.9 142.6 204.4 189.0 172.8 156.9 135.5 206.8 a. Do the suppliers appear to differ in their levels of variability? Use a = .05. b. Is there significant evidence of a difference in the mean deviations for the five suppliers? Use an a = .05 AOV F test. c. Apply the Kruskal-Wallis test to evaluate differences in the distributions of the deviations for the five suppliers? Use a = .05. d. Suppose a difference in mean deviations of 20 units would have commercial con- sequences for the manufacturer of the lenses. Does there appear to be a practical difference in the materials from the five suppliers?

Answer 1

a) and b)

Null and Alternative Hypothesis:

H₀: µ_{Supplier 1} = µ_{Supplier 2} = µ_{Supplier 3} = µ_{Supplier 4}= µ_{Supplier 5}

H₁: Not all Means are equal

Alpha = 0.05

N = 45

n = 9

Degress of Freedom:

df_Between = a – 1 = 5-1 =4

df_Within = N-a = 45-5 = 40

df_Total = N-1 = 45-1 = 44

Critical Value:

(df_Between, df_Within): (4,40) = 2.61

Decision Rule:

If F is greater than 2.61, reject the null hypothesis

Test Statistics:

SS_Betwen = ∑(∑a_i)²/n - T²/N = 28,024.35

SS_Within = ∑(Y)² - ∑(∑a_i)²/n = 1053.79

SS_Total = SS_{Betwen ­­}+ SS_Within = 29078.14

MS = SS/df

               

F = MS_effect / MS_error

Hence,

F = 7006.09/26.34 = 265.94

Source of Variation	SS	df	MS	F
Between Groups	28,024.35	4	7,006.09	265.94
Within Groups	1,053.79	40	26.34
Total	29,078.14	44

Result:

Our F = 265.94, we reject the null hypothesis

Conclusion:

Not all means are equal.

c)

Kruskal Wallis test:

Null and Alternate Hypothesis:

H0: Means for all groups are same

Ha: Means for all groups are not same

Alpha = 0.05

Supplier 1	Rank	Supplier 2	Rank	Supplier 3	Rank	Supplier 4	Rank	Supplier 5	Rank
191.9	36	178.2	27	156.6	13	125.8	1	218.6	45
189.1	33	174.1	23	158.4	17	132.4	4	208.4	43
190.9	34	170.3	19	157.7	15	132.2	3	187.1	30
183.8	28	171.6	20	154.1	12	133	6	199.5	39
185.5	29	171.7	21	152.3	11	133.2	7	202	40
190.9	34	174.7	24	161.5	18	125.9	2	211.1	44
192.8	37	176	25	158.1	16	132.9	5	197.6	38
188.4	31	176.6	26	150.9	10	142.6	9	204.4	41
189	32	172.8	22	156.9	14	135.5	8	206.8	42
Total	294		207		126		45		362

H = (12/n*(n+1))* ∑R_i²/n_i – 3*(n+1)

n = 45

n_i = 9

H = (12 /(45*46))*(294²/9 + 207²/9 +126²/9 +45²/9 +362²/9) – 3 * 46 = 41.21

df = a-1 = 5-1 = 4

Chi Square critical = 9.488

Results and Conclusion

Since H > Chi Square critical, we reject the null hypothesis ie the means are not same.

d)

	Supplier 1	Supplier 2	Supplier 3	Supplier 4	Supplier 5
Std Dev	2.95	2.62	3.30	5.00	8.96
Mean	189.14	174.00	156.28	132.61	203.94

Since none of the Std Deviations is more than 10 and the limit for concern is 20, hence there is no practical difference in the 5 suppliers.