Question

PROBLEM-1) (100 Pts) At an aircraft manufacturing company six test systems are used in the final test of the XYZ part. The test systems are being compared concerning the sensitivity measure of the product. A random sample of four sensitivity measures from each test system is used to determine whether the mean sensitivity measurement varies among test systems. The followings are the measurements in gr per square millimeter x 10-1. 1 5 6 18.6 15.8 16.9 17.5 2 15.4 17.7 19.2 16.4 Test System 3 4 18.9 18.9 17.8 20.8 15.7 16.7 20.3 14.6 20.5 16.5 19.2 17.5 20.1 17.5 16.2 18.3 a. (30 pts.) Perform the analysis of variance at the 0.05 level of significance and indicate whether or not the mean sensitivity measures of the test systems differ significantly for the six systems. b. (40 pts.) Find whether the systems differ significantly from each other by using Tukey's test. (a = 0.05) C. (40 pts.) Find whether the systems differ significantly from each other by using Duncan's test. (a = 0.05)

Answer 1

(A)

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ₁ = μ₂ = μ₃ = μ₄ = μ₅ = μ₆

Ha: Not all means are equal

The above hypotheses will be tested using an F-ratio for a One-Way ANOVA.

(2) Rejection Region

Based on the information provided, the significance level is \alpha = 0.05α=0.05, and the degrees of freedom are df_1 = 5df1​=5 and df_2 = 5df2​=5, therefore, the rejection region for this F-test is R = \{F: F > F_c = 2.773\}R={F:F>Fc​=2.773}.

(3) Test Statistics

The following table is obtained:

	Group 1	Group 2	Group 3	Group 4	Group 5	Group 6
	18.6	15.4	18.9	18.9	20.5	20.1
	15.8	17.7	17.8	20.8	16.5	17.5
	16.9	19.2	15.7	16.7	19.2	16.2
	17.5	16.4	20.3	14.6	17.5	18.3
Sum =	68.8	68.7	72.7	71	73.7	72.1
Average =	17.2	17.175	18.175	17.75	18.425	18.025
\sum_i X_{ij}^2 =∑i​Xij2​=	1187.46	1188.05	1332.63	1281.9	1367.39	1307.59
St. Dev. =	1.169	1.646	1.941	2.686	1.776	1.632
SS =	4.0999999999997	8.1275000000003	11.3075	21.65	9.4674999999997	7.9875000000004
n =	4	4	4	4	4	4

The total sample size is N = 24N=24. Therefore, the total degrees of freedom are:

dftotal​=24−1=23

Also, the between-groups degrees of freedom are dfbetween​=6−1=5, and the within-groups degrees of freedom are:

dfwithin​=dftotal​−dfbetween​=23−5=18

First, we need to compute the total sum of values and the grand mean. The following is obtained

∑​Xij​=68.8+68.7+72.7+71+73.7+72.1=427

Also, the sum of squared values is

∑​Xij2​=1187.46+1188.05+1332.63+1281.9+1367.39+1307.59=7665.02

Based on the above calculations, the total sum of squares is computed as follows

The within sum of squares is computed as shown in the calculation below:

F=MSwithin​MSbetween​​=3.481.068​=0.307

(4) Decision about the null hypothesis

Since it is observed that F=0.307≤Fc​=2.773, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.9024, and since p=0.9024≥0.05, it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that not all 6 population means are equal, at the α=0.05 significance level the mean sensitivity for all the groups is same.

(B)

Tukey HSD results

treatments pair	Tukey HSD Q statistic	Tukey HSD p-value	Tukey HSD inferfence
A vs B	0.0268	0.8999947	insignificant
A vs C	1.0453	0.8999947	insignificant
A vs D	0.5897	0.8999947	insignificant
A vs E	1.3133	0.8999947	insignificant
A vs F	0.8845	0.8999947	insignificant
B vs C	1.0721	0.8999947	insignificant
B vs D	0.6165	0.8999947	insignificant
B vs E	1.3401	0.8999947	insignificant
B vs F	0.9113	0.8999947	insignificant
C vs D	0.4556	0.8999947	insignificant
C vs E	0.2680	0.8999947	insignificant
C vs F	0.1608	0.8999947	insignificant
D vs E	0.7237	0.8999947	insignificant
D vs F	0.2948	0.8999947	insignificant
E vs F	0.4288	0.8999947	insignificant

Hence the mean difference among the 6 groups is insignificant at 0.05 significance level.