Respuesta :
Answer:
Part a: At-least one circuit is different.
Part b: Circuit 2 is different.
Part c: Scaled t plots agree with part a and b.
Part d: Circuit 2 is different from average of Circuit 1 and Circuit 3
Part e: Either type 1 or type 3 can be selected as they have the lowest response time and are not different.
Part f: The normal and equal variance assumptions on the model errors are satisfied.
Step-by-step explanation:
For test purposes the hypothesis are formed as
H0=All circuits are not different
H1=At-least 1 circuit is different.
Using following procedure in Minitab, the output is as attached with the solution.
Procedure at Minitab
Step 01: Enter Data as in following pattern
- In column 1 add heading as Circuit Type and write the number of circuit for each response time.
- In column 2 add the heading as Response time and write each entry in a new row.
Step 02: Select and Configure the specific method of Analysis
- Choose Stat>ANOVA>One-Way from the Menu.
- In Response option, Enter the Column of Response Time.
- In Factor option, Enter the Column of Circuit Type.
- In Options, Set value of Confidence Level as 99 (100-1%)
- In Comparisons
Select Tukey.
Set value of error rate of comparison as 1.
- In Graphs
Select Normal Probability plot of Residuals
Select Residuals vs Fits
Set Value of Residual vs the Variable as the Circuit Type.
- Select OK.
Part a:
- The corresponding P-value is 0.000 which is less than α = 0.01.
This indicates that the Null Hypothesis is not valid i.e. At-least 1 circuit is different.
Part b:
Scale is calculated as
[tex]S=\sqrt{\frac{MS_E}{n}}\\S=\sqrt{\frac{16.9}5}}\\S=1.8385[/tex]
Now the value of MSD is given as
[tex]MSD=q_{0.01}_{(f,n)}S\\MSD=q_{0.01}_{(3,12)} \times 1.8385\\MSD=5.04 \times 1.8385\\MSD=9.266\\[/tex]
For comparison with the mean of individual circuit response times it is observed that
1 vs. 2: |10.8-22.2|=11.4 > MSD
1 vs. 3: |10.8-8.4|=2.4 < MSD
2 vs. 3: |22.2-8.4|=13.8 > MSD
1 and 2 are different. 2 and 3 are different.
Notice that the results indicate that the mean of treatment 2 differs from the means of both treatments 1 and 3, and that the means for treatments 1 and 3 are the same. This is also evident in the grouping output of the Minitab.
Part c:
The scaled-t plot agrees with part (b) and part (a). In this case, the large difference between the mean of treatment 2 and the other two treatments is very obvious.
Part d:
[tex]H0 \,\, is\, \, \mu_1-2\mu_2+\mu_3=0\\H1 \, \,is\, \, \mu_1-2\mu_2+\mu_3\neq 0[/tex]
So
[tex]C_1=y_1-2y_2+y_3\\Here\\y_1=9+12+10+8+15 =54\\y_2=20+21+23+17+30 =111\\y_3=6+5+8+16+7 = 42\\\\C_1=y_1-2y_2+y_3\\C_1=54-2\times 111+42\\C_1=-126[/tex]
Now the standard error and Fc are given as
[tex]SS_c_1=\frac{C_1^2}{5 \times 6}\\SS_c_1=\frac{(-126)^2}{5 \times 6}\\SS_c_1=529.2\\\\F_c_1=\frac{SS_c_1}{MS_E}\\F_c_1=\frac{529.2}{16.9}\\F_c_1=31.31[/tex]
From this it is concluded that the it differs from the average of type 1 and type 3.
Part e
Either type 1 or type 3 can be selected as they have the lowest response time and are not different.
Part f
The residual plot as in the output of MInitab has some unequal variances.
- However, the Levene’s test shows that the equal variance assumption is valid (P-value = .6145).
The normal probability plot has some points that do not lie along the line in the upper region. These are outliers in the data.
- However, the standardized residuals at those points are still within ±2.
It is therefore concluded that the normal and equal variance assumptions on the model errors are satisfied.