Question

The response time in milliseconds was determined for three different types of circuits that could be used in an automatic valve shutoff mechanism.
The results are shown in the following table.

Circuit Type Response Time
1 9 12 10 8 15
2 20 21 23 17 30
3 6 5 8 16 7

(a) Test the hypothesis that the three circuit types have the same response time. Use α = 0.01
(b) Use Tukey’s test to compare pairs of treatment means. Use α = 0.01.
(c) Use the graphical procedure to compare the treatment means. What conclusions can you draw? How do they compare with the conclusions from part (a)?
(d) Construct a set of orthogonal contrasts, assuming that at the outset of the experiment you suspected the response time of circuit type 2 to be different from the other two.
(e) If you were a design engineer and you wished to minimize the response time, which circuit type would you select?
(f) Analyze the residuals from this experiment. Are the basic analysis of variance assumptions satisfied?

Answer 1

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:

1. 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

$S=\sqrt{\frac{MS_E}{n}}\\S=\sqrt{\frac{16.9}5}}\\S=1.8385$

Now the value of MSD is given as

$MSD=q_{0.01}_{(f,n)}S\\MSD=q_{0.01}_{(3,12)} \times 1.8385\\MSD=5.04 \times 1.8385\\MSD=9.266$

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:

              $H0 \,\, is\, \, \mu_1-2\mu_2+\mu_3=0\\H1 \, \,is\, \, \mu_1-2\mu_2+\mu_3\neq 0$

So

$C_1=y_1-2y_2+y_3\\Here\\y_1=9+12+10+8+15 =54\\y_2=20+21+23+17+30 =111\\y_3=6+5+8+16+7 = 42\\\\C_1=y_1-2y_2+y_3\\C_1=54-2\times 111+42\\C_1=-126$

Now the standard error and Fc are given as

$SS_c_1=\frac{C_1^2}{5 \times 6}\\SS_c_1=\frac{(-126)^2}{5 \times 6}\\SS_c_1=529.2\\\\F_c_1=\frac{SS_c_1}{MS_E}\\F_c_1=\frac{529.2}{16.9}\\F_c_1=31.31$

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.

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