University of Texas at El Paso IE 3352 Design of Experiments 2. Four different d

ID: 3052155 • Letter: U

Question

University of Texas at El Paso IE 3352 Design of Experiments 2. Four different designs for a digital computer circuit are being studied in order to compare the amount of noise present. The following data have been obtained: Circuit Design Noise Observed 2 80 3 47 4 1 17 20 19 18 18 50 40 97 73 25 83 61 26 56 66 35 75 a. Is the amount of noise present the same for all four designs? Use 0.05. i. State the hypotheses, ii. What is the p-value? ili. What is the decision? iv. Make conclusions. b. Analyze the residuals from this experiment. Are the basic analysis of variance assumptions satisfied? i. State the hypotheses i. What is the p-value? ii. What is the decision? iv. Make conclusions. c. Based on your answer on part (b) conduct another analysis of the observed noise data with a transformation and draw the appropriate conclusions. d. Use Tukey's method to compare all pair of Noise transformation means. What conclusions can you draw e. Use Kruskal Wallis test to analyze the observed noise and draw conclusions. i. State the hypotheses, ii. What is the test statistic? ii. What is the p-value? iv. What is the decision? v. Make conclusions.

Explanation / Answer

Question 2.a.

Part i

For the given test, the null and alternative hypotheses are given as below:

Null hypothesis: H0: The average amounts of noise present in four circuit designs are same.

Alternative hypothesis: Ha: The average amounts of noise present in four circuit designs are not same.

Alternative hypothesis: Ha: At least two average amounts of noise present in four circuit designs are not same.

In symbolic form, it is given as below:

H0: µ1 = µ2 = µ3 = µ4 versus Ha: at least two µ’s are not same.

Part ii

Here, we have to find p-value for the given test. We have to use one way analysis of variance or ANOVA test for checking the above null hypothesis. The ANOVA table for this test by using excel is given as below:

ANOVA: Single Factor

SUMMARY

Groups

Count

Sum

Average

Variance

Design 1

18.4

1.3

Design 2

330

131.5

Design 3

173

34.6

87.3

Design 4

379

75.8

238.7

ANOVA

Source of Variation

P-value

F crit

Between Groups

10753

3584.333

31.24964

6.36E-07

3.238872

Within Groups

1835.2

114.7

Total

12588.2

From above ANOVA table, the p-value for this test is given as 6.36E-07 or 0.000000636.

Part iii

We reject the null hypothesis if the p-value is less than the given level of significance. Here, we are given a level of significance = = 0.05

P-value = 0.000000636

= 0.05

P-value < = 0.05

So, we reject the null hypothesis that the average amounts of noise present in four circuit designs are same.

Part iv

For the above test, we reject the null hypothesis that the average amounts of noise present in four circuit designs are same. This means, there is sufficient evidence to conclude that the average amounts of noise present in four circuit designs are not same.