Q2 ANSWER ALL PARTS PLEASE The response time in milliseconds was determined for
ID: 3181963 • Letter: Q
Question
Q2 ANSWER ALL PARTS PLEASE
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. a. Test the hypothesis that the three circuit types have the same response time. Use alpha = 0.01. b. Use Fisher LSD to compare all pairs of treatment means. Use alpha = 0.01. c. Use Tukey's method to compare all pairs of treatment means. Use alpha = 0.01. d. If you were a design engineer and you wished to minimize the response time, which circuit type would you select? Use Dunnett's test. e. If you are minimizing the response time, compare the results from parts b, c, and d.Explanation / Answer
Question 2
Part a
Here, we have to test the hypothesis that the three circuit types have the same average response time or not. For checking this hypothesis we have to use the one way analysis of variance or ANOVA test. The null and alternative hypothesis for this test is given as below:
Null hypothesis: H0: The three circuit types have the same average response time.
Alternative hypothesis: Ha: The three circuit types have not the same average response time.
The level of significance or alpha value is given as 0.01.
The ANOVA table for calculation of F test statistic is given as below:
ANOVA
Response Time
Sum of Squares
df
Mean Square
F
Sig.
Between Groups
784.133
2
392.067
84.619
.000
Within Groups
55.600
12
4.633
Total
839.733
14
From this ANOVA table we get the test statistic value F = 84.619 with the p-value of 0.00 which is less than the given alpha value 0.01. So, we reject the null hypothesis that the three circuit types have the same average response time. This means we conclude that there is sufficient evidence that the three circuit types do not have the same average response time.
Part b
Now, we have to use the Fisher LSD test for the multiple comparisons of the treatment means. The level of significance or alpha value is given as 0.01. The Fisher LSD test for compare all pairs of treatment means is given as below:
Multiple Comparisons
Dependent Variable: Response Time
(I) Circuit Type
(J) Circuit Type
Mean Difference (I-J)
Std. Error
p-value
LSD
1.00
2.00
-12.80000
1.36137
.000
3.00
4.20000
1.36137
.009
2.00
1.00
12.80000
1.36137
.000
3.00
17.00000
1.36137
.000
3.00
1.00
-4.20000
1.36137
.009
2.00
-17.00000
1.36137
.000
From the above test, it is observed that all the p-values are less than the given level of significance or alpha value 0.01, so we conclude that all differences are statistically significant. There is no any pair having same average value.
Part c
Here, we have to use the Tukey’s method for the multiple comparisons of the all treatment means. For this test, the level of significance or alpha value is given as 0.01. The Tukey’s test for comparing the pairs of treatment means is summarised as below:
Multiple Comparisons
Dependent Variable: Response Time
(I) Circuit Type
(J) Circuit Type
Mean Difference (I-J)
Std. Error
Sig.
Tukey HSD
1.00
2.00
-12.80000*
1.36137
.000
3.00
4.20000*
1.36137
.024
2.00
1.00
12.80000*
1.36137
.000
3.00
17.00000*
1.36137
.000
3.00
1.00
-4.20000*
1.36137
.024
2.00
-17.00000*
1.36137
.000
From the above test, it is observed that the pair 1 and 3 does not have significant difference in the mean as the p-value for this paired comparison is given as 0.024 which is greater than the given level of significance as 0.01. For all other pairs, the p-values are less than the given level of significance or alpha value 0.01, so we reject the null hypothesis that there is no any statistically significant difference in the average values for the given pairs. This means there is a statistically significant difference exists in these pairs of means.
Part d
For the given scenario for this particular case, we will select the circuit type as 1 and 3 because it would be helpful in minimizing the response time due to approximate equal mean response times.
ANOVA
Response Time
Sum of Squares
df
Mean Square
F
Sig.
Between Groups
784.133
2
392.067
84.619
.000
Within Groups
55.600
12
4.633
Total
839.733
14
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