can you use Minitab and please especially d Design Noise 1 19 1 22 1 15 1 23 1 2
ID: 3067958 • Letter: C
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can you use Minitab and please especially d
Design Noise 1 19 1 22 1 15 1 23 1 26 1 31 1 25 1 28 2 54 2 65 2 68 2 59 2 61 2 67 2 70 2 64 3 35 3 39 3 35 3 31 3 39 3 42 3 40 3 37 4 68 4 73 4 70 4 68 4 69 4 64 4 71 4 72 The data set noise.xlsx (also in .txt format: noise.txt) is the data about the noise level of computers from four different designs. Analyze the data to answer the following questions: (a) Report the average and standard deviation of noise level for each design. (b) Report the ANOVA table. (c) Test if the amount of noise for the four different designs are significant different using the level of significance 5%. [Hint: use the p-value of the F-test provided in the output.] (d) Use Tukey's Multiple Comparisons method to perform pairwise comparisons. Use the multiple comparison results to select your 'best' design? [note: the lower the noise the better].Explanation / Answer
a) The Table-1 gives the average (mean) and Standard deviation of noise level for each design. The average (mean) and Standard deviation of noise level for design-1 is 23.63 and 5.069 respectively. The average (mean) and Standard deviation of noise level for design-2, design-3 and design-4 are given below.
Table 1: Descriptive
Noise
Design
N
Mean
Std. Deviation
Std. Error
95% Confidence Interval for Mean
Minimum
Maximum
Lower Bound
Upper Bound
1
8
23.63
5.069
1.792
19.39
27.86
15
31
2
8
63.50
5.264
1.861
59.10
67.90
54
70
3
8
37.25
3.495
1.236
34.33
40.17
31
42
4
8
69.38
2.825
0.999
67.01
71.74
64
73
Total
32
48.44
19.480
3.444
41.41
55.46
15
73
b) The Table -2 gives the ANOVA table,
Table -2: ANOVA
Noise
Sum of Squares
df
Mean Square
F
Sig.
Between Groups
11248.625
3
3749.542
203.760
0.000
Within Groups
515.250
28
18.402
Total
11763.875
31
c) Hypothesis,
H0 : The mean amount of noise for four different design are same.
Vs
H1 : At least one pair of mean amount of noise is differs significantly.
The p-value of F-test provided by table-2(ANOVA) is 0.000. This p-value is less than 0.05. Thus, The null hypothesis H0 is rejected. Hence we can conclude that, at least one pair of mean amount of noise is differs significantly. To find differ pairs we use Tukey’s multiple comparison method.
d) The Table-3 (Multiple Comparisons), shows which groups differed from each other. To make pair wise comparison we use Tukey’s multiple comparison method. We can see that from Table-3, there is significant difference in mean amount of noise between the design-1 & deign-2 (P-value = 0.000), design-1 & deign-3 (P-value = 0.000), design-1 & deign-4 (P-value = 0.000), design-2 & deign-3 (P-value = 0.000), design-2 & deign-4 (P-value = 0.049) and design-3 & deign-4 (P-value = 0.000).
Table -3: Multiple Comparisons
Dependent Variable: Noise
Tukey HSD
(I)
Design
(J)
Design
Mean Difference
(I-J)
Std. Error
Sig.
95% Confidence Interval
Lower Bound
Upper Bound
1
2
-39.875*
2.145
0.000
-45.73
-34.02
3
-13.625*
2.145
0.000
-19.48
-7.77
4
-45.750*
2.145
0.000
-51.61
-39.89
2
1
39.875*
2.145
0.000
34.02
45.73
3
26.250*
2.145
0.000
20.39
32.11
4
-5.875*
2.145
0.049
-11.73
-0.02
3
1
13.625*
2.145
0.000
7.77
19.48
2
-26.250*
2.145
0.000
-32.11
-20.39
4
-32.125*
2.145
0.000
-37.98
-26.27
4
1
45.750*
2.145
0.000
39.89
51.61
2
5.875*
2.145
0.049
0.02
11.73
3
32.125*
2.145
0.000
26.27
37.98
*. The mean difference is significant at the 0.05 level.
The design -1 has mean amount of nose is 23.63(See table-1). This is lower as compared to other designs. Hence design -1 is best design.
Note: SPSS software is used for analysis.
Noise
Design
N
Mean
Std. Deviation
Std. Error
95% Confidence Interval for Mean
Minimum
Maximum
Lower Bound
Upper Bound
1
8
23.63
5.069
1.792
19.39
27.86
15
31
2
8
63.50
5.264
1.861
59.10
67.90
54
70
3
8
37.25
3.495
1.236
34.33
40.17
31
42
4
8
69.38
2.825
0.999
67.01
71.74
64
73
Total
32
48.44
19.480
3.444
41.41
55.46
15
73
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