Reference no: EM13866309
1. A manager is interested in testing whether three populations of interest have equal population means. Simple random samples of size 10 were selected from each population. The following ANOVA table and related statistics were computed:
ANOVA: Single Factor
|
Summary
|
Groups
|
Count
|
Sum
|
Average
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Variance
|
Sample 1
|
10
|
507.18
|
50.72
|
35.06
|
Sample 2 |
10 |
405.79 |
40.58 |
30.08 |
Sample 3 |
10 |
487.64 |
48.76 |
23.13 |
ANOVA
|
Source
|
SS
|
df
|
MS
|
F
|
p-value
|
F-crit
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Between Groups
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578.78
|
2
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289.39
|
9.84
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0.0006
|
3.354
|
Within Groups
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794.36
|
27
|
29.42
|
|
|
|
Total
|
1,373.14
|
29
|
|
|
|
|
a. State the appropriate null and alternative hypotheses.
b. Based on your answer to part a, what conclusions can be reached about the null and alternative hypotheses. Use a 0.05 level of significance.
c. If warranted, use the Tukey-Kramer procedure for multiple comparisons to determine which populations have different means. (Assume a = 0.05.)
2. Respond to each of the following questions using this partially completed one-way ANOVA table:
Source of Variation
|
SS
|
df
|
MS
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F-ratio
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Between Samples
|
|
3
|
|
|
Within Samples
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405
|
|
|
|
Total
|
888
|
51
|
|
|
a. How many different populations are being considered in this analysis?
b. Fill in the ANOVA table with the missing values.
c. State the appropriate null and alternative hypotheses.
d. Based on the analysis of variance F-test, what conclusion should be reached regarding the null hypothesis? Test using a = 0.05.
3. Given the following sample data
Item
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Group 1
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Group 2
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Group 3
|
Group 4
|
1
|
20.9
|
28.2
|
17.8
|
21.2
|
2
|
27.2
|
26.2
|
15.9
|
23.9
|
3
|
26.6
|
21.6
|
18.4
|
19.5
|
4
|
22.1
|
29.7
|
20.2
|
17.4
|
5
|
25.3
|
30.3
|
14.1
|
|
6
|
30.1
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25.9
|
|
|
7
|
23.8
|
|
|
|
a. Based on the computations for the within- and between-sample variation, develop the ANOVA table and test the appropriate null hypothesis using a = 0.05. Use the p-value approach.
b. If warranted, use the Tukey-Kramer procedure to determine which populations have different means. Use a = 0.05.
4. Descent, Inc., produces a variety of climbing and mountaineering equipment. One of its products is a traditional three-strand climbing rope. An important characteristic of any climbing rope is its tensile strength. Descent produces the three-strand rope on two separate production lines: one in Bozeman and the other in Challis. The Bozeman line has recently installed new production equipment. Descent regularly tests the tensile strength of its ropes by randomly selecting ropes from production and subjecting them to various tests. The most recent random sample of ropes, taken after the new equipment was installed at the Bozeman plant, revealed the following:
Bozeman
|
Challis
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x1‾ = 7,200 lb
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x2‾ =7,087 lb
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s1 =425
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s2 =415
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n1 =25
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n2 =20
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Descent's production managers are willing to assume that the population of tensile strengths for each plant is approximately normally distributed with equal variances. Based on the sample results, can Descent's managers conclude that there is a difference between the mean tensile strengths of ropes produced in Bozeman and Challis? Conduct the appropriate hypothesis test at the 0.05 level of significance.
5. One of the advances that helped to diminish carpal tunnel syndrome is ergonomic keyboards. The ergonomic keyboards may also increase typing speed. Ten administrative assistants were chosen to type on both standard and ergonomic keyboards. The resulting word-per-minute typing speeds follow:
Ergonomic:
|
69
|
80
|
60
|
71
|
73
|
64
|
63
|
70
|
63
|
74
|
Standard:
|
70
|
68
|
54
|
56
|
58
|
64
|
62
|
51
|
64
|
53
|
a. Were the two samples obtained independently? . Support your assertion.
b. Conduct a hypothesis test to determine if the ergonomic keyboards increase the average words per minute attained while typing. Use a p-value approach with a significance level of 0.01.
6. An article in The American Statistician (M. L. R.Ernst, et al., "Scatterplots for Unordered Pairs," 50 (1996), pp. 260-265) reports on the difference in the measurements by two evaluators of the cardiac output of 23 patients using Doppler echocardiography. Both observers took measurements from the same patients. The measured outcomes were as follows:
Patient
|
1
|
2
|
3
|
4
|
5
|
6
|
7
|
8
|
9
|
10
|
11
|
12
|
Evaluator 1
|
4.8
|
5.6
|
6.0
|
6.4
|
6.5
|
6.6
|
6.8
|
7.0
|
7.0
|
7.2
|
7.4
|
7.6
|
Evaluator 2
|
5.8
|
6.1
|
7.7
|
7.8
|
7.6
|
8.1
|
8.0
|
8.21
|
6.6
|
8.1
|
9.5
|
9.6
|
Patient
|
13
|
14
|
15
|
16
|
17
|
18
|
19
|
20
|
21
|
22
|
23
|
|
Evaluator 1
|
7.7
|
7.7
|
8.2
|
8.2
|
8.3
|
8.5
|
9.3
|
10.2
|
10.4
|
10.6
|
11.4
|
|
Evaluator 2
|
8.5
|
9.5
|
9.1
|
10.0
|
9.1
|
10.8
|
11.5
|
11.5
|
11.2
|
11.5
|
12.0
|
|
a. Conduct a hypothesis test to determine if the average cardiac outputs measured by the two evaluators differ. Use a significance level of 0.02.
b. Calculate the standard error of the difference between the two average outputs assuming that the sampling was done independently. Compare this with the standard error obtained in part a.
A16.6ml sample of a0.500m aqueousnitrous acidsolution
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Properties does li have a larger value than potassium
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The volume of the football is held constant
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The company expects that the machine could be used
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