Turn-In Hard Copy Problem (42 total points). The Environmental Protection Agency
ID: 3053306 • Letter: T
Question
Turn-In Hard Copy Problem (42 total points). The Environmental Protection Agency (EPA) advocates a maximum arsenic level in water of 10 micrograms per liter. The state agency dealing with water quality in Nevada is currently testing wells in a rural county known to have naturally occurring arsenic in the water table. The agency wants to determine if the age of the well affects the levels of arsenic in the water. The file for the data collected from randomly chosen wells in the rural county can be accessed in the file Arsenic Levels_1 (click on the paper clip icon in your Connect assignment). You are asked to perform the appropriate analysis to determine if the mean arsenic level differs depending on the age of the well. Part a) Perform the appropriate test using an alpha of .05. Be sure to include all required steps of the hypothesis testing process........... Part b) Perform the appropriate post hoc analysis, if required, to discuss differences or non-differences using the confidence interval approach outlined in class. Note: You must provide the statistical analysis using Minitab. Make sure you print out a hard copy of Minitab and attach the results to your written answers. Failure to provide a hard copy printout will result in loss of points. Do not email your Minitab output - attach to your submission.
Age of Well: Under 10 10 to 19 20 and Over 5.4 6.1 6.8 4.3 4.1 5.4 6.1 5.8 5.7 3.4 5.1 4.5 3.7 3.7 5.5 4.3 4.4 4.6 2.4 3.8 3.9 2.9 2.7 2.9 2.7 3.4 4 1.5 6 6.5 5 5.2 5.5 2.5 4.1 5.4 2 3.8 4 3.1 4.2 6 3 4.2 4.2Explanation / Answer
Under 10
10 to 19
20 and Over
5.4
6.1
6.8
4.3
4.1
5.4
6.1
5.8
5.7
3.4
5.1
4.5
3.7
3.7
5.5
4.3
4.4
4.6
2.4
3.8
3.9
2.9
2.7
2.9
2.7
3.4
4
1.5
6
6.5
5
5.2
5.5
2.5
4.1
5.4
2
3.8
4
3.1
4.2
6
3
4.2
4.2
X1
X2
X3
3.486666667
4.44
4.993333333
X
4.306666667
SSA=15(3.49-4.31)^2+15(4.44-4.31)^2+15(4.99-4.31)^2=17.43
SSW=(5.4-3.49)^2+……..+(4.2-4.99)^2=54.16
MSA=17.43/(3-1)=8.71
MSW=54.16/(45-3)=1.29
F=MSA/MSW=8.71/1.29=6.76
Fcrit=3.22
Since F>Fcrit, reject the null hypothesis
Anova: Single Factor
SUMMARY
Groups
Count
Sum
Average
Variance
Under 10
15
52.3
3.486666667
1.701238095
10 to 19
15
66.6
4.44
0.991142857
20 and Over
15
74.9
4.993333333
1.176380952
ANOVA
Source of Variation
SS
df
MS
F
P-value
F crit
Between Groups
17.42533333
2
8.712666667
6.756166609
0.002857959
3.219942293
Within Groups
54.16266667
42
1.289587302
Total
71.588
44
Under 10
10 to 19
20 and Over
5.4
6.1
6.8
4.3
4.1
5.4
6.1
5.8
5.7
3.4
5.1
4.5
3.7
3.7
5.5
4.3
4.4
4.6
2.4
3.8
3.9
2.9
2.7
2.9
2.7
3.4
4
1.5
6
6.5
5
5.2
5.5
2.5
4.1
5.4
2
3.8
4
3.1
4.2
6
3
4.2
4.2
X1
X2
X3
3.486666667
4.44
4.993333333
X
4.306666667
SSA=15(3.49-4.31)^2+15(4.44-4.31)^2+15(4.99-4.31)^2=17.43
SSW=(5.4-3.49)^2+……..+(4.2-4.99)^2=54.16
MSA=17.43/(3-1)=8.71
MSW=54.16/(45-3)=1.29
F=MSA/MSW=8.71/1.29=6.76
Fcrit=3.22
Since F>Fcrit, reject the null hypothesis
Anova: Single Factor
SUMMARY
Groups
Count
Sum
Average
Variance
Under 10
15
52.3
3.486666667
1.701238095
10 to 19
15
66.6
4.44
0.991142857
20 and Over
15
74.9
4.993333333
1.176380952
ANOVA
Source of Variation
SS
df
MS
F
P-value
F crit
Between Groups
17.42533333
2
8.712666667
6.756166609
0.002857959
3.219942293
Within Groups
54.16266667
42
1.289587302
Total
71.588
44
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