Fit one-way ANOVA. Find the null and alternative hypotheses for the F-test and the corresponding p-values. Compare means using Bonferroni and Tukey confidence intervals.
Amount (milligrams) | Volume (milliliters) |
0 | 720 |
0 | 785 |
0 | 810 |
0 | 620 |
0 | 860 |
0 | 810 |
0 | 920 |
0 | 745 |
0 | 885 |
0 | 820 |
1 | 860 |
1 | 1055 |
1 | 960 |
1 | 805 |
1 | 915 |
1 | 845 |
1 | 950 |
1 | 750 |
1 | 1040 |
1 | 920 |
2 | 970 |
2 | 920 |
2 | 1020 |
2 | 830 |
2 | 650 |
2 | 1055 |
2 | 780 |
2 | 850 |
2 | 870 |
2 | 1035 |
3 | 820 |
3 | 735 |
3 | 880 |
3 | 870 |
3 | 945 |
3 | 1015 |
3 | 920 |
3 | 845 |
3 | 975 |
3 | 775 |
4 | 700 |
4 | 855 |
4 | 920 |
4 | 830 |
4 | 755 |
4 | 830 |
4 | 875 |
4 | 870 |
4 | 765 |
4 | 815 |
One-way ANOVA: Volume (milliliters) versus Amount (milligrams)
Source
DF SS
MS F P
Amount (milligrams) 4 96515
24129 2.65 0.045
Error
45 409035 9090
Total
49 505550
From ANOVA table we see that p-value<0.05 so we reject H0 and conclude that at least one is different from others.
Grouping Information Using Tukey Method
Amount
(Grouping) N Mean
1
10 910.00
2
10 898.00
3
10 878.00
4
10 821.50
0
10 797.50
Tukey 95% Simultaneous Confidence Intervals
All Pairwise Comparisons among Levels of Amount (milligrams)
Individual confidence level = 99.33%
Amount (milligrams) = 0 subtracted from:
Amount
(milligrams) Lower Center Upper
1
-8.70 112.50
233.70
2
-20.70 100.50
221.70
3
-40.70 80.50
201.70
4
-97.20 24.00
145.20
From above confidence intervals we see that all intervals contain zero hence mean volume at Amount 0 is insignificantly different from other mean volumns at amounts 1-4.
Amount (milligrams) = 1 subtracted from:
Amount
(milligrams) Lower Center Upper
2
-133.20 -12.00
109.20
3
-153.20 -32.00 89.20
4
-209.70 -88.50 32.70
From above confidence intervals we see that all intervals contain zero hence mean volume at Amount 1 is insignificantly different from other mean volumns at amounts 2-4.
Amount (milligrams) = 2 subtracted from:
Amount
(milligrams) Lower Center Upper
3
-141.20 -20.00 101.20
4
-197.70 -76.50 44.70
From above confidence intervals we see that all intervals contain zero hence mean volume at Amount 2 is insignificantly different from other mean volumns at amounts 3-4.
Amount (milligrams) = 3 subtracted from:
Amount
(milligrams) Lower Center Upper
4
-177.70 -56.50 64.70
From above confidence interval we see that the above interval contains zero hence mean volume at Amount 3 is insignificantly different from mean volumn at amount 4.
Hence using Tukey's method we do not get any significant
difference between means. Hence after Tukey's test we accept null
hypothesis.
Grouping Information Using Fisher Method (using Bonferroni's
correction)
Amount
(milligrams) N Mean
Grouping
1
10 910.00
A
2
10 898.00 A B
3
10 878.00 A B C
4
10 821.50 B C
0
10 797.50 C
Means that do not share a letter are significantly different.
Fisher 95% Individual Confidence Intervals
All Pairwise Comparisons among Levels of Amount (milligrams)
Simultaneous confidence level = 72.40%
Amount (milligrams) = 0 subtracted from:
Amount
(milligrams) Lower Center Upper
1
26.62 112.50
198.38
2
14.62 100.50
186.38
3
-5.38 80.50
166.38
4
-61.88 24.00
109.88
Here we see that Amount 0 and Amount 1 are significantly different
since the corresponding interval does not contain zero.
Similarly Amount 0 and Amount 2 are significantly different since the corresponding interval does not contain zero.
Amount (milligrams) = 1 subtracted from:
Amount
(milligrams) Lower Center Upper
2
-97.88 -12.00
73.88
3
-117.88 -32.00 53.88
4
-174.38 -88.50 -2.62
From above confidence intervals we see that all intervals contain
zero hence mean volume at Amount 1 is insignificantly different
from other mean volumns at amounts
2-4.
Amount (milligrams) = 2 subtracted from:
Amount
(milligrams) Lower Center Upper
3
-105.88 -20.00 65.88
4
-162.38 -76.50 9.38
From above confidence interval we see that the above interval
contains zero hence mean volume at Amount 3 is insignificantly
different from mean volumn at amount 4.
Hence here we get different conclusion. -
Amount (milligrams) = 3 subtracted from:
Amount
(milligrams) Lower Center Upper
-------+---------+---------+---------+--
4
-142.38 -56.50 29.38
(-------*--------)
-------+---------+---------+---------+--
-100
0
100 200
Fit one-way ANOVA. Find the null and alternative hypotheses for the F-test and the corresponding p-values. Compare means using Bonferroni and Tukey confidence intervals. Amount (milligrams) Vol...
Using the chart of accounts in Figure 2-1, determine the changes to the balance sheet, income statement, job cost ledger, and equipment ledger as the result of purchasing a new loader (Loader 3) to replace an existing loader (Loader 2). The new loader costs $115,200. The new loader will be paid for by trading in the existing loader for a credit of $15,200 and the remaining $100,000 will be financed through the dealership. The existing loader was purchased for $95,000...
Create a program that will use the attached input file and perform the following operations. Read the file into an appropriate JCF data structure. Look up a (list of) names and numbers matching a last name or the first letters of a last name, ignoring case. Look up a (list of) names and numbers matching a number or the first digits of a number. Add a name and number to the list. Sort the list by first name, last name...