Question

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

Answer 1

Let μ,-true mean volume (milliliters) at amount iis 0,1,2,3,4 Null hypothesis, Ho : μ,-/1112|13 vs Alternative hypothesis, H1 . At least one μ¡ is different from others

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 $\mu_i$ 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