(A)
(1) Null and Alternative Hypotheses
The following null and alternative hypotheses need to be tested:
Ho: μ1 = μ2 = μ3 = μ4 = μ5 = μ6
Ha: Not all means are equal
The above hypotheses will be tested using an F-ratio for a One-Way ANOVA.
(2) Rejection Region
Based on the information provided, the significance level is \alpha = 0.05α=0.05, and the degrees of freedom are df_1 = 5df1=5 and df_2 = 5df2=5, therefore, the rejection region for this F-test is R = \{F: F > F_c = 2.711\}R={F:F>Fc=2.711}.
(3) Test Statistics
The following table is obtained:
Group 1 | Group 2 | Group 3 | Group 4 | Group 5 | Group 6 | |
14.1 | 12.8 | 13.5 | 13.2 | 16.8 | 18.1 | |
13.6 | 12.5 | 13.4 | 12.7 | 17.2 | 17.2 | |
14.4 | 13.4 | 14.1 | 12.6 | 16.4 | 18.7 | |
14.3 | 13.0 | 14.3 | 14.1 | 17.3 | 18.4 | |
12.3 | 18.0 | |||||
Sum = | 56.4 | 64 | 55.3 | 52.6 | 85.7 | 72.4 |
Average = | 14.1 | 12.8 | 13.825 | 13.15 | 17.14 | 18.1 |
\sum_i X_{ij}^2 =∑iXij2= | 795.62 | 819.94 | 765.11 | 693.1 | 1470.33 | 1311.7 |
St. Dev. = | 0.356 | 0.43 | 0.443 | 0.686 | 0.598 | 0.648 |
SS = | 0.37999999999988 | 0.74000000000001 | 0.58750000000009 | 1.4099999999999 | 1.4319999999998 | 1.2599999999998 |
n = | 4 | 5 | 4 | 4 | 5 | 4 |
The total sample size is N = 26N=26. Therefore, the total degrees of freedom are:
dftotal=26−1=25
Also, the between-groups degrees of freedom are dfbetween=6−1=5, and the within-groups degrees of freedom are:
dfwithin=dftotal−dfbetween=25−5=20
First, we need to compute the total sum of values and the grand mean. The following is obtained
∑Xij=56.4+64+55.3+52.6+85.7+72.4=386.4
Also, the sum of squared values is
∑Xij2=795.62+819.94+765.11+693.1+1470.33+1311.7=5855.8
Based on the above calculations, the total sum of squares is computed as follows
The within sum of squares is computed as shown in the calculation below:
F=74.011
(4) Decision about the null hypothesis
Since it is observed that F=74.011>Fc=2.711, it is then concluded that the null hypothesis is rejected.
Using the P-value approach: The p-value is p=0(option(D) p<0.001) and since p=0<0.05, it is concluded that the null hypothesis is rejected.
(5) Conclusion
It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that not all 6 population means are equal, at the α=0.05 significance level.
option(B) is correct.
(B)
A=Imperial D=Chiffon
B=Parkey E=Mazola
C=Blue Bonnet F=Fleischmaan's
Tukey HSD results
treatments pair |
Tukey HSD Q statistic |
Tukey HSD p-value |
Tukey HSD inferfence |
A vs B | 5.0851 | 0.0192747 | * p<0.05 |
A vs C | 1.0205 | 0.8999947 | insignificant |
A vs D | 3.5253 | 0.1733858 | insignificant |
A vs E | 11.8913 | 0.0010053 | ** p<0.01 |
A vs F | 14.8435 | 0.0010053 | ** p<0.01 |
B vs C | 4.0094 | 0.0922096 | insignificant |
B vs D | 1.3691 | 0.8999947 | insignificant |
B vs E | 18.0061 | 0.0010053 | ** p<0.01 |
B vs F | 20.7315 | 0.0010053 | ** p<0.01 |
C vs D | 2.5048 | 0.5038367 | insignificant |
C vs E | 12.9670 | 0.0010053 | ** p<0.01 |
C vs F | 15.8640 | 0.0010053 | ** p<0.01 |
D vs E | 15.6073 | 0.0010053 | ** p<0.01 |
D vs F | 18.3688 | 0.0010053 | ** p<0.01 |
E vs F | 3.7551 | 0.1292520 | insignificant |
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