Answer:
Part 1
a. Sum of Squares, Treatment= 61
b. Sum of Squares, Error= 7.5
c. Mean Squares, Treatment = 30.5
d. Mean Squares, Error= 0.5
2. the F value lies in the rejection region > 3.6823
3. The value of the test statistic = 61
4. The p-value is < 0.00001
5. Conclusion
Since p-value < α, H0 is rejected.
6. Between x`2 and x`3
7. Fisher's Least Significant Difference value almost 0.869
8.There is a significant difference between the means
Step-by-step explanation:
Summary of Data
<u> Treatments</u>
1 2 3 Total
n 6 6 6 18
∑x 42 57 30 129
Mean 7 9.5 5 7.167
<u>∑x2 298 543 152 993</u>
<u>Sd.D 0.8944 0.5477 0.6325 2.0073</u>
ANOVA Table
<u>Source SS df MS </u>
Between-treatments 61 2 30.5 F = 61
<u>Error 7.5 15 0.5 </u>
<u>Total 6 8.5 17 </u>
a. Sum of Squares, Treatment= 61
b. Sum of Squares, Error= 7.5
c. Mean Squares, Treatment = 30.5
d. Mean Squares, Error= 0.5
2. Using alpha= 0.05 the F value lies in the rejection region i.e F > 3.6823
x1` -x2`= 7-9.5= -2.5 Not significant as difference <3.68
x1`- x3`= 7-5= 2 Not significant as difference <3.68
x2` -x3`= 9.5-5= 4.5 Significant as difference > 3.68
3. The value of the F test statistic = 61
4. The p-value is < 0.00001
5. Conclusion
<em>Since p-value < α, H0 is rejected.</em>
6. Using alpha= .05, differences occurs between x2` and x3` as their difference is greater than 3.68
7. Fisher's Least Significant Difference value almost 0.869
Least Significant Difference= t( 0.025,15) √2s²/r s²= 0.50 r= 6 =n1=n2=n3
Least Significant Difference= 2.13 √ 2*0.50/ 6
=0.869
8.There is a significant difference between the means
x1` -x2`= 7-9.5= -2.5 Significant as difference > Least Significant Difference
x1`- x3`= 7-5= 2 Significant as difference > Least Significant Difference
x2` -x3`= 9.5-5= 4.5 Significant as difference > Least Significant Difference
