Use the appropriate calculations in SF1 – SF6 of “Videos-Topics In Stat 230” to
ID: 3128503 • Letter: U
Question
Use the appropriate calculations in SF1 – SF6 of “Videos-Topics In Stat 230” to compute F(test) = MST/MSE. Then enter the data for each treatment into EXCEL and compute the ANOVA table and compare your results. State Ho, Ha and your conclusions. The data in the table resulted from an experiement that used a completely randomized design. treatment 1(3.8, 1.2, 4.1, 5.5 2.3) treatment 2 (5.4,2.0,4.8,3.8) treatment 3 (1.3,0.7,2.2)
(a) Use statistical software (or the appropriate calculation formulas in Appendix C) to complete the following Anova Table.
Source df ss ms f
Treatment error
Total
(b) Test the null hypothesis 1 =2 =3 where 1 represents the true mean for treatment i is against the alternative that at least two of the mean differ.Use =.01.
Explanation / Answer
Use the appropriate calculations in SF1 – SF6 of “Videos-Topics In Stat 230” to compute F(test) = MST/MSE. Then enter the data for each treatment into EXCEL and compute the ANOVA table and compare your results. State Ho, Ha and your conclusions. The data in the table resulted from an experiement that used a completely randomized design. treatment 1(3.8, 1.2, 4.1, 5.5 2.3) treatment 2 (5.4,2.0,4.8,3.8) treatment 3 (1.3,0.7,2.2)
(a) Use statistical software (or the appropriate calculation formulas in Appendix C) to complete the following Anova Table.
Source df ss ms f
Treatment error
Total
ANOVA: Single Factor
SUMMARY
Groups
Count
Sum
Average
Variance
T1
5
16.9
3.38
2.7770
T2
4
16
4
2.2133
T3
3
4.2
1.4
0.5700
ANOVA
Source of Variation
SS
df
MS
F
P-value
F crit
Between treatments
12.3012
2
6.1506
2.9307
0.1047
8.02
Error
18.8880
9
2.0987
Total
31.1892
11
Level of significance
0.01
(b) Test the null hypothesis 1 =2 =3 where 1 represents the true mean for treatment i is against
the alternative that at least two of the mean differ. Use =.01.a
Table value of F(2,9) at 0.01 level = 8.02
Calculated F=2.93 < 8.02, the table value.
The null hypothesis is not rejected.
There is no evidence to conclude that at least two of the mean differ.
ANOVA: Single Factor
SUMMARY
Groups
Count
Sum
Average
Variance
T1
5
16.9
3.38
2.7770
T2
4
16
4
2.2133
T3
3
4.2
1.4
0.5700
ANOVA
Source of Variation
SS
df
MS
F
P-value
F crit
Between treatments
12.3012
2
6.1506
2.9307
0.1047
8.02
Error
18.8880
9
2.0987
Total
31.1892
11
Level of significance
0.01
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.