STA100 GDB 3 Solution
2020-Spring Semester-VU-General Mathematics and Biostatistics
 |
| The solution of STAT100 GDB No 3 Spring semester 2020-Virtual University |
Opening date: 21st August 2020
Closing Date: 26th August 2020
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| STA100 GDB 3 Solution 2020-Spring Semester-VU-General Mathematics and Biostatistics |
Solution
F(2,9)=4.26 positively
skewed that lies on the right tail of the curve
Calculating
d.f for Errors
Degree of freedom for Error=Total degree of
freedom-d.f for Treatments
d.f for Error = 11-2=9
Calculating
Mean Squares
Mean Square of Treatments = SSTr/df of Treatments
Mean Square of Treatments = 54.5/2=27.25
Mean Square of Error = SSE/df of Error
Mean Square of Error = 217.75/9=24.19
Computing F
F(2,9) = Mean Square of Treatment/Mean Square of
Error
F(2,9) = 27.25/24.19
F(2,9) = 1.1265
Final table after placing all calculated missing values
is
|
Source of Variation |
Degree of freedom |
Sum of Squares |
Mean Squares |
Computed F |
|
Treatments |
2 |
54.5 |
27.25 |
1.1265 |
|
Errors |
9 |
217.75 |
24.19 |
|
|
Total |
11 |
272.25 |
|
|
Conclusion
As the value of Computed F(2,9) 1.1265 is less than 4.26, it
means we fail to reject the null hypothesis.