STA3701 ASSINGMENT 4
QUESTION 1
1.1
The model of the experiment is:
𝑌𝑖𝑗 = 𝜇 + 𝛼𝑖 + 𝜀𝑖𝑗 for i=1,2,3,4 and j=1,2, ...,24
With
• I=1,2,3,4 indicating the 4 training methods
• J=1,2,…,24 indicating the 24 output of improved score
• 𝑌𝑖𝑗 is the j’th output of score under I’th training
• µ is the overall mean for the out output of scores
• 𝛼𝑖 denoting a random effect of the I’th training
• 𝜀𝑖𝑗 is the I’th experiment error associated with the ith training
Model assumption:
• The training effects 𝛼𝑖 are assumed to be independent and normally distributed with mean 0
and variance 𝜎 2
• The random error 𝜀𝑖𝑗 are assumed to be independent and normally distributed with mean 0
and variance 𝜎 2
• 𝛼𝑖 and 𝜀𝑖𝑗 are independent for all I and j
1.2
Figure 1.1 Boxplot of Student scores versus Training methods
, The boxplot in Figure 1.1 shows that the variances are not equal in the four Training methods.
Further, the boxplot shows evidence of skewness showing a lack of normality which might be due to
the outlier in no training method.
1.3
Analysis of Variance Table
Response: score
Df Sum Sq Mean Sq F value Pr(>F)
training 3 498.38 166.125 14.415 8.952e-08 ***
Residuals 92 1060.25 11.524
---
𝐻𝑂 : 𝜇1 = 𝜇2 = 𝜇3 = 𝜇4
𝐻: 𝜇𝑖 ≠ 𝜇𝑘 for at least one pair i≠k where i=1,2,3,4 representing the four training methods
Critical region
Critical region
We reject 𝐻𝑂 if 𝐹0 > 𝐹0.05,3,92=2.703594
Test statistics
𝐹0 =14.415
Conclusion
Since
𝐹0 = 14.415 > 𝐹0.05,3,92 = 2.703594 we reject null hypothesis with 5% significance value. Hence
there is a difference between training methods.
1.4
QUESTION 1
1.1
The model of the experiment is:
𝑌𝑖𝑗 = 𝜇 + 𝛼𝑖 + 𝜀𝑖𝑗 for i=1,2,3,4 and j=1,2, ...,24
With
• I=1,2,3,4 indicating the 4 training methods
• J=1,2,…,24 indicating the 24 output of improved score
• 𝑌𝑖𝑗 is the j’th output of score under I’th training
• µ is the overall mean for the out output of scores
• 𝛼𝑖 denoting a random effect of the I’th training
• 𝜀𝑖𝑗 is the I’th experiment error associated with the ith training
Model assumption:
• The training effects 𝛼𝑖 are assumed to be independent and normally distributed with mean 0
and variance 𝜎 2
• The random error 𝜀𝑖𝑗 are assumed to be independent and normally distributed with mean 0
and variance 𝜎 2
• 𝛼𝑖 and 𝜀𝑖𝑗 are independent for all I and j
1.2
Figure 1.1 Boxplot of Student scores versus Training methods
, The boxplot in Figure 1.1 shows that the variances are not equal in the four Training methods.
Further, the boxplot shows evidence of skewness showing a lack of normality which might be due to
the outlier in no training method.
1.3
Analysis of Variance Table
Response: score
Df Sum Sq Mean Sq F value Pr(>F)
training 3 498.38 166.125 14.415 8.952e-08 ***
Residuals 92 1060.25 11.524
---
𝐻𝑂 : 𝜇1 = 𝜇2 = 𝜇3 = 𝜇4
𝐻: 𝜇𝑖 ≠ 𝜇𝑘 for at least one pair i≠k where i=1,2,3,4 representing the four training methods
Critical region
Critical region
We reject 𝐻𝑂 if 𝐹0 > 𝐹0.05,3,92=2.703594
Test statistics
𝐹0 =14.415
Conclusion
Since
𝐹0 = 14.415 > 𝐹0.05,3,92 = 2.703594 we reject null hypothesis with 5% significance value. Hence
there is a difference between training methods.
1.4
