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THE MODULE IS STA 3701 - APPLIED STATISTICS ,USING R PROGRAMMING TO ANALYSE DATA GIVEN,ANSWERIG ASSIGNMENT 4 2022 WITH DETAILED EXPLANATION, ALSO INCLUDING THE CODES WRITTEN IN R
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THE DOCUMENTS CONTAINS DETAILED EXPLANATION OF STA3701 ASSIGNMENT 4 2022, WITH CODES FROM R PROGRAMMING

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  • January 13, 2023
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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

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