In the Protocol ‘modern biotechnology’ is referred to as the application of fusion of cells beyond the taxonomic family (called ‘in vitro nucleic acid’ techniques) that overcome natural physiological reproductive or recombination barriers and are not techniques used in traditional breeding and selection. The term Living modified organism (LMO) is defined as the processing of any material that is of LMO origin, containing detectable novel combinations of replicable genetic material obtained through the use of modern biotechnology (for instance, flour from genetically modified maize). The LMOs intended for direct use as food or feed, or for processing are agricultural commodities from genetically modified crops (http://www.gmofreeregions.org/glossary.html Accessed 26.11.16.). Although the Protocol did not use the term genetically modified organism, the LMO is almost equivalent to the GMO, which is any organism whose genetic material has been altered using genetic engineering techniques (i.e., a genetically engineered organism). A more popular name of the GMO is ‘transgenic organism’
SOLVED STA643 Assignment 1 Solution and Discussion

Course Title Experimental Designs
Course Code STA643
Activity Assignment # 1
Submission Deadline Date: Monday, Nov 22, 2019
Time: 23:59 Hours (Midnight)Instructions to attempt assignment:
You can use/make a single table for all necessary computations.
Use calculator for calculations.
Write the complete formulas and all steps, wherever necessary.
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Make your own attempt, any copy/past; cheating from other student/website may result in ZERO marks and/or assignment cancelation.Assignment 1
Question no. 1:
Test the hypothesis that the mean of a normal population with known variance 70 is 31, if a sample of size 13 gave x ̅ = 34. Let the alternative hypothesis be H1: µ > 31, and let α = 0.10.Questions no. 2:
Explain in detail why it is not a good statistical procedure to perform several ttest on pairs of means , when several means are to be compared. Suggest the alternative statistical procedure and also give its assumptions. 
Solution:
Question no. 1:
Test the hypothesis that the mean of a normal population with known variance 70 is 31, if a sample of size 13 gave x ̅ = 34. Let the alternative hypothesis be H1: µ > 31, and let α = 0.10.
Solution:
Formulation of Hypothesis:
Ho: µ = 31
H1: µ > 31
Level of Significance:
α = 0.10
Test Statistics:
z=(x ̅μ)/(σ⁄√n)
Calculation:
z=(3431)/(√70⁄√13)z = 1.29.
Critical Region:
Reject Ho, if
Z > Zα
1.29 > Z0.10
1.29 > 1.28
We reject Ho.
Conclusion:
Since 1.29 > 1.28 fall in the critical region, so we reject Ho.
Questions no. 2:
Explain in detail why it is not a good statistical procedure to perform several ttest on pairs of means , when several means are to be compared. Suggest the alternative statistical procedure and also give its assumptions.
Solution:
Whenever we compare more than two population means, we apply the twosample ttest to all possible pairwise comparisons of means. For example, if we wish to compare 4 population means, there will be 6 pairs and to test the hypothesis that all four population means are equal, would require six twosample ttest. This type of multiple twosample ttest has two disadvantages. First, the procedure is difficult and time consuming and secondly, the level of significance increases as the number of ttest increases. Thus, a series of twosample ttest is not a good procedure. ANOVA is a technique that measure the variations between the means.
Assumptions of ANOVA:
1: Experimental errors are normally distributed.
2: Equal variance between the treatments.
3: Samples are independent. 
Solution:
Question no. 1:
Test the hypothesis that the mean of a normal population with known variance 70 is 31, if a sample of size 13 gave x ̅ = 34. Let the alternative hypothesis be H1: µ > 31, and let α = 0.10.
Solution:
Formulation of Hypothesis:
Ho: µ = 31
H1: µ > 31
Level of Significance:
α = 0.10
Test Statistics:
z=(x ̅μ)/(σ⁄√n)
Calculation:
z=(3431)/(√70⁄√13)z = 1.29.
Critical Region:
Reject Ho, if
Z > Zα
1.29 > Z0.10
1.29 > 1.28
We reject Ho.
Conclusion:
Since 1.29 > 1.28 fall in the critical region, so we reject Ho.
Questions no. 2:
Explain in detail why it is not a good statistical procedure to perform several ttest on pairs of means , when several means are to be compared. Suggest the alternative statistical procedure and also give its assumptions.
Solution:
Whenever we compare more than two population means, we apply the twosample ttest to all possible pairwise comparisons of means. For example, if we wish to compare 4 population means, there will be 6 pairs and to test the hypothesis that all four population means are equal, would require six twosample ttest. This type of multiple twosample ttest has two disadvantages. First, the procedure is difficult and time consuming and secondly, the level of significance increases as the number of ttest increases. Thus, a series of twosample ttest is not a good procedure. ANOVA is a technique that measure the variations between the means.
Assumptions of ANOVA:
1: Experimental errors are normally distributed.
2: Equal variance between the treatments.
3: Samples are independent. 
@zareen said in STA643 Assignment 1 Solution and Discussion:
Test the hypothesis that the mean of a normal population with known variance 70 is 31, if a sample of size 13 gave x ̅ = 34. Let the alternative hypothesis be H1: µ > 31, and let α = 0.10.
@zareen said in STA643 Assignment 1 Solution and Discussion:
Questions no. 2:
Explain in detail why it is not a good statistical procedure to perform several ttest on pairs of means , when several means are to be compared. Suggest the alternative statistical procedure and also give its assumptions.