Recent Topics

What is Unitary State?
General Discussion4 
What precaution you can take to protect yourself from the Coronavirus?
General Discussion3 
Lucknow Pact
HIS  History3 
Do You Know About Coronavirus? Myths And Facts! Get Certified
General Discussion2 
The lucknow Pact 1916
HIS  History1 
What are the merits of Unitary State?
General Discussion1 
Ecommerce SEO 101: Why Is Your Store Not On Google?
SEO (Search Engine Optimization)1 
What are the reasons behind Lucknow Pact?
HIS  History1 
Should people avoid all goods and products from China?
General Discussion1 
What are the best ways to guard yourself against the coronavirus?
General Discussion1 
Examples of Unitary State
POL  301 Principles of Political Science1 
FRL101 Assignment 2 Solution and Discussion Batch 6
Freelancing1 
What precaution you can take to protect yourself from the Coronavirus?
General Discussion0
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.
For writing symbols, use MathType software. If you already did not download it, get it from here.
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.