STA643 Assignment 2 Solution and Discussion

[zareen]

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 t-test on pairs of means , when several means are to be compared. Suggest the alternative statistical procedure and also give its assumptions.

[zareen]

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=(34-31)/(√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 t-test 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 two-sample t-test 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 two-sample t-test. This type of multiple two-sample t-test has two disadvantages. First, the procedure is difficult and time consuming and secondly, the level of significance increases as the number of t-test increases. Thus, a series of two-sample t-test 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]

@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.

Solution Idea

@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 t-test on pairs of means , when several means are to be compared. Suggest the alternative statistical procedure and also give its assumptions.

Ideas Solution

[zareen]

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=(34-31)/(√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 t-test 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 two-sample t-test 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 two-sample t-test. This type of multiple two-sample t-test has two disadvantages. First, the procedure is difficult and time consuming and secondly, the level of significance increases as the number of t-test increases. Thus, a series of two-sample t-test 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.