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)

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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.

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.

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 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