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    zareenZ
    Q. 1 Varieties A B C 23 18 16 26 28 25 20 17 12 17 21 14 Test the hypothesis that the three verities of potatoes are not different in the yielding capabilities. Solution Q.1: [image: dzN6VLQ.png] [image: SO55ms8.png] [image: XZEmXSQ.png] [image: nfGQY25.png] Q.2 Solution: Complete the ANOVA table. Source of variation Sum of squares Degrees of freedom Mean square F B/W treatments 79.44 4 19.860 6.90 Error 57.600 20 2.88 Total 137.040 24
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    zareenZ
    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.
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