Test the hypothesis that the three verities of potatoes are not different in the yielding capabilities.

**Solution Q.1:**

**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 ]]>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.

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.