Answer to Question #307535 in Management for MANISH

Question

Four observers determine the moisture content of samples of a powder, each man taking a sample from each of six consignments. Their assessments are given below: Observers Consignments 1 2 3 4 5 6 1 2 3 4 9 12 11 12 10 11 10 11 9 9 10 11 10 11 12 14 11 10 11 12 11 10 10 10 Discuss whether there is any significant difference between consignments. (Useful data: Ftab(5,15):2.96, Ftab(3,15):3.29)

Answer 1

From the data we understand that there are two factors. One is consignment and other is observer. So, we apply ANOV A two way classification.

Step 1: Null Hypothesis Ho (I): There is no difference in the yield between treatments (rows). Null Hypothesis Ho (II): There is no difference in the yield between blocks(columns).

Step 2: Test Statistic (Calculating F ratio) As the data given is numerically larger, let us subtract all values given by 10 make them simpler for calculation ease.

Total = 4 + 4 + -1 + 7 + 4 +1 = 19

The square of the above table values are,

Ix 2 =10+10+3+21+6+1 = 51

Correction factor = -15.04

SST= Ix - -=51-15.04 = 35.96

SSR = 15.04 SSE = SST - SSC-SSR = 35.96 - 13.13 - 9.71 = 13.12

Step 3: Level of significance: Not given. So, let us take 5 % = 0.05

Step 4: Degrees of Freedom : for rows (3,15) for columns (5,15)

Step 5: Table value : Fo.os(3,15)= 3.29 Fo.os(5,15) = 5.05

Step 6: Conclusion: Comparing the FR- value calculated at step 2 with table value of step 5 FR>Ftab,we Reject the Null hypothesis 1. That is there is significant difference in the moisture content of the powder between the observers(rows).

Comparing the Fe - value calculated at step 2 with table value of step 5 Fe< Ftab, we accept the Null hypothesis II. That is there is no difference in the moisture content between the consignments( columns)