Table given below depicts the data on production rate by five workmen on four machines. Test whether the rate is significantly different due to workers and machines.
2 a. Explain The concept of One Way ANOVA.
- Table given below depicts the data on production rate by five workmen on four machines. Test whether the rate is significantly different due to workers and machines.
| Machines | Workmen | ||||
| I | II | III | IV | V | |
| 1 | 46 | 48 | 36 | 35 | 40 |
| 2 | 40 | 42 | 38 | 40 | 44 |
| 3 | 49 | 54 | 46 | 48 | 51 |
| 4 | 38 | 45 | 34 | 35 | 41 |
Answer: – The concept of One Way ANOVA
Procedure for carrying out the One-way ANOVA
| *Compute the sum of all values ‘T’.
*Find the correction factor: 2 Correction factor = T N *Find Total sum of squares: T 2 / N SST = Sum of squares of all observations – |
| *Decision: If the computed value of F > Table (critical) value of F for degrees of freedom (k-1, n – k) at α% (5% or 1%), then we reject H0 and conclude that all the population means are unequal. Otherwise accept H0 and conclude that the population means are not unequal.
*Sum of the squares of the Error within columns (samples): SSE =SST -SSC -SSR *Variance between samples: MSC=SSC/(k-1) *Variance within the samples: MSE=SSE/(n-k)
|
Explanation of ANOVA Numerical Solution
Solutions
Solution
The null hypothesis ‘H0’ is given as: 0Η : There is no difference between the replications or the varieties
| Machines | Workmen | |||||
| I | II | III | IV | V | ||
| 1 | 46 | 48 | 36 | 35 | 40 | 205
204 |
| 2 | 40 | 42 | 38 | 40 | 44 | 204 |
| 3 | 49 | 54 | 46 | 48 | 51 | 248 |
| 4 | 38 | 45 | 34 | 35 | 41 | 193 |
| 173 | 189 | 154 | 158 | 176 | 850 | |
sum of all values = 850
Correction factor = T2/N
| 2 |
=850/20
=722500/20
=36125
Total sum of squares: SST = Sum of squares of all observations T2/N
SST=
2116+1600+2401+1444=7561
2304+1764+2916+2025=9009
1296+1444+2116+1156=6012
1225+1600+2304+1225=6354
1600+1936+2601+1681=7818
T=36754-36125
=629
SSC= (173*2+189*2+154*2+158*2+176*2 / 4)-36125
=29929+35721+23716+24964+30976/4
=(145306/4)-36125
=36326.5-36125
=201.5
MSC=SSC/(C-1)
201.5/(5-1) =201.5/4 = 50.375
SSR = 205*2+204*2+248*2+193*2 / 5
=36478.8-36125
=353.8
MSR = SSC/(C-1)
=36478.8/(5-1)
=353.8/4
=88.45
SS residual or error: SSE = SST – SSC – SSR
SSE = 629-201.5-353.8
=73.7
=88.45/12
=7.37
| Source of Variation | Sum of Squares | Degrees of Freedom | Mean Square | F-Ratio |
| Between Columns Between Rows
Residual |
SSC = 201.5
SSR = 353.8 SSE = 73.7 |
c – 1 = (5-1) = 4
r – 1 =(4-1) = 3 (c-1) x (r -1) = 12 |
MSC = 50.375
MSR = 88.45 MSE = 7.37 |
Fcal = 8.20 for workman
Fcal = 19.20 for machines Both are not significant
|
| Total | SST = 629 | n – 1 = 20-1= 19 |
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