# SMU MBA ASSIGNMENTS

## 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.

December 02, 2017 By: Meliza Category: 1st SEM

2          a.         Explain The concept of One Way ANOVA.

1. 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