Uncetrtainty Assignment

Assignment2: Analytical measurement uncertainty and method validation

Question 1:

i)

Source of variation | Sum of squares | Degrees of freedom | Mean square |

Between-cells | 1.00 | 3 | 0.3333 |

Within cells | 0.68 | 16 | 0.0425 |

Total | 1.68 | 19 | - |

ii) we are going to test this significance between lab at 95%confidnece interval using the one sided F-test consider the hypothesis that sigma2b=0

Null hypothesis: there is no long term variation between laboratories

Alternative hypothesis: there is long term variation between laboratories

Test statistic: between labs mean square/within lab means square=0.3333/0.0425=7.84

Critical values: from the one sided F-table with 3 and 16 degree of freedom are: 3.25 at 95% confidence interval

Decision: as 7.84>5.32, we reject the null hypothesis at 95% confidence interval

Conclusion: we conclude that there are long- term differences between laboratories

iii)Least Significant Difference (LSD) = t2 ×WLMSequal replicate in each laboratory

= t162 ×0.04255 = 2.122 ×0.04255 = 0.276

X Largest - X smallest = XC - XB=0.9-0.3=0.6>0.276 significant difference

X2nd Largest - X smallest = Xd - XB=0.7-0.3=0.4>0.276 significant difference

X3d Largest - X smallest = XA - XB=0.5-0.3=0.2<0.276 not significant difference

XC - Xd=0.9-0.7=0.2<0.276 not significant difference

XC - XA=0.9-0.5=0.3>0.276 significant difference

Xd - XA=0.7-0.5=0.2<0.276 not significant difference

iv) Within laboratories estimate standard deviation=Σ [(d.f)*(SD)]’2/Σ (d.f)= 4*(0.187)’2+4*(0.224)’2+4*(0.212)’2+4*(0.2)’2/4+4+4+4=0.0425

V) Between laboratories estimate standard deviation=between lab mean square- within lab mean square/number of determinations per lab= (0.3333-0.0425)/5=0.0581

vi)The within laboratory degree of freedom=16

t16 estimate ofσ2ᴡ / no of determination in that lab=2.120.0425/5=0.19

Laboratory | Mean and its 95% confidence interval |

A | 25%±0.19 |

B | 24.8%±0.19 |...