An appropriate probability model could be a Poisson distribution on the basis of following assumptions:
malfunctions occur independently,
simultaneous malfunctions are impossible,
malfunctions occur randomly in time,
malfunctions occur uniformly (mean number for time period
proportional to the period length).
A Poisson distribution has one parameter, λ, which is the mean (and also the variance).
mean=xˉ=∑fi∑fixi
∑fixi=77(0)+90(1)+55(2)+30(3)+5(4)
+3(5)=325
∑fi=77+90+55+30+5+3=260
mean=xˉ=260325=1.25
∑fixi2=77(0)2+90(1)2+55(2)2+30(3)2
+5(4)2+3(5)2=735
Var(X)=s2=260735−1.252=1.264423 Hence with λ=1.25
P(X=x)=x!e−1.25(1.25)x,x=0,1,2,3,...which gives the following probabilities and expected frequencies
(260× probability)
xi012345≥6P(X=xi)0.28650.35810.22380.09330.02910.00730.00191.0000Ei74.593.158.224.27.61.90.5260.0Since a Poisson distribution is valid for all positive values of X, the additional class X≥6 is necessary, and
P(X≥6)=1−P(X≤6)
xi0123≥4Oi=fi779055308P(X=xi)0.28650.35810.22350.09330.03831.0000Ei74.593.158.224.210.0
xi0123≥4Oi−Ei2.5−3.1−3.25.8−2.0(Oi−Ei)26.259.6110.2433.644.00Ei(Oi−Ei)20.0840.1030.1761.3900.4002.153 H0: number of daily malfunctions is ∼ Poisson
H1: number of daily malfunctions is not∼ Poisson
Significance level, α=0.05
Degrees of freedom, ν=5−2=3
5 classes: 0,1,2,3,≥4
2 constrants: ∑Ei=∑Oi and ∑Eixi=∑Oixi
from estimation of λ
Critical region χ2>7.815
Test statistic is
χ2=∑Ei(Oi−Ei)2=2.153This value does not lie in the critical region.
There is no evidence, at the 5% significant level, to suggest that the number
of computed malfunctions per day does not have a Poisson distribution.
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