Question #160383

The number of computer malfunctions per day is recorded for 260 days with the

following results.


Number of malfunctions (x i ) 0 1 2 3 4 5


Number of days(f i ) 77 90 55 30 5 3


Test the goodness of fit of an appropriate probability model.


1
Expert's answer
2021-02-03T03:55:05-0500

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,\lambda, which is the mean (and also the variance).


mean=xˉ=fixifimean=\bar{x}=\dfrac{\sum f_ix_i}{\sum f_i}

fixi=77(0)+90(1)+55(2)+30(3)+5(4)\sum f_ix_i=77(0)+90(1)+55(2)+30(3)+5(4)


+3(5)=325+3(5)=325




fi=77+90+55+30+5+3=260\sum f_i=77+90+55+30+5+3=260


mean=xˉ=325260=1.25mean=\bar{x}=\dfrac{325}{260}=1.25

fixi2=77(0)2+90(1)2+55(2)2+30(3)2\sum f_ix_i^2=77(0)^2+90(1)^2+55(2)^2+30(3)^2


+5(4)2+3(5)2=735+5(4)^2+3(5)^2=735

Var(X)=s2=7352601.252=1.264423Var(X)=s^2=\dfrac{735}{260}-1.25^2=1.264423

Hence with λ=1.25\lambda=1.25


P(X=x)=e1.25(1.25)xx!,x=0,1,2,3,...P(X=x)=\dfrac{e^{-1.25}(1.25)^x}{x!}, x=0,1,2,3,...

which gives the following probabilities and expected frequencies

(260×\times probability)


xiP(X=xi)Ei00.286574.510.358193.120.223858.230.093324.240.02917.650.00731.960.00190.51.0000260.0\begin{matrix} x_i & P(X=x_i) & E_i \\ \hline \\ 0 & 0.2865 & 74.5 \\ 1 & 0.3581 & 93.1 \\ 2 & 0.2238 & 58.2 \\ 3 & 0.0933 & 24.2 \\ 4 & 0.0291 & 7.6 \\ 5 & 0.0073 & 1.9 \\ \geq6 & 0.0019 & 0.5 \\ & 1.0000 & 260.0 \end{matrix}

Since a Poisson distribution is valid for all positive values of X,X, the additional class X6X\geq6 is necessary, and

P(X6)=1P(X6)P(X\geq6)=1-P(X\leq6)

xiOi=fiP(X=xi)Ei0770.286574.51900.358193.12550.223558.23300.093324.2480.038310.01.0000\def\arraystretch{1.5} \begin{array}{c:c:c: c} x_i & O_i=f_i & P(X=x_i) & E_i \\ \hline 0 & 77 & 0.2865 & 74.5 \\ \hdashline 1 & 90 & 0.3581 & 93.1 \\ \hdashline 2 & 55 & 0.2235 & 58.2 \\ \hdashline 3 & 30 & 0.0933 & 24.2 \\ \hdashline \geq4 & 8 & 0.0383 & 10.0 \\ \hdashline & & 1.0000 & \\ \end{array}

xiOiEi(OiEi)2(OiEi)2Ei02.56.250.08413.19.610.10323.210.240.17635.833.641.39042.04.000.4002.153\def\arraystretch{1.5} \begin{array}{c:c:c: c} x_i & O_i-E_i & (O_i-E_i)^2 & \dfrac{(O_i-E_i)^2}{E_i} \\ \hline 0 & 2.5 &6.25 & 0.084 \\ \hdashline 1 & -3.1 & 9.61 & 0.103 \\ \hdashline 2 & -3.2 & 10.24 & 0.176 \\ \hdashline 3 & 5.8 & 33.64 & 1.390 \\ \hdashline \geq4 & -2.0 & 4.00 & 0.400 \\ \hdashline & & & 2.153\\ \end{array}

H0:H_0: number of daily malfunctions is \sim Poisson

H1:H_1: number of daily malfunctions is not\sim Poisson


Significance level, α=0.05\alpha=0.05

Degrees of freedom, ν=52=3\nu=5-2=3

5 classes: 0,1,2,3,40,1,2,3,\geq4

2 constrants: Ei=Oi\sum E_i=\sum O_i and Eixi=Oixi\sum E_ix_i=\sum O_ix_i

from estimation of λ\lambda

Critical region χ2>7.815\chi^2>7.815


Test statistic is


χ2=(OiEi)2Ei=2.153\chi^2=\sum\dfrac{(O_i-E_i)^2}{E_i}=2.153

This 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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