Answer to Question #160383 in Statistics and Probability for Abu Ubayda

Question

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.

Answer 1

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

$$mean=\bar{x}=\dfrac{\sum f_ix_i}{\sum f_i}$$

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

$$+3(5)=325$$

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

$$mean=\bar{x}=\dfrac{325}{260}=1.25$$

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

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

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

Hence with $\lambda=1.25$

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

which gives the following probabilities and expected frequencies

(260$\times$ probability)

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

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

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

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

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

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

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

Significance level, $\alpha=0.05$

Degrees of freedom, $\nu=5-2=3$

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

2 constrants: $\sum E_i=\sum O_i$ and $\sum E_ix_i=\sum O_ix_i$

from estimation of $\lambda$

Critical region $\chi^2>7.815$

Test statistic is

$$\chi^2=\sum\dfrac{(O_i-E_i)^2}{E_i}=2.153$$

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.