EGN 2440 Quiz 7 (OPTIONAL Show your work. Do not just write the final answer. A

Question

EGN 2440 Quiz 7 (OPTIONAL Show your work. Do not just write the final answer. A traffic engineer needs your help. The engineer needs to find a distribution (random variable) that can describe the number rear-end crashes per month at an expressway exit ramp. The following data was collected Number of crashes Observed Observed Theoretical Chi-Sq frequency probability probability value 0 2 12 15 10 Totals Average value of the observed [You need this value to compute the theoretical probabilities] distribution () Conduct a Chi-Sq Goodness of Fit Test at 0.05 significance level to determine whether the Poisson distribution is a good fit for this data. Hint: state the null and alternate hypotheses, compute the test statistic, compute the critical value, state whether the claim is true or false]

Explanation / Answer

H0=The given data are consistent with Poisson distribution

H1= The given data are not condsistent with poisson distribution

The estimator of the poisson parameter is the mean of the given data i.e 4.23

For the Poisson distribytion the probability of occuring x events is given by

P[X=x]=(e^-l)*(l^x)/x! ;x=0,1,2,3...

#####No. of clashes
> x=c(0,seq(1:10));x
[1] 0 1 2 3 4 5 6 7 8 9 10
> ###Observed Frequency
> oi=c(2,4,8,12,15,11,6,4,2,3,1);oi;length(oi)
[1] 2 4 8 12 15 11 6 4 2 3 1
[1] 11
> ####Observed Probability
> opi=oi/68;opi
[1] 0.02941176 0.05882353 0.11764706 0.17647059 0.22058824 0.16176471
[7] 0.08823529 0.05882353 0.02941176 0.04411765 0.01470588
> ##expected probabilities
> epi=exp(-4.23)*((4.23)^x)/factorial(x);epi
[1] 0.01455239 0.06155661 0.13019223 0.18357105 0.19412639 0.16423092
[7] 0.11578280 0.06996589 0.03699447 0.01738740 0.00735487
> ##expected frequency
> ei=68*epi;ei
[1] 0.9895626 4.1858496 8.8530719 12.4828314 13.2005942 11.1677027
[7] 7.8732304 4.7576807 2.5156237 1.1823431 0.5001311
> approxei=c(1,4,9,12,13,11,8,5,3,1,1);approxei
[1] 1 4 9 12 13 11 8 5 3 1 1
> e1i
Error: object 'e1i' not found
> ###Chiquare Value
> chi=(oi-approxei)^2/approxei;chi
[1] 1.0000000 0.0000000 0.1111111 0.0000000 0.3076923 0.0000000 0.5000000
[8] 0.2000000 0.3333333 4.0000000 0.0000000

> cbind(x,oi,opi,epi,approxei,chi)
x oi opi epi approxei chi
[1,] 0 2 0.02941176 0.01455239 1 1.0000000
[2,] 1 4 0.05882353 0.06155661 4 0.0000000
[3,] 2 8 0.11764706 0.13019223 9 0.1111111
[4,] 3 12 0.17647059 0.18357105 12 0.0000000
[5,] 4 15 0.22058824 0.19412639 13 0.3076923
[6,] 5 11 0.16176471 0.16423092 11 0.0000000
[7,] 6 6 0.08823529 0.11578280 8 0.5000000
[8,] 7 4 0.05882353 0.06996589 5 0.2000000
[9,] 8 2 0.02941176 0.03699447 3 0.3333333
[10,] 9 3 0.04411765 0.01738740 1 4.0000000
[11,] 10 1 0.01470588 0.00735487 1 0.0000000

> > chisquare=sum(chi);chisquare
[1] 6.452137

The number of degrees of freedom is n-2 here n=11 tabulated value for chisquare(9,0.05) is 16.92

Conclusion: Since the calculated value 6.452137 is less than the tabulated value 16.92 we can say that the calculated valuue is not significant. So we do not have enough evidence to reject the null hypothesis. We conclude that the data is consistent with Poisson Distribution with parameter 4.23.

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