The Poisson distribution is a discrete probability distribution that is used to

Question

The Poisson distribution is a discrete probability distribution that is used to model count data, such as the number of events occurring over a time interval or region. For a random variable X that is Poisson-distributed, Fe- r! where > 0 is the rate parameter. The mean of the Poisson distribution is, in fact, so X is an unbiased estimator of A scientist is measuring the number of radioactive a particles emitted from a sample of an unknown radioactive material. The following data indicate the number of such particles emitted during observed 0.5 second intervals for 50 seconds (i.e., 100 total observations). For example, there were 22 intervals during which 2 particles were detected. (This data set is also available on Blackboard as RadEmit.txt.) Particles uenc 6 14 2 21 17 6 9 Determine, at the 5% significance level, if these data follow a Poisson distribution. You are encouraged to use Excel or other technology to perform your computations; if you do so, please attach your output and results. (a) State the u and alternative hypotheses. (b) Compute the test statistic. (c) Determine the rejection rule. (d) State your conclusion

Explanation / Answer

a]

Using MINITAB

Use Goodness-of-Fit Test for Poisson to test the hypotheses:

null hypothesis: H0: Data follow a Poisson distribution

alternative hypothesis : H1: Data do not follow a Poisson distribution

Choose Stat > Basic Statistics > Goodness-of-Fit Test for Poisson.

   In Variable, enter Particles.

   In Frequency, enter Frequency

   Click OK.

Session window output

MTB > PGoodness 'Particles';
SUBC>   Frequencies 'Frequency';
SUBC>   GBar;
SUBC>   GChiSQ;
SUBC>     Pareto;
SUBC>   RTable.

Goodness-of-Fit Test for Poisson Distribution

Data column: Particles
Frequency column: Frequency

Poisson mean for Particles = 2.98

                                  
Particles Observed Poisson Probability Expected   Contribution to Chi-Sq
0                 6                0.050793                5.0793      0.166897
1                14       0.151363              15.1363      0.085298
2                22       0.225530 22.5530      0.013561
3                21       0.224027              22.4027      0.087825
4                17               0.166900              16.6900      0.005758
5                11       0.099472   9.9472      0.111419
>=6             9    0.081915    8.1915      0.079798

N    DF    Chi-Sq       P-Value
100      5     0.550557    0.990

b]

test statistic Chi-Sq 0.550557

c]

Decision rule:

1) If p-value < level of significance (alpha) then we reject null hypothesis

2) If p-value > level of significance (alpha) then we fail to reject null hypothesis.

Here p value = 0.990 > 0.05 so we used 2nd rule.

That is we fail to reject null hypothesis

d]

Conclusion:

At 5% level of significance there is sufficient evidence that to say the given data follows the Poisson distribution.

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