Data: Treatment A -8,7,6,6,3,4,7,2,3,4 Treatment B-9,9,8,14,8,13,11,5,7,6 A.) co

Question

Data:

Treatment A -8,7,6,6,3,4,7,2,3,4

Treatment B-9,9,8,14,8,13,11,5,7,6

A.) conduct wald and score 95% confidence intervals fort he mean number of imperfections for Treatment A and Bassuming that teh counts have a Poisson distribution

B.) For each treatment, compute the sample mean and sample vairnace. Use these to comment on the assumption that tehse are random samples from the Poisson distribution. Then test fort he goodness of fit for each treatment to the Poisson distribution using the test statistic Z=(S2/x)-1)sqrt(n-1/2

This statistic has approximately a standard normal distribution when the data come from a Poisson distribution. You reject the Poisson distribution for large values of Z.

Explanation / Answer

Solution

95% Confidence Interval for mean of Poisson Distribution is given by:

Sample Mean ± Z/2{(Sample mean/Sample size)}

Calculations

Treatment 1: n = 10, = 0.05, Sample mean = 50/10 = 5; Z/2 = 1.96; CI: 5 ± 1.96x(5/10)

= 5 ± 1.96x0.7071 = 5 ± 1.39 ANSWER 1

Treatment 2: n = 10, = 0.05, Sample mean = 90/10 = 9; Z/2 = 1.96; CI: 9 ± 1.96x(9/10)

= 9 ± 1.96x0.9487 = 9 ± 1.86 ANSWER 2

Variance

Treatment 1: 3.8 and Treatment 2: 7.6

In both cases, variance is fairly close to the mean. => the data may be considered to be Poisson distributed.

For testing for goodness of fit, the test statistics specified, Z=(S2/x)-1)sqrt(n-1/2 is not clearly typed – what does S stand for? What is S2? And is it sq.rt of {(n - 1)/2} or (sq.rt of n) – ½ ?

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