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Poisson distribution Confidence Interval

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Find 95% confidence intervals for the population percent dying based on these data: (1) 199 of 13,078 electronics technicians died of disease; (2) 100 of 13,078 electronics technicians died of circulatory disease; (3) 308 of 10,116 radarmen died (of any cause); (4) 441 of 13,078 electronics technicians died (of any cause); (5) 103 of 10,116 radarmen died of an accidental death.
(a) Use the normal approximation to the Poisson distribution (which is approximating a binomial distribution).
(b) Use the large-sample binomial confidence intervals (of Section 6.2.6). Do you think the intervals are similar to those calculated in part (a)?

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This solution helps with finding 95% confidence intervals using the normal approximation to the Poisson distribution and the large-sample binomial confidence intervals.

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Solution
a) Let pi be the population percent dying based on a particular disease. Then, lambda = n*pi denotes the mean number of people dying based on that particular disease.
Using Normal approximation, an approximate 100(1-alpha)% confidence interval for lambda is given by,
(see attachment for equation)
where z_(1-alpha/2) is a standard normal deviate at two-sided significance level alpha and Y is the observed number of people dying based on a particular disease.
For 95% confidence interval z_(1-alpha/2) = 1.96.
1) 199 of 13,078 electronics technicians died of disease.
Here, Y = 199, n = 13,078.
Therefore, 95% confidence interval for the mean, lambda , is
199 (plus or minus) (1.96)(√199) = 199 (plus or minus) 27.65 = (171.35, 226.65)
Thus, 95% confidence interval for the population percent dying, , is given by
(171.35/13078, 226.65/13078) or (0.0131, 0.0173) or (1.31%, 1.73%)
2) 100 of 13,078 electronics technicians ...

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