Example The superintendent of a large school district, having once had a course in probability and statistics, believes that the number of teachers absent on any given day has a Poisson distribution with parameter A. Using the accompanying data on absences for 50 days to derive a large-sample 95% confidence interval for A. Number of absences | 0 1 2 Frequency 3 4 5 6 7 8 9 10 1 4 8 10 8 7 5 3 2 1 1 [Hint: The Poisson population has mean A and variance d, so basically speaking, we are still trying to estimate the population mean using sample mean. And, by the central limit theorem, approximately X ~ N ( A, n

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Chapter1: Combinatorial Analysis
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Example 2.2: The superintendent of a large school district, having once had a course
in probability and statistics, believes that the number of teachers absent on any given
day has a Poisson distribution with parameter A. Using the accompanying data on
absences for 50 days to derive a large-sample 95% confidence interval for A.
Number of absences | 0 1 2
Frequency
3 4 5 6 7 8 9 10
1 4 8 10 8 7 5 3 2 1
1
[Hint: The Poisson population has mean A and variance d, so basically speaking, we are
still trying to estimate the population mean using sample mean. And, by the central
limit theorem, approximately
X ~ N (x).
i.e., approximately
~ N (0,1).
Then, follow the derivation in section 2 to find the margin of error, MOE, such that
Pr(X – MOE < <X+MOE) = 0.95. This MOE will contain A, which is unknown,
but we have an estimate for it.
Transcribed Image Text:Example 2.2: The superintendent of a large school district, having once had a course in probability and statistics, believes that the number of teachers absent on any given day has a Poisson distribution with parameter A. Using the accompanying data on absences for 50 days to derive a large-sample 95% confidence interval for A. Number of absences | 0 1 2 Frequency 3 4 5 6 7 8 9 10 1 4 8 10 8 7 5 3 2 1 1 [Hint: The Poisson population has mean A and variance d, so basically speaking, we are still trying to estimate the population mean using sample mean. And, by the central limit theorem, approximately X ~ N (x). i.e., approximately ~ N (0,1). Then, follow the derivation in section 2 to find the margin of error, MOE, such that Pr(X – MOE < <X+MOE) = 0.95. This MOE will contain A, which is unknown, but we have an estimate for it.
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