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 µ.

Use the accompanying data on absences for 50 days to obtain a 95% large sample CI for µ.

Absences: 0 1 2 3 4 5 6 7 8 9 10

Frequency: 2 3 8 11 8 7 6 2 1 1 1

[Hint: The mean and variance of a Poisson variable both equal m, so

Z = (X - µ) / √ (µ/n),

has approximately a standard normal distribution. Now proceed as in the derivation of the interval for p by making a probability statement (with probability 1 - a) and solving the resulting inequalities for µ.

Question

Akira Nunez

Mathematics

24 March, 19:38

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 µ.

Use the accompanying data on absences for 50 days to obtain a 95% large sample CI for µ.

Absences: 0 1 2 3 4 5 6 7 8 9 10

Frequency: 2 3 8 11 8 7 6 2 1 1 1

[Hint: The mean and variance of a Poisson variable both equal m, so

Z = (X - µ) / √ (µ/n),

has approximately a standard normal distribution. Now proceed as in the derivation of the interval for p by making a probability statement (with probability 1 - a) and solving the resulting inequalities for µ.

+2

Answers (1)

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Serenity Carlson · Answer 1

The confidence interval required is (3.37,4.47)

Step-by-step explanation:

From the given data

Let X represents the absence and f is the frequency

Summation of f = 50

Summation of fX=196

Mean of a Poisson distribution = 196/50 = 3.92

Critical value

At 5% level of significance for a two tailed z-distribution

Z (0.05/2) = + or - 1.96

The 95% confidence interval sample confidence interval

Cl (95%) = mean+or - z (at 0.05/2) * √ (mean/n

Where n=50

Cl (95%) = 3.92+/-1.96 (√3.92/50)

= (3.37,4.47)