Question

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.

Answer 1

Answer:

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)