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 m. Use the accompanying data on absences for 50 days to obtain a large sample CI for m. (Hint: The mean and variance of a Poisson variable both equal m, so

\({\rm{Z = }}\frac{{{\rm{\bar X - \mu }}}}{{\sqrt {{\rm{\mu /n}}} }}\)

has approximately a standard normal distribution. Now proceed as in the derivation of the interval for p by making a probability statement and solving the resulting inequalities for m

Short answer

\(95\% \)confidence interval: \((3.5015,4.6185)\)

Step-by-step solution

Derivation formula confidence interval

Given:

\({\rm{Z - }}\frac{{{\rm{\bar X - \mu }}}}{{{\rm{\mu /}}\sqrt {\rm{n}} }}{\rm{ n = 50}}\)

Multiply each side of the given equation by \(\sqrt {\frac{{\rm{\mu }}}{{\rm{n}}}} \)

\({\rm{Z}}\sqrt {\frac{{\rm{\mu }}}{{\rm{n}}}} {\rm{ - \bar X - \mu }}\)

Add \({\rm{\mu }}\)to each side of the equation:

\({\rm{\mu + Z}}\sqrt {\frac{{\rm{\mu }}}{{\rm{n}}}} {\rm{ - \bar X}}\)

Subtract \({\rm{Z = }}\sqrt {\frac{{\rm{\mu }}}{{\rm{n}}}} \)from each side of the equation:

\({\rm{\mu - \bar X - Z}}\sqrt {\frac{{\rm{\mu }}}{{\rm{n}}}} \)

By replacing Z by the critical z-scores \({\rm{ \pm }}{{\rm{z}}_{{\rm{\alpha /2}}}}\)we then obtain a confidence interval for\({\rm{\bar X}}\)

\({\rm{\mu - \bar X \pm }}{{\rm{z}}_{{\rm{\alpha /2}}}}\sqrt {\frac{{\rm{\mu }}}{{\rm{n}}}} \)

Calculation confidence interval

Note: I assume that we need to determine a \(95\% \)confidence interval, you can determine other confidence intervals similarly.

The sample mean is then the sum of the products of the midpoints and the frequencies, divided by the total frequency:

\({{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ - 1}}{\rm{.96}}\)

The boundaries of the confidence interval then become:

\(\begin{array}{l}{\rm{\bar x - }}{{\rm{z}}_{{\rm{\alpha /2}}}} = \sqrt {\frac{{\rm{\mu }}}{{\rm{n}}}} {\rm{ - 4}}{\rm{.06 - 1}}{\rm{.96}}\sqrt {\frac{{{\rm{4}}{\rm{.06}}}}{{{\rm{50}}}}} {\rm{\gg 3}}{\rm{.5015}}\\{\rm{\bar x + }}{{\rm{z}}_{{\rm{\alpha /2}}}} = \sqrt {\frac{{\rm{\mu }}}{{\rm{n}}}} {\rm{ - 4}}{\rm{.06 + 1}}{\rm{.96}}\sqrt {\frac{{{\rm{4}}{\rm{.06}}}}{{{\rm{50}}}}} {\rm{\gg 4}}{\rm{.6185}}\end{array}\)

Hence \(95\% \)confidence interval: \((3.5015,4.6185)\)