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Chapter 7: Q26E (page 294)

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

Expert verified

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

Step by step solution

01

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\({\rm{\bar X}}\)

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

02

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: \((3.5015,4.6185)\)

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Most popular questions from this chapter

A study of the ability of individuals to walk in a straight line reported the accompanying data on cadence (strides per second) for a sample of n =\({\rm{20}}\) randomly selected healthy men.

\({\rm{.95 }}{\rm{.85 }}{\rm{.92 }}{\rm{.95 }}{\rm{.93 }}{\rm{.86 1}}{\rm{.00 }}{\rm{.92 }}{\rm{.85 }}{\rm{.81 }}{\rm{.78 }}{\rm{.93 }}{\rm{.93 1}}{\rm{.05 }}{\rm{.93 1}}{\rm{.06 1}}{\rm{.06 }}{\rm{.96 }}{\rm{.81 }}{\rm{.96}}\)

A normal probability plot gives substantial support to the assumption that the population distribution of cadence is approximately normal. A descriptive summary of the data from Minitab follows:

Variable N Mean Median TrMean StDev SEMean cadence

\({\rm{20 0}}{\rm{.9255 0}}{\rm{.9300 0}}{\rm{.9261 0}}{\rm{.0809 0}}{\rm{.0181}}\)

Variable Min Max Q1 Q3 cadence

\({\rm{0}}{\rm{.7800 1}}{\rm{.0600 0}}{\rm{.8525 0}}{\rm{.9600}}\)

a. Calculate and interpret a \({\rm{95\% }}\) confidence interval for population mean cadence.

b.Calculate and interpret a \({\rm{95\% }}\)prediction interval for the cadence of a single individual randomly selected from this population.

c. Calculate an interval that includes at least \({\rm{99\% }}\)of the cadences in the population distribution using a confidence level of \({\rm{95\% }}\)

a. Under the same conditions as those leading to the interval\({\rm{(7}}{\rm{.5),p((}}\overline {\rm{X}} {\rm{ - \mu )/(\sigma /}}\sqrt {\rm{n}} {\rm{) < 1}}{\rm{.645 = }}{\rm{.95}}{\rm{.}}\)Use this to derive a one-sided interval for\({\rm{\mu }}\)that has infinite width and provides a lower confidence bound on m. What is this interval for the data in Exercise 5(a)?

b. Generalize the result of part (a) to obtain a lower bound with confidence level\({\rm{100(1 - \alpha )\% }}\)

c. What is an analogous interval to that of part (b) that provides an upper bound on\({\rm{\mu }}\)? Compute this 99% interval for the data of Exercise 4(a).

The following observations are lifetimes (days) subsequent to diagnosis for individuals suffering from blood cancer (鈥淎 Goodness of Fit Approach to the Class of Life Distributions with Unknown Age,鈥� Quality and Reliability Engr. Intl., \({\rm{2012: 761--766):}}\)

\(\begin{array}{*{20}{l}}{{\rm{115 181 255 418 441 461 516 739 743 789 807}}}\\{{\rm{865 924 983 1025 1062 1063 1165 1191 1222 1222 1251}}}\\{{\rm{1277 1290 1357 1369 1408 1455 1478 1519 1578 1578 1599}}}\\{{\rm{1603 1605 1696 1735 1799 1815 1852 1899 1925 1965}}}\end{array}\)

a. Can a confidence interval for true average lifetime be calculated without assuming anything about the nature of the lifetime distribution? Explain your reasoning. (Note: A normal probability plot of the data exhibits a reasonably linear pattern.)

b. Calculate and interpret a confidence interval with a \({\rm{99\% }}\)confidence level for true average lifetime.

Each of the following is a confidence interval for \({\rm{\mu = }}\) true average (i.e., population mean) resonance frequency (Hz) for all tennis rackets of a certain type: \({\rm{(114}}{\rm{.4,115}}{\rm{.6)(114}}{\rm{.1,115}}{\rm{.9)}}\) a. What is the value of the sample mean resonance frequency? b. Both intervals were calculated from the same sample data. The confidence level for one of these intervals is \({\rm{90\% }}\) and for the other is \({\rm{99\% }}\). Which of the intervals has the \({\rm{90\% }}\) confidence level, and why?

A state legislator wishes to survey residents of her district to see what proportion of the electorate is aware of her position on using state funds to pay for abortions.

a. What sample size is necessary if the \({\rm{95\% }}\)CI for p is to have a width of at most . \({\rm{10}}\)irrespective of p?

b. If the legislator has strong reason to believe that at least \({\rm{2/3}}\)of the electorate know of her position, how large a sample size would you recommend?

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