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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

Consider a normal population distribution with the value of \({\rm{\sigma }}\) known. a. What is the confidence level for the interval \({\rm{\bar x \pm 2}}{\rm{.81\sigma /}}\sqrt {\rm{n}} \)? b. What is the confidence level for the interval \({\rm{\bar x \pm 1}}{\rm{.44\sigma /}}\sqrt {\rm{n}} \)? c. What value of \({{\rm{z}}_{{\rm{\alpha /2}}}}\) in the CI formula (\({\rm{7}}{\rm{.5}}\)) results in a confidence level of \({\rm{99}}{\rm{.7\% }}\)? d. Answer the question posed in part (c) for a confidence level of \({\rm{75\% }}\).

The article 鈥淢easuring and Understanding the Aging of Kraft Insulating Paper in Power Transformers鈥� contained the following observations on degree of polymerization for paper specimens for which viscosity tim\({\rm{418 421 421 422 425 427 431 434 437 439 446 447 448 453 454 463 465}}\)es concentration fell in a certain middle range:

a. Construct a boxplot of the data and comment on any interesting features.

b. Is it plausible that the given sample observations were selected from a normal distribution?

c. Calculate a two-sided \({\rm{95\% }}\)confidence interval for true average degree of polymerization (as did the authors of the article). Does the interval suggest that \({\rm{440}}\) is a plausible value for true average degree of polymerization? What about \({\rm{450}}\)?

A CI is desired for the true average stray-load loss \({\rm{\mu }}\) (watts) for a certain type of induction motor when the line current is held at \({\rm{10 amps}}\) for a speed of \({\rm{1500 rpm}}\). Assume that stray-load loss is normally distributed with \({\rm{\sigma = 3}}{\rm{.0}}\). a. Compute a \({\rm{95\% }}\) CI for \({\rm{\mu }}\) when \({\rm{n = 25}}\) and \({\rm{\bar x = 58}}{\rm{.3}}\). b. Compute a \({\rm{95\% }}\) CI for \({\rm{\mu }}\) when \({\rm{n = 100}}\) and \({\rm{\bar x = 58}}{\rm{.3}}\). c. Compute a \({\rm{99\% }}\) CI for \({\rm{\mu }}\) when \({\rm{n = 100}}\) and \({\rm{\bar x = 58}}{\rm{.3}}\). d. Compute an \({\rm{82\% }}\) CI for \({\rm{\mu }}\) when \({\rm{n = 100}}\) and \({\rm{\bar x = 58}}{\rm{.3}}\). e. How large must n be if the width of the \({\rm{99\% }}\) interval for \({\rm{\mu }}\) is to be \({\rm{1}}{\rm{.0}}\)?

High concentration of the toxic element arsenic is all too common in groundwater. The article 鈥淓valuation of Treatment Systems for the Removal of Arsenic from Groundwater鈥� (Practice Periodical of Hazardous, Toxic, and Radioactive Waste Mgmt., 2005: 152鈥�157) reported that for a sample of n = 5 water specimens selected for treatment by coagulation, the sample mean arsenic concentration was 24.3 碌g/L, and the sample standard deviation was 4.1. The authors of the cited article used t-based methods to analyze their data, so hopefully had reason to believe that the distribution of arsenic concentration was normal.

a.Calculate and interpret a 95% CI for true average arsenic concentration in all such water specimens.

b.Calculate a 90% upper confidence bound for the standard deviation of the arsenic concentration distribution.

c.Predict the arsenic concentration for a single water specimen in a way that conveys information about precision and reliability.

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