/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q26E The superintendent of a large sc... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

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: \((3.5015,4.6185)\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Determine the values of the following quantities

\(\begin{array}{l}{\rm{a}}{\rm{.}}{{\rm{x}}^{\rm{2}}}{\rm{,1,15}}\\{\rm{b}}{\rm{.}}{{\rm{X}}^{\rm{3}}}{\rm{,125}}\\{\rm{c}}{\rm{.}}{{\rm{X}}^{\rm{7}}}{\rm{01,25}}\\{\rm{d}}{\rm{.}}{{\rm{X}}^{\rm{2}}}{\rm{00525}}\\{\rm{e}}{\rm{.}}{{\rm{X}}^{\rm{7}}}{\rm{9925}}\\{\rm{f}}{\rm{.}}{{\rm{X}}^{\rm{7}}}{\rm{995,25}}\end{array}\)

Let X1, X2,鈥�, Xn be a random sample from a continuous probability distribution having median \(\widetilde {\bf{\mu }}\) (so that \({\bf{P(}}{{\bf{X}}_{\bf{i}}} \le \widetilde {\bf{\mu }}{\bf{) = P(}}{{\bf{X}}_{\bf{i}}} \le \widetilde {\bf{\mu }}{\bf{) = }}{\bf{.5}}\)).

a. Show that

\({\bf{P(min(}}{{\bf{X}}_{\bf{i}}}{\bf{) < }}\widetilde {\bf{\mu }}{\bf{ < max(}}{{\bf{X}}_{\bf{i}}}{\bf{)) = 1 - }}{\left( {\frac{{\bf{1}}}{{\bf{2}}}} \right)^{{\bf{n - 1}}}}\)

So that \({\bf{(min(}}{{\bf{x}}_{\bf{i}}}{\bf{),max(}}{{\bf{x}}_{\bf{i}}}{\bf{))}}\)is a \({\bf{100(1 - \alpha )\% }}\) confidence interval for \(\widetilde {\bf{\mu }}\) with (\({\bf{\alpha = 1 - }}{\left( {\frac{{\bf{1}}}{{\bf{2}}}} \right)^{{\bf{n - 1}}}}\).Hint :The complement of the event \(\left\{ {{\bf{min(}}{{\bf{X}}_{\bf{i}}}{\bf{) < \mu < max(}}{{\bf{X}}_{\bf{i}}}{\bf{)}}} \right\}\)is \({\bf{\{ max(}}{{\bf{X}}_{\bf{i}}}{\bf{)}} \le \widetilde {\bf{\mu }}{\bf{\} }} \cup {\bf{\{ min(}}{{\bf{X}}_{\bf{i}}}{\bf{)}} \ge \widetilde {\bf{\mu }}{\bf{\} }}\). But \({\bf{max(}}{{\bf{X}}_{\bf{i}}}{\bf{)}} \le \widetilde {\bf{\mu }}\) iff \({\bf{(}}{{\bf{X}}_{\bf{i}}}{\bf{)}} \le \widetilde {\bf{\mu }}\)for all i.

b.For each of six normal male infants, the amount of the amino acid alanine (mg/100 mL) was determined while the infants were on an isoleucine-free diet,

resulting in the following data:

2.84 3.54 2.80 1.44 2.94 2.70

Compute a 97% CI for the true median amount of alanine for infants on such a diet(鈥淭he Essential Amino Acid Requirements of Infants,鈥� Amer. J. Of Nutrition, 1964: 322鈥�330).

c.Let x(2)denote the second smallest of the xi鈥檚 and x(n-1) denote the second largest of the xi鈥檚. What is the confidence level of the interval(x(2), x(n-1)) for \(\widetilde \mu \)?

TV advertising agencies face increasing challenges in reaching audience members because viewing TV programs via digital streaming is gaining in popularity. The Harris poll reported on November 13, 2012, that 53% of 2343 American adults surveyed said they have watched digitally streamed TV programming on some type of device.

a. Calculate and interpret a confidence interval at the 99% confidence level for the proportion of all adult Americans who watched streamed programming up to that point in time.

b. What sample size would be required for the width of a 99% CI to be at most .05 irrespective of the value of p 藛?

The alternating current (AC) breakdown voltage of an insulating liquid indicates its dielectric strength. The article 鈥淭esting Practices for the AC Breakdown Voltage Testing of Insulation Liquids'' (IEEE)gave the accompanying sample observations on breakdown voltage (kV) of a particular circuit under certain conditions.

\(\begin{array}{l}{\rm{62\;50\;53\;57\;41\;53\;55\;61\;59\;64\;50\;53\;64\;62}}\\{\rm{\;50\;68 54\;55\;57\;50\;55\;50\;56\;55\;46\;55\;53\;54\;}}\\{\rm{52\;47\;47\;55 57\;48\;63\;57\;57\;55\;53\;59\;53\;52\;}}\\{\rm{50\;55\;60\;50\;56\;58}}\end{array}\)

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

b. Calculate and interpret a \({\rm{95\% }}\)CI for true average breakdown voltage m. Does it appear that m has been precisely estimated? Explain.

c. Suppose the investigator believes that virtually all values of breakdown voltage are between \({\rm{40 and 70}}\). What sample size would be appropriate for the \({\rm{95\% }}\)CI to have a width of 2 kV (so that m is estimated to within 1 kV with \({\rm{95\% }}\)confidence)

Determine the confidence level for each of the following large-sample one-sided confidence bounds:

\(\begin{array}{l}{\rm{a}}{\rm{.Upperbound:\bar x + }}{\rm{.84s/}}\sqrt {\rm{n}} \\{\rm{b}}{\rm{.Lowerbound:\bar x - 2}}{\rm{.05s/}}\sqrt {\rm{n}} \\{\rm{c}}{\rm{.Upperbound:\bar x + }}{\rm{.67\;s/}}\sqrt {\rm{n}} \end{array}\)
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Mechanics Maths

Read Explanation

Pure Maths

Read Explanation

Discrete Mathematics

Read Explanation

Geometry

Read Explanation

Applied Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.