/*! 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} Problem 59 An ordinance requiring that a sm... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An ordinance requiring that a smoke detector be installed in all previously constructed houses has been in effect in a particular city for 1 year. The fire department is concerned that many houses remain without detectors. Let \(p=\) the true proportion of such houses having detectors, and suppose that a random sample of 25 homes is inspected. If the sample strongly indicates that fewer than \(80 \%\) of all houses have a detector, the fire department will campaign for a mandatory inspection program. Because of the costliness of the program, the department prefers not to call for such inspections unless sample evidence strongly argues for their necessity. Let \(X\) denote the number of homes with detectors among the 25 sampled. Consider rejecting the claim that \(p \geq .8\) if \(x \leq 15\). a. What is the probability that the claim is rejected when the actual value of \(p\) is \(.8\) ? b. What is the probability of not rejecting the claim when \(p=.7 ?\) When \(p=.6 ?\) c. How do the "error probabilities" of parts (a) and (b) change if the value 15 in the decision rule is replaced by 14 ?

Short Answer

Expert verified
Probabilities for each scenario change with the decision criterion; part (c) shows increased probabilities for errors with \( x = 14 \).

Step by step solution

01

Understand the Problem

We need to calculate probabilities related to a binomial distribution, where we determine the likelihood of rejecting or not rejecting a claim based on a sample size.
02

Define Parameters

Set the parameters for the binomial distribution. The number of trials, \( n \), is 25 (sample size), and the probability of success, \( p \), is provided for each part of the question.
03

Find Probabilities for Part (a)

For part (a), we need the probability of \( X \leq 15 \) when \( p = 0.8 \). Use the cumulative binomial distribution function to calculate \( P(X \leq 15 | p = 0.8) \).
04

Find Probabilities for Part (b)

For part (b), calculate the probabilities \( P(X > 15) \) for \( p = 0.7 \) and \( p = 0.6 \). Use \( P(X > 15) = 1 - P(X \leq 15) \) for both cases.
05

Adjust Decision Criterion in Part (c)

In part (c), repeat the calculations from part (a) and (b) using \( x = 14 \) for the decision rule. Calculate \( P(X \leq 14 | p = 0.8) \), \( P(X > 14 | p = 0.7) \), and \( P(X > 14 | p = 0.6) \).

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 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Calculation
Probability calculation is an essential part of statistics, especially when dealing with binomial distributions. Here, it involves finding the likelihood of certain outcomes based on predefined probabilities.
For instance, when determining if a certain situation, like the presence of smoke detectors in houses, meets expectations or not, a probability calculation helps determine the decision's confidence level.
  • The probability of rejecting a claim (in this case, that 80% of homes have detectors) can be calculated by considering outcomes less than 15 detectors when the true probability is 0.8.
  • This calculation requires using the cumulative probability formula, summing probabilities of having 0 to 15 detectors out of 25 homes.
    Calculating these probabilities provides a sense of how likely it is that the reality (e.g., proportion of homes with detectors) deviates from the expected outcome.
Understanding this calculation provides a foundation for evaluating scenarios and making decisions based on data rather than assumptions.
Error Probabilities
Error probabilities describe the likelihood of making errors in statistical decisions. In hypothesis testing, like the decision regarding smoke detectors, two types of common errors are examined:
  • Type I Error (False Positive): Rejecting a true claim, in this case, suggesting fewer homes have detectors when they actually do meet the 80% threshold.
  • Type II Error (False Negative): Failing to reject a false claim, implying homes meet the threshold when fewer actually have detectors. In part (b) of the exercise, this is highlighted by determining the likelihood of not rejecting the claim when the true probabilities are 0.7 or 0.6.
Altering decision criteria, such as changing the threshold from 15 to 14, influences these error probabilities. Lowering the acceptable threshold reduces the chance of a Type I error but could increase the Type II error probability. This balance is crucial to consider for effective decision-making.
Binomial Distribution Function
The binomial distribution function is a statistical method used to model the number of successful outcomes (like the presence of a smoke detector) in a fixed number of trials (25 homes) with a given probability of success (such as 0.8). It's represented as: \ \[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\]\This formula helps in various calculations, such as deciding how likely it is to observe certain data outcomes under specific assumptions.
The cumulative distribution function (CDF) is particularly useful, summing probabilities from 0 up to a desired number of successes, to give the probability of observing a number of successes less than or equal to a specific value.
In our context, for example, when it calculates the chance of having 0 to 15 homes with detectors in a sample of 25, if 80% is the expected success rate.
The cumulative function is crucial for evaluating hypotheses by comparing observed data with expected probabilities and deciding whether discrepancies are due to random variation or indicate a significant difference from the expectation.

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

Suppose that the number of drivers who travel between a particular origin and destination during a designated time period has a Poisson distribution with parameter \(\lambda=20\) (suggested in the article "Dynamic Ride Sharing: Theory and Practice," J. of Transp. Engr., 1997: 308-312). What is the probability that the number of drivers will a. Be at most 10 ? b. Exceed 20 ? c. Be between 10 and 20 , inclusive? Be strictly between 10 and 20 ? d. Be within 2 standard deviations of the mean value?

Suppose that \(30 \%\) of all students who have to buy a text for a particular course want a new copy (the successes!), whereas the other \(70 \%\) want a used copy. Consider randomly selecting 25 purchasers. a. What are the mean value and standard deviation of the number who want a new copy of the book? b. What is the probability that the number who want new copies is more than two standard deviations away from the mean value? c. The bookstore has 15 new copies and 15 used copies in stock. If 25 people come in one by one to purchase this text, what is the probability that all 25 will get the type of book they want from current stock? [Hint: Let \(X=\) the number who want a new copy. For what values of \(X\) will all 15 get what they want?] d. Suppose that new copies cost \(\$ 100\) and used copies cost \(\$ 70\). Assume the bookstore currently has 50 new copies and 50 used copies. What is the expected value of total revenue from the sale of the next 25 copies purchased? Be sure to indicate what rule of expected value you are using.

A small market orders copies of a certain magazine for its magazine rack each week. Let \(X=\) demand for the magazine, with pmf $$ \begin{array}{c|cccccc} x & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline p(x) & \frac{1}{15} & \frac{2}{15} & \frac{3}{15} & \frac{4}{15} & \frac{3}{15} & \frac{2}{15} \end{array} $$ Suppose the store owner actually pays \(\$ 1.00\) for each copy of the magazine and the price to customers is \(\$ 2.00\). If magazines left at the end of the week have no salvage value, is it better to order three or four copies of the magazine?

A particular type of tennis racket comes in a midsize version and an oversize version. Sixty percent of all customers at a certain store want the oversize version. a. Among ten randomly selected customers who want this type of racket, what is the probability that at least six want the oversize version? b. Among ten randomly selected customers, what is the probability that the number who want the oversize version is within 1 standard deviation of the mean value? c. The store currently has seven rackets of each version. What is the probability that all of the next ten customers who want this racket can get the version they want from current stock?

A company that produces fine crystal knows from experience that \(10 \%\) of its goblets have cosmetic flaws and must be classified as "seconds." a. Among six randomly selected goblets, how likely is it that only one is a second? b. Among six randomly selected goblets, what is the probability that at least two are seconds? c. If goblets are examined one by one, what is the probability that at most five must be selected to find four that are not seconds?

See all solutions

Recommended explanations on Math Textbooks

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.