/*! 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

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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 \( P(X > 14 | p = 0.6) \).

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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.

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Most popular questions from this chapter

A very large batch of components has arrived at a distributor. The batch can be characterized as acceptable only if the proportion of defective components is at most .10. The distributor decides to randomly select 10 components and to accept the batch only if the number of defective components in the sample is at most \(2 .\) a. What is the probability that the batch will be accepted when the actual proportion of defectives is \(.01 ? .05 ? .10\) ? \(.20 ? .25 ?\) b. Let \(p\) denote the actual proportion of defectives in the batch. A graph of \(P\) (batch is accepted) as a function of \(p\), with \(p\) on the horizontal axis and \(P\) (batch is accepted) on the vertical axis, is called the operating characteristic curve for the acceptance sampling plan. Use the results of part (a) to sketch this curve for \(0 \leq p \leq 1\). c. Repeat parts (a) and (b) with " 1 " replacing " 2 " in the acceptance sampling plan. d. Repeat parts (a) and (b) with " 15 " replacing " 10 " in the acceptance sampling plan. e. Which of the three sampling plans, that of part (a), (c), or (d), appears most satisfactory, and why?

A manufacturer of flashlight batteries wishes to control the quality of its product by rejecting any lot in which the proportion of batteries having unacceptable voltage appears to be too high. To this end, out of each large lot ( 10,000 batteries), 25 will be selected and tested. If at least 5 of these generate an unacceptable voltage, the entire lot will be rejected. What is the probability that a lot will be rejected if a. \(5 \%\) of the batteries in the lot have unacceptable voltages? b. \(10 \%\) of the batteries in the lot have unacceptable voltages? c. \(20 \%\) of the batteries in the lot have unacceptable voltages? d. What would happen to the probabilities in parts (a) - (c) if the critical rejection number were increased from 5 to 6 ?

Customers at a gas station pay with a credit card \((A)\), debit card \((B)\), or cash \((C)\). Assume that successive customers make independent choices, with \(P(A)=.5, P(B)=.2\), and \(P(C)=.3 .\) a. Among the next 100 customers, what are the mean and variance of the number who pay with a debit card? Explain your reasoning. b. Answer part (a) for the number among the 100 who don't pay with cash.

Consider a collection \(A_{1}, \ldots, A_{k}\) of mutually exclusive and exhaustive events, and a random variable \(X\) whose distribution depends on which of the \(A_{i}\) 's occurs (e.g., a commuter might select one of three possible routes from home to work, with \(X\) representing the commute time). Let \(E\left(X \mid A_{i}\right)\) denote the expected value of \(X\) given that the event \(A_{i}\) occurs. Then it can be shown that \(E(X)=\) \(\Sigma E\left(X \mid A_{i}\right) \cdot P\left(A_{i}\right)\), the weighted average of the individual "conditional expectations" where the weights are the probabilities of the partitioning events. a. The expected duration of a voice call to a particular telephone number is 3 minutes, whereas the expected duration of a data call to that same number is 1 minute. If \(75 \%\) of all calls are voice calls, what is the expected duration of the next call? b. A deli sells three different types of chocolate chip cookies. The number of chocolate chips in a type \(i\) cookie has a Poisson distribution with parameter \(\lambda_{i}=i+1\) \((i=1,2,3)\). If \(20 \%\) of all customers purchasing a chocolate chip cookie select the first type, \(50 \%\) choose the second type, and the remaining \(30 \%\) opt for the third type, what is the expected number of chips in a cookie purchased by the next customer?

a. Show that \(b(x ; n, 1-p)=b(n-x ; n, p)\). b. Show that \(B(x ; n, 1-p)=1-B(n-x-1 ; n, p)\). [Hint: At most \(x S\) 's is equivalent to at least \((n-x) F\) 's.] c. What do parts (a) and (b) imply about the necessity of including values of \(p\) greater than \(.5\) in Appendix Table A.1?

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