/*! 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 63 A city commissioner claims that ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A city commissioner claims that \(80 \%\) of all people in the city favor garbage collection by contract to a private concern (in contrast to collection by city employees). To check the theory that the proportion of people in the city favoring private collection is .8 , you randomly sample 25 people and find that \(x\), the number of people who support the commissioner's claim, is \(22 .\) a. What is the probability of observing at least 22 who support the commissioner's claim if, in fact, \(p=.8 ?\) b. What is the probability that \(x\) is exactly equal to \(22 ?\) c. Based on the results of part a, what would you conclude about the claim that \(80 \%\) of all people in the city favor private collection? Explain.

Short Answer

Expert verified
If these probabilities are high, can we conclude that the claim is reasonable? Answer: The probabilities for observing at least 22 and exactly 22 people supporting the city commissioner's claim can be calculated using the binomial probability distribution. If the probabilities are high (e.g., higher than a predefined threshold like 0.05), we can conclude that the claim is reasonable. Otherwise, we cannot support the commissioner's claim that 80% of the people in the city favor private collection.

Step by step solution

01

Identify the variables

We are given the following variables: - The proportion of people in favor of private garbage collection (p): 0.8 - Sample size (n): 25 - Number of people supporting the commissioner's claim (x): 22
02

Calculate the probability for observing at least 22 supporters using the binomial distribution

Using the binomial distribution formula, we can determine the probability of observing at least 22 people supporting the commissioner's claim: \(P(x \geq 22) = P(x=22) + P(x=23) + P(x=24) + P(x=25)\) For each term, we use the binomial distribution formula: \(P(x=k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k}\) Calculating the probabilities for x from 22 to 25: \(P(x=22) = \binom{25}{22} \cdot 0.8^{22} \cdot 0.2^{3}\) \(P(x=23) = \binom{25}{23} \cdot 0.8^{23} \cdot 0.2^{2}\) \(P(x=24) = \binom{25}{24} \cdot 0.8^{24} \cdot 0.2^{1}\) \(P(x=25) = \binom{25}{25} \cdot 0.8^{25} \cdot 0.2^{0}\) Adding up the probabilities, we get the probability of observing at least 22 supporters.
03

Calculate the probability for observing exactly 22 supporters

Using the binomial distribution formula, we calculate the probability of observing exactly 22 people who support the commissioner's claim: \(P(x=22) = \binom{25}{22} \cdot 0.8^{22} \cdot 0.2^{3}\)
04

Provide conclusions based on the results of part a

Based on the probability of observing at least 22 people supporting the commissioner's claim given the proportion of favoring private collection is 0.8, we can judge if the claim is reasonable or not: - If the probability of observing at least 22 supporters is high (e.g., higher than a predefined threshold like 0.05), then we can conclude that the claim is reasonable. - Otherwise, if the probability is low (e.g., lower than the threshold), we can't support the commissioner's claim that 80% of the people in the city favor private collection.

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
To solve problems related to real-world events like the commissioner's claim, we often use probability calculations. In this exercise, we're dealing with a _binomial distribution_.
The binomial distribution is ideal for situations where there are two possible outcomes, like support or no support. We can specify a probability of an event occurring, noted as \(p\). In this case, \(p = 0.8\), indicating that the claimed proportion of the city's support for private garbage collection is 80%.
The first step is to determine the probability of observing at least 22 out of 25 supporters. To find this, we sum up the probabilities of observing exactly 22, 23, 24, and 25 supporters using:
  • \( P(x = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k} \)
where \( \binom{n}{k} \) is a binomial coefficient, showing the number of ways to choose \(k\) successes from \(n\) trials. By computing and adding these probabilities, we get the overall probability for the desired event.
Statistical Inference
Statistical inference helps us make conclusions about a population based on sample data. In this scenario, we're using the sample of 25 people to infer about the entire city's opinion on private garbage collection.
In our exercise, we aim to determine if the observed data, where 22 out of 25 favor the proposal, aligns with the commissioner's claim of 80% support. Through probability calculation, a core aspect of statistical inference, we can evaluate how probable it is to observe such results if the true proportion is 0.8.
The calculated probability gives us an idea of how likely the observed sample results would occur under given assumptions.
  • If the probability for the sample matches or exceeds a certain threshold (e.g., 0.05), it suggests the sample is consistent with the population parameter p = 0.8.
  • If it's less, we may infer that the claim may not be accurate.
Thus, statistical inference enables us to make educated assumptions or reject claims about a larger group based on sample data.
Hypothesis Testing
Hypothesis Testing is a systematic method used to evaluate claims or hypotheses about a population. In our example, the hypothesis in question is that 80% of the city supports privatization of garbage collection.
Here's how hypothesis testing works in this context:
  • **Null Hypothesis \((H_0)\)**: The proportion of support is 80%, or \( p = 0.8 \).
  • **Alternative Hypothesis \((H_a)\)**: The proportion of support is not 80%, \( p eq 0.8 \).
We then use our sample data and calculate the probability of observing a result as extreme as, or more extreme than, the actual sample (in this case, 22 supporters). This probability is central in determining if our null hypothesis can hold.
If the calculated probability is less than a predetermined significance level (often 0.05), there is sufficient evidence to reject the null hypothesis, suggesting the population proportion differs from 80%.
Conversely, if the probability is higher, we fail to reject the null hypothesis, implying there's not enough evidence to dispute the commissioner's claim.

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

In southern California, a growing number of persons pursuing a teaching credential are choosing paid internships over traditional student teaching programs. A group of eight candidates for three local teaching positions consisted of five candidates who had enrolled in paid internships and three candidates who had enrolled in traditional student teaching programs. Let us assume that all eight candidates are equally qualified for the positions. Let \(x\) represent the number of internship- trained candidates who are hired for these three positions. a. Does \(x\) have a binomial distribution or a hypergeometric distribution? Support your answer. b. Find the probability that three internship-trained candidates are hired for these positions. c. What is the probability that none of the three hired was internship- trained? d. Find \(P(x \leq 1)\).

In \(2006,\) the average combined SAT score (reading \(+\) verbal \(+\) writing) for collegebound students in the United States was 1518 (out of 2400). Suppose that approximately \(45 \%\) of all high school graduates took this test, and that 100 high school graduates are randomly selected from throughout the United States. \({ }^{1}\) Which of the following random variables has an approximate binomial distribution? If possible, give the values for \(n\) and \(p\). a. The number of students who took the SAT b. The scores of the 100 students on the SAT

Security Systems A home security system is designed to have a \(99 \%\) reliability rate. Suppose that nine homes equipped with this system experience an attempted burglary. Find the probabilities of these events: a. At least one of the alarms is triggered. b. More than seven of the alarms are triggered. c. Eight or fewer alarms are triggered.

According to the Humane Society of the United States, there are approximately 65 million owned dogs in the United States, and approximately \(40 \%\) of all U.S. households own at least one dog. \({ }^{4}\) Suppose that the \(40 \%\) figure is correct and that 15 households are randomly selected for a pet ownership survey. a. What is the probability that exactly eight of the households have at least one dog? b. What is the probability that at most four of the households have at least one dog? c. What is the probability that more than 10 households have at least one dog?

Let \(x\) be a binomial random variable with \(n=7\), \(p=.3 .\) Find these values: a. \(P(x=4)\) b. \(P(x \leq 1)\) c. \(P(x>1)\) d. \(\mu=n p\) e. \(\sigma=\sqrt{n p q}\)

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.