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6.89 Suppose that \(20 \%\) of the 10,000 signatures on a certain recall petition are invalid. Would the number of invalid signatures in a sample of size 2000 have (approximately) a binomial distribution? Explain.

Short Answer

Expert verified
In conclusion, the number of invalid signatures in a sample of size 2000 would approximately follow a binomial distribution. This is because, although the trials are not completely independent (since sampling is done without replacement), the sample size is small enough compared to the total population for it to have an approximately negligible effect, and the other conditions for the binomial distribution are satisfied.

Step by step solution

01

Understanding the Problem

We have been provided with the information that 20% (2,000) of the 10,000 signatures on a recall petition are invalid. We want to see if the number of invalid signatures in a sample of size 2000 follows a binomial distribution.
02

Recall Properties of Binomial Distribution

A random variable follows a binomial distribution if the following conditions are met: 1. The number of trials (n) is fixed. 2. Each trial has only two possible outcomes: success (in this case, an invalid signature) and failure (a valid signature). 3. The probability of success (p) is constant for each trial. 4. The trials are independent of each other.
03

Checking the Conditions

Now, let's check if the given scenario satisfies each of these conditions: 1. In the given problem, the number of trials is fixed at n = 2000 signatures. 2. There are only two outcomes: the signature is either invalid (success) or valid (failure). 3. The probability of success (p) is 20%, and the probability of failure (q = 1 - p) is 80%. These probabilities remain constant throughout the trials. 4. Since we are sampling without replacement (as we are not putting back the chosen signature and re-selecting it), the trials are not completely independent. However, the sample size is small compared to the total number of signatures (2000 out of 10,000), and we can apply the Finite Population Correction (FPC) factor and approximate the distribution as a binomial distribution.
04

Conclusion

In conclusion, the number of invalid signatures in a sample of size 2000 would approximately follow a binomial distribution. This is because, although the trials are not completely independent (since sampling is done without replacement), the sample size is small enough compared to the total population for it to have an approximately negligible effect, and the other conditions for the binomial distribution are satisfied.

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Key Concepts

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

Sampling
Sampling is a fundamental concept in statistics. It involves selecting a subset, or sample, from a larger group, known as the population. Here, the population consists of the 10,000 signatures, while we take a sample of 2,000 signatures for further analysis.

This method is vital for gaining information when dealing with large datasets. Sampling allows us to make inferences about the entire population without examining every single item. This saves both time and resources.

There are several methods of sampling, such as:
  • Random Sampling: Every member of the population has an equal chance of being selected.
  • Systematic Sampling: Members are selected at regular intervals.
  • Stratified Sampling: The population is divided into different subgroups, and samples are taken from each.
In the exercise, the sample of 2000 signatures involves random selection, which helps in maintaining the randomness required for statistical analysis.
Probability
Probability is the study of uncertainty and chance, which is essential in determining the likelihood of various outcomes. In our context, we are interested in the probability of selecting an invalid signature from the sample.

The probability is expressed as a number between 0 and 1, where 0 implies an impossible event and 1 indicates certainty. For this exercise:
  • The probability of success (invalid signature) is 20% or 0.2.
  • The probability of failure (valid signature) is 80% or 0.8.
Using these probabilities, we can predict the number of invalid signatures in the sample of 2,000.

Understanding probability is crucial for determining the expected outcomes and making decisions based on statistical data. It enables us to evaluate the feasibility of certain hypotheses, like the assumption of a binomial distribution in this case.
Finite Population Correction
The Finite Population Correction (FPC) is a useful adjustment made when sampling from a finite population without replacement. It helps in ensuring more accurate results when the sample size is not negligible compared to the population size.

In our example with the 10,000 signatures, we sample 2,000. While this number might not be very small, it is still only a fifth of the total, which keeps the sampling influence relatively minor. However, technically, such sampling without replacement makes the trials less than perfectly independent.

Normally, to account for this lack of independence, we apply the FPC, which slightly adjusts our calculations compared to cases involving infinite populations, or where the sample size is very small. The correction factor is given by:\[ \text{FPC} = \sqrt{\frac{N-n}{N-1}} \] where \( N \) is the population size, and \( n \) is the sample size.

Applying FPC allows the approximation of the distribution as binomial, as it ensures that small deviations due to sampling without replacement do not significantly affect the analysis.

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Most popular questions from this chapter

Suppose that fund-raisers at a university call recent graduates to request donations for campus outreach programs. They report the following information for last year's graduates: $$\begin{array}{lllll}\text { Size of donation } & \$ 0 & \$ 10 & \$ 25 & \$\end{array}$$ 0.30 0 Proportion of calls 0.45 .20 0.05 Three attempts were made to contact each graduate. A donation of $$\$ 0$$ was recorded both for those who were contacted but declined to make a donation and for those who were not reached in three attempts. Consider the variable \(x=\) amount of donation for a person selected at random from the population of last year's graduates of this university. a. Write a few sentences describing what donation amounts you would expect to see if the value of \(x\) was observed for each of 1000 graduates. b. What is the most common value of \(x\) in this population? c. What is \(P(x \geq 25)\) ? d. What is \(P(x>0)\) ?

A business has six customer service telephone lines. Consider the random variable \(x=\) number of lines in use at a randomly selected time. Suppose that the probability distribution of \(x\) is as follows: \(\begin{array}{lccccccc}x & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ p(x) & 0.10 & 0.15 & 0.20 & 0.25 & 0.20 & 0.06 & 0.04\end{array}\) a. Calculate the mean value and standard deviation of \(x\). b. What is the probability that the number of lines in use is farther than 3 standard deviations from the mean value?

Consider the variable \(x=\) time required for a college student to complete a standardized exam. Suppose that for the population of students at a particular university, the distribution of \(x\) is well approximated by a normal curve with mean 45 minutes and standard deviation 5 minutes. a. If 50 minutes is allowed for the exam, what proportion of students at this university would be unable to finish in the allotted time? b. How much time should be allowed for the exam if you wanted \(90 \%\) of the students taking the test to be able to finish in the allotted time? c. How much time is required for the fastest \(25 \%\) of all students to complete the exam?

Industrial quality control programs often include inspection of incoming materials from suppliers. If parts are purchased in large lots, a typical plan might be to select 20 parts at random from a lot and inspect them. Suppose that a lot is judged acceptable if one or fewer of these 20 parts are defective. If more than one part is defective, the lot is rejected and returned to the supplier. Find the probability of accepting lots that have each of the following (Hint: Identify success with a defective part.): a. \(5 \%\) defective parts b. \(10 \%\) defective parts c. \(20 \%\) defective parts

A coin is flipped 25 times. Let \(x\) be the number of flips that result in heads (H). Consider the following rule for deciding whether or not the coin is fair: Judge the coin fair if \(8 \leq x \leq 17\). Judge the coin biased if either \(x \leq 7\) or \(x \geq 18\). a. What is the probability of judging the coin biased when it is actually fair? b. Suppose that a coin is not fair and that \(P(\mathrm{H})=0.9\) What is the probability that this coin would be judged fair? What is the probability of judging a coin fair if \(P(\mathrm{H})=0.1 ?\) c. What is the probability of judging a coin fair if \(P(\mathrm{H})=0.6 ?\) if \(P(\mathrm{H})=0.4 ?\) Why are these probabilities large compared to the probabilities in Part (b)? d. What happens to the "error probabilities" of Parts (a) and \((b)\) if the decision rule is changed so that the coin is judged fair if \(7 \leq x \leq 18\) and unfair otherwise? Is this a better rule than the one first proposed? Explain.

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