/*! 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 57 Finite populations The formula \... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Finite populations The formula \(\sigma / \sqrt{n}\) for the standard deviation of \(\bar{x}\) actually is an approximation that treats the population size as infinitely large relative to the sample size \(n\). The exact formula for a finite population size \(N\) is Standard deviation \(=\sqrt{\frac{N-n}{N-1}} \frac{\sigma}{\sqrt{n}}\) The term \(\sqrt{(N-n) /(N-1)}\) is called the finite population correction. a. When \(n=300\) students are selected from a college student body of size \(\mathrm{N}=30,000\), show that the standard deviation equals \(0.995 \sigma / \sqrt{n}\). (When \(n\) is small compared to the population size \(\mathrm{N},\) the approximate formula works very well.) b. If \(n=\mathrm{N}\) (that is, we sample the entire population), show that the standard deviation equals \(0 .\) In other words, no sampling error occurs, since \(\bar{x}=\mu\) in that case.

Short Answer

Expert verified
a. Standard deviation is \(0.995 \frac{\sigma}{\sqrt{n}}\). b. Standard deviation is 0, indicating no sampling error.

Step by step solution

01

Understanding the given problem

The problem involves finding the exact standard deviation of the sample mean \( \bar{x} \) when drawn from a finite population. We need to use the finite population correction (FPC) to solve it.
02

Identifying the correct formula

For a finite population size \( N \) and a sample size \( n \), the standard deviation of \( \bar{x} \) is given by \( \sqrt{\frac{N-n}{N-1}} \frac{\sigma}{\sqrt{n}} \). This is different from the approximation used for infinite populations.
03

Apply the formula for part (a)

Given: \( n = 300 \) and \( N = 30,000 \). Substitute into FPC: \( \sqrt{\frac{N-n}{N-1}} = \sqrt{\frac{30,000-300}{30,000-1}} = \sqrt{\frac{29,700}{29,999}} \). Compute the correction factor.
04

Calculate the finite population correction factor for part (a)

Calculate \( \sqrt{\frac{29,700}{29,999}} \). This approximates to \( \sqrt{1 - \frac{300}{30,000}} \). Since \( n \) is very small compared to \( N \), this value should be close to \( 1 \).
05

Conclusion for part (a)

The factor \( \sqrt{\frac{29,700}{29,999}} \) is approximately \( 0.995 \), so the standard deviation can be expressed as \( 0.995 \frac{\sigma}{\sqrt{n}} \), confirming the statement in part (a).
06

Apply the formula for part (b)

When \( n = N \), then \( \sqrt{\frac{N-n}{N-1}} = \sqrt{\frac{0}{N-1}} = \sqrt{0} = 0 \). This means that the standard deviation is \( 0 \), indicating no sampling error.

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Key Concepts

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

Standard Deviation
Standard deviation is a key concept in statistics, measuring how much individual data points differ from the sample mean. Think of it as a way to measure the spread or variability of data around the average value. When dealing with a finite population, the standard deviation becomes slightly more complex. In typical cases with large or infinite populations, you can estimate the standard deviation of a sample mean using the formula \( \sigma / \sqrt{n} \). Here, \( \sigma \) represents the standard deviation of the entire population, and \( n \) is the sample size. However, this assumes that the population is infinitely large, which often isn't the case in real-world scenarios.For finite populations, we adjust with the finite population correction factor. This adjusts our standard deviation formula to accurately reflect the limited size of the population. Without this correction, our estimates could be slightly off, especially when the sample size is a significant fraction of the population.
Sample Mean
The sample mean, often represented as \( \bar{x} \), is the average value of a set of observations drawn from a population. It's essentially a snapshot of the "average" behavior of a subset of that population. Calculating the sample mean is straightforward: add up all the sample values and divide by the number of values, like this:
  • Add all sample values together.
  • Divide this total by the number of values present in the sample.
For instance, if you have selected 300 students as a sample out of 30,000, the average behavior or characteristic of these 300 students is captured by the sample mean. The closer your sample matches the population in size and diversity, the closer your sample mean is to the true population mean (\( \mu \)).In sampling, especially with finite populations, the sample mean serves as an estimate of the population mean. If you sample the entire population (where \( n = N \)), your sample mean is exactly the same as the population mean. There's no error at all, as you've captured every member's data in the population.
Finite Population
A finite population refers to a population of a known, manageable size. Unlike infinite populations, where size is assumed to be so large that small samples have minimal impact, finite populations require special consideration for their smaller size.When working with a finite population:
  • The finite population correction (FPC) must often be applied to adjust calculations. This is because sampling a large portion of a finite population influences the calculated statistics more significantly.
  • The FPC term, \( \sqrt{\frac{N-n}{N-1}} \), adjusts the standard deviation, accounting for the finite size. As the sample size \( n \) approaches the population size \( N \), this correction becomes more pronounced.
  • If your entire population is sampled (i.e., \( n = N \)), the standard deviation of the sample mean becomes zero, as there's no variation between your sample mean and the population mean.
Understanding finite populations and the necessary corrections ensures more accurate statistical analyses, especially in fields where entire populations are considered, such as in small community studies or specific demographic research.

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Most popular questions from this chapter

Canada lottery In one lottery option in Canada (Source: Lottery Canada), you bet on a six-digit number between 000000 and \(999999 .\) For a \(\$ 1\) bet, you win \(\$ 100,000\) if you are correct. The mean and standard deviation of the probability distribution for the lottery winnings are \(\mu=0.10\) (that is, 10 cents) and \(\sigma=100.00\). Joe figures that if he plays enough times every day, eventually he will strike it rich, by the law of large numbers. Over the course of several years, he plays 1 million times. Let \(\bar{x}\) denote his average winnings. a. Find the mean and standard deviation of the sampling distribution of \(\bar{x}\). b. About how likely is it that Joe's average winnings exceed \(\$ 1,\) the amount he paid to play each time? Use the central limit theorem to find an approximate answer.

Sample versus sampling \(\quad\) Each student should bring 10 coins to class. For each coin, observe its age, the difference between the current year and the year on the coin. a. Using all the students' observations, the class should construct a histogram of the sample ages. What is its shape? b. Now each student should find the mean for that student's 10 coins, and the class should plot the means of all the students. What type of distribution is this, and how does it compare to the one in part a? What concepts does this exercise illustrate?

Sample = population Let \(X=\) GPA for students in your school. a. What would the sampling distribution of the sample mean look like if you sampled every student in the school, so the sample size equals the population size? (Hint: The sample mean then equals the population mean.) b. How does the sampling distribution compare to the population distribution if we take a sample of size \(n=1 ?\)

Basketball shooting In college basketball, a shot made from beyond a designated arc radiating about 20 feet from the basket is worth three points, instead of the usual two points given for shots made inside that arc. Over his career, University of Florida basketball player Lee Humphrey made \(45 \%\) of his three-point attempts. In one game in his final season, he made only 3 of 12 three-point shots, leading a TV basketball analyst to announce that Humphrey was in a shooting slump. a. Assuming Humphrey has a \(45 \%\) chance of making any particular three-point shot, find the mean and standard deviation of the sampling distribution of the proportion of three-point shots he will make out of 12 shots. b. How many standard deviations from the mean is this game's result of making 3 of 12 three-point shots? c. If Humphrey was actually not in a slump but still had a \(45 \%\) chance of making any particular three-point shot, explain why it would not be especially surprising for him to make only 3 of 12 shots. Thus, this is not really evidence of a shooting slump.

Income of farm workers For the population of farm workers in New Zealand, suppose that weekly income has a distribution that is skewed to the right with a mean of \(\mu=\$ 500(\mathrm{~N} . Z .)\) and a standard deviation of \(\sigma=\$ 160 .\) A researcher, unaware of these values, plans to randomly sample 100 farm workers and use the sample mean annual income \(\bar{x}\) to estimate \(\mu\). a. Show that the standard deviation of \(\bar{x}\) equals 16.0 . b. Explain why it is almost certain that the sample mean will fall within \(\$ 48\) of \(\$ 500\). c. The sampling distribution of \(\bar{x}\) provides the probability that \(\bar{x}\) falls within a certain distance of \(\mu,\) regardless of the value of \(\mu\). Show how to calculate the probability that \(\bar{x}\) falls within \(\$ 20\) of \(\mu\) for all such workers. (Hint: Using the standard deviation, convert the distance 20 to a \(z\) -score for the sampling distribution.)

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