91Ó°ÊÓ

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 $$\text { Standard deviation }=\sqrt{\frac{N-n}{N-1}} \frac{\sigma}{\sqrt{n}}$$ 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 \(N=30,000\), show that the standard deviation equals \(0.995 \sigma / \sqrt{n}\). (When \(n\) is small compared to the population size \(N\), the approximate formula works very well.) b. If \(n=N\) (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.

Step by step solution

01

Calculate the finite population correction for part (a)

First, calculate the finite population correction term for part (a). We use the formula \( \sqrt{\frac{N-n}{N-1}} \) with \( N = 30,000 \) and \( n = 300 \):\[\sqrt{\frac{30,000 - 300}{30,000 - 1}} = \sqrt{\frac{29,700}{29,999}}.\]

02

Simplify and approximate the correction for part (a)

Calculate the approximate value:\[\sqrt{\frac{29,700}{29,999}} \approx 0.995. \]Thus, the standard deviation becomes \( 0.995 \times \frac{\sigma}{\sqrt{n}} \).

03

Calculate the finite population correction for part (b)

For part (b), when \( n = N \) (i.e., the sample is the whole population), the finite population correction becomes:\[\sqrt{\frac{N-N}{N-1}} = \sqrt{\frac{0}{N-1}} = \sqrt{0} = 0.\]

04

Conclusion for part (b)

Since the correction factor is \( 0 \), the exact standard deviation becomes \( 0 \times \frac{\sigma}{\sqrt{n}} = 0 \). This confirms that there is no sampling error if the sample includes the entire population.

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.

Standard Deviation

Standard deviation is a measure of the spread of data points in a data set. It shows how much variation there is from the mean. In the context of samples, it helps in understanding the variation within sample means when samples are drawn from a population. For a finite population, the modification using the finite population correction makes it more precise.
The formula for the standard deviation of a sample mean approximates to \( \frac{\sigma}{\sqrt{n}} \). However, this assumes an infinite population size. The finite correction is applied as \( \sqrt{\frac{N-n}{N-1}} \). This term adjusts the calculated standard deviation to better reflect the reduced variability that actually occurs when dealing with a finite population.

Sample Size

Sample size, denoted by \( n \), is the number of observations in a sample. It plays a crucial role in determining the precision of sample estimates, such as the sample mean. Larger sample sizes generally lead to more accurate estimates of the population characteristics.
The sample size affects the calculation of the standard deviation of the sample mean \( \bar{x} \), as seen in the formula \( \sigma/\sqrt{n} \). A larger sample size reduces the standard error, making the estimate of the population parameter more reliable. When the sample size reaches the population size \( N \), the standard deviation becomes 0, eliminating sampling error.

Population Size

Population size, \( N \), refers to the total number of elements or individuals in the group being studied. It affects the accuracy of sample estimates, particularly when dealing with finite populations.
In the context of the finite population correction, a larger population size compared to the sample size implies that the standard deviation of the sample mean is very close to the theoretical approximation without the correction. As shown in the exercise, if \( N \) is significantly larger than \( n \) (e.g., 30,000 compared to 300), the correction factor approaches 1, meaning the simpler formula suffices.

Sampling Error

Sampling error is the difference between a sample statistic (like the sample mean) and the actual population parameter (like the population mean). It is attributed to the fact that only a portion of the population is being analyzed.
In the exercise, when the sample size equals the population size \( n = N \), the finite population correction factor becomes zero, resulting in a standard deviation of 0. This illustrates that sampling error disappears because the sample includes every individual in the population, meaning the sample mean exactly equals the population mean.