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Chapter 7: Problem 30

To estimate the mean age for a population of 4000 employees, a simple random sample of 40 employees is selected. a. Would you use the finite population correction factor in calculating the standard error of the mean? Explain. b. If the population standard deviation is \(\sigma=8.2\) years, compute the standard error both with and without the finite population correction factor. What is the rationale for ignoring the finite population correction factor whenever \(n / N \leq .05 ?\) c. What is the probability that the sample mean age of the employees will be within 卤2 years of the population mean age?

Short Answer

Expert verified
a) No, because \(n/N = 0.01\) is less than 0.05. b) \(SE \approx 1.296\) without FPC; \(\approx 1.295\) with FPC. c) Probability is approximately 87.6%.

Step by step solution

01

Check the Need for Finite Population Correction

The finite population correction factor is used in the calculation of the standard error when the sample size is a significant fraction of the population. Specifically, it is commonly applied when the sample size \(n\) over the population size \(N\) satisfies \(n/N > 0.05\). In this exercise, \(n = 40\) and \(N = 4000\), so \(n / N = 0.01\). Since 0.01 is not greater than 0.05, the finite correction factor is generally not necessary.
02

Calculate Standard Error Without Finite Population Correction

The standard error of the mean without the finite population correction factor is calculated using the formula: \(SE = \frac{\sigma}{\sqrt{n}}\). Here, \(\sigma = 8.2\) and \(n = 40\). Therefore, the standard error is \(SE = \frac{8.2}{\sqrt{40}} \approx 1.296\).
03

Calculate Standard Error With Finite Population Correction

The standard error with the finite population correction factor is calculated by \(SE_{FPC} = \frac{\sigma}{\sqrt{n}} \times \sqrt{\frac{N-n}{N-1}}\). Plugging in the values: \(SE_{FPC} = \frac{8.2}{\sqrt{40}} \times \sqrt{\frac{4000-40}{4000-1}} \approx 1.295\). The difference here is minimal.
04

Explain Rationale for Ignoring Finite Population Correction

The rationale for ignoring the finite population correction factor when \(n/N \leq 0.05\) is because the impact on the standard error is negligible. This simplifies calculations without significantly affecting accuracy.
05

Compute Probability of Sample Mean within 卤2 Years

To find the probability that the sample mean is within 卤2 years of the true population mean, we assume a normal distribution. The standard normal variable is: \(Z = \frac{X - \mu}{SE}\). Here, \(SE \approx 1.296\). \(Z\) values for 卤2 years will be: \(Z = \frac{2}{1.296} \approx 卤1.544\). Using a Z-table, the probability of \(Z\) between -1.544 and 1.544 is approximately 0.876 (0.437 on each side from the mean), or 87.6%.

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Key Concepts

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

Understanding the Standard Error in Sampling
Standard error is a concept essential to understanding the precision of a sample statistic, primarily the sample mean. Imagine you pick a sample from a larger population to learn about, say, average age: this sampled group must represent the larger population accurately.
Here鈥檚 how it works:
  • **Standard Error (SE)**: It shows how much variance there is likely to be in a sample's mean compared to the actual population mean.
  • **Formula**: For any given sample, the standard error of the mean is calculated by \(SE = \frac{\sigma}{\sqrt{n}}\). Here, \(\sigma\) is the population standard deviation, and \(n\) is the sample size.
  • **Purpose**: A smaller SE means that the sample mean is closer to the population mean, indicating a more accurate representation.
In our exercise, the standard error for a sample of 40 employees from 4000 is calculated as about 1.296, assuming normal distribution. This SE suggests that the sample mean should closely reflect the population mean, within a predictable deviation.
Finite Population Correction and Its Usage
Finite population correction (FPC) becomes relevant when working with a sample size that's a substantial part of the entire population. Here's the essence:
  • **Concept**: If the sample size \(n\) is significant in relation to the total population \(N\), FPC is required to adjust the standard error appropriately.
  • **Applicability**: It's generally applied when \(\frac{n}{N} > 0.05\), meaning more than 5% of the total population is included in the sample.
  • **Formula**: Introduces an additional factor in SE calculation: \(SE_{FPC} = \frac{\sigma}{\sqrt{n}} \times \sqrt{\frac{N-n}{N-1}}\).
  • **Why Ignore**: In cases where the sample covers 5% or less of the population, the correction has a negligible effect and doesn鈥檛 materially influence the error margin.
For the sample of 40 from 4000 employees, the ratio \(\frac{40}{4000} = 0.01\) is much less than 0.05, so FPC is not necessary. However, when used, the difference is minimal: 1.296 versus 1.295. This supports omitting FPC for simplicity without losing precision.
The Concept of Probability Distribution in Statistics
Probability distribution is a cornerstone of statistics, helping to predict how likely outcomes are, such as the average age in a sample. In simple terms:
  • **Definition**: A mathematical function that provides the probabilities of occurrence of different possible outcomes in an experiment.
  • **Normal Distribution**: When talking about sample means, they鈥檙e typically assumed to fit a bell-shaped curve, which means they follow a normal distribution.
  • **Application**: In our example, we're interested in the probability of the sample mean being within a certain range of the population mean.
  • **Example**: To find the probability of the sample mean lying within 卤2 years of the mean age, we calculate the standard normal variable: \(Z = \frac{X - \mu}{SE}\). Here, \(Z\) values give us a measure of deviation from the mean.
The calculation shows that the chance of the average age being within 2 years of the true mean is around 87.6%, suggesting a high concentration of data points in that specified range. This is vital for making informed statistical inferences about populations.

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Most popular questions from this chapter

Assume a finite population has 350 elements. Using the last three digits of each of the following five-digit random numbers (e.g.; \(601,022,448, \ldots),\) determine the first four elements that will be selected for the simple random sample. \\[ \begin{array}{lllllll} 98601 & 73022 & 83448 & 02147 & 34229 & 27553 & 84147 & 93289 & 14209 \end{array} \\]

Americans have become increasingly concerned about the rising cost of Medicare. In \(1990,\) the average annual Medicare spending per enrollee was \(\$ 3267 ;\) in \(2003,\) the average annual Medicare spending per enrollee was \(\$ 6883\) (Money, Fall 2003 ). Suppose you hired a consulting firm to take a sample of fifty 2003 Medicare enrollees to further investigate the nature of expenditures. Assume the population standard deviation for 2003 was \(\$ 2000\). a. Show the sampling distribution of the mean amount of Medicare spending for a sample of fifty 2003 enrollees. b. What is the probability the sample mean will be within \(\pm \$ 300\) of the population mean? c. What is the probability the sample mean will be greater than \(\$ 7500 ?\) If the consulting firm tells you the sample mean for the Medicare enrollees they interviewed was \(\$ 7500,\) would you question whether they followed correct simple random sampling procedures? Why or why not?

Suppose a simple random sample of size 50 is selected from a population with \(\sigma=10\) Find the value of the standard error of the mean in each of the following cases (use the finite population correction factor if appropriate). a. The population size is infinite. b. The population size is \(N=50,000\). c. The population size is \(N=5000\). d. The population size is \(N=500\).

The president of Doerman Distributors, Inc., believes that \(30 \%\) of the firm's orders come from first-time customers. A simple random sample of 100 orders will be used to estimate the proportion of first-time customers. a. Assume that the president is correct and \(p=.30 .\) What is the sampling distribution of \(\bar{p}\) for this study? b. What is the probability that the sample proportion \(\bar{p}\) will be between .20 and \(.40 ?\) c. What is the probability that the sample proportion will be between .25 and \(.35 ?\)

The College Board American College Testing Program reported a population mean SAT score of \(\mu=1020\) (The World Almanac 2003). Assume that the population standard deviation is \(\sigma=100\) a. What is the probability that a random sample of 75 students will provide a sample mean SAT score within 10 of the population mean? b. What is the probability a random sample of 75 students will provide a sample mean SAT score within 20 of the population mean?
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