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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 ±2 years of the population mean age?

Short Answer

Expert verified
a) No, the correction factor is not needed because \(n/N = 0.01 \leq 0.05\). b) With correction: 1.30, Without: 1.30 (minimal difference). c) Probability is 83.84%.

Step by step solution

01

Determine if Finite Population Correction is Needed

First, check the ratio of the sample size to the population size. The correction factor is used if this ratio is larger than 0.05.The sample size \(n = 40\) and the population size \(N = 4000\). Compute \(\frac{n}{N} = \frac{40}{4000} = 0.01\).Since \(0.01 \leq 0.05\), we do not need to apply the finite population correction factor.
02

Calculate Standard Error Without Correction Factor

The standard error of the mean without the correction factor is calculated using the formula:\[ SE = \frac{\sigma}{\sqrt{n}} \] Given \(\sigma = 8.2\) and \(n = 40\), we calculate:\[ SE = \frac{8.2}{\sqrt{40}} = \frac{8.2}{6.32} \approx 1.30 \]
03

Calculate Standard Error With Correction Factor

Although we established it is not necessary, we will calculate it for thoroughness. The formula incorporating the correction is:\[ SE = \frac{\sigma}{\sqrt{n}} \times \sqrt{\frac{N-n}{N-1}} \]Substitute \(\sigma = 8.2\), \(n = 40\), and \(N = 4000\):\[ SE = \frac{8.2}{\sqrt{40}} \times \sqrt{\frac{4000-40}{4000-1}} \approx 1.30 \times 0.997 = 1.299 \approx 1.30 \] Notice there is negligible difference because \(n/N\) is small.
04

Explain Rationale for Ignoring Correction Factor

When \(\frac{n}{N} \leq 0.05\), the finite population correction factor makes a very minimal impact on the calculation of standard error, simplifying analysis without significantly affecting accuracy.
05

Calculate Probability of Sample Mean Within ±2 Years

Use the standard error calculated without the correction factor to find the probability that the sample mean is within ±2 of the population mean.Calculate the z-scores for ±2:\[ z = \frac{\pm 2}{SE} = \frac{\pm 2}{1.30} \approx \pm 1.54 \]Use a standard normal distribution table to find the probability associated with \(z = 1.54\), which is approximately 0.9192.The probability that the sample mean is within ±2 years is:\[ P(-1.54 < Z < 1.54) = 0.9192 - (1 - 0.9192) \approx 0.8384 \]Thus, the probability is approximately 83.84%.

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Key Concepts

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

Finite Population Correction
When dealing with statistical sampling, it's important to understand when to apply the finite population correction (FPC). This correction factor becomes relevant in calculating the standard error of a sample mean when you're dealing with a relatively small population.
The goal of the finite population correction is to adjust the standard error to account for the fact that as the sample size becomes a more significant fraction of the total population, the sample information becomes more reliable. **When is FPC used?**
  • The finite population correction is necessary when the ratio of the sample size to the population size, denoted as \( \frac{n}{N} \), exceeds 0.05, or 5%.
  • In cases where the sample is less than or equal to 5% of the population, the impact of FPC is minimal, and its effect on the standard error is negligible.
By adjusting calculations only when \( n/N > 0.05 \), statisticians can maintain a balance between complexity and accuracy. In many real-world applications, ignoring FPC for small ratios leads to only slight differences, making it a practical choice.
Standard Error
The standard error (SE) is a critical measure in statistics that reflects how much a sample mean is expected to vary from the actual population mean. It's essentially the standard deviation of the sampling distribution of the sample mean, providing an indication of the precision of the sample mean as an estimate of the population mean. **How to Calculate SE?**
  • Without finite population correction, the formula for standard error is: \[SE = \frac{\sigma}{\sqrt{n}} \]where \( \sigma \) is the population standard deviation, and \( n \) is the sample size.
  • With finite population correction, the formula becomes: \[SE = \frac{\sigma}{\sqrt{n}} \times \sqrt{\frac{N-n}{N-1}} \]where \( N \) is the population size.
Essentially, incorporating finite population correction slightly reduces the standard error when \( n/N \) is larger, tightening the spread of the sampling distribution.
This correction, however, is usually ignored when \( n/N \leq 0.05 \) due to its minimal impact. This approach simplifies calculations without compromising statistical accuracy in most practical scenarios.
Z-scores
Z-scores are a fundamental concept in statistics for understanding how much a given data point deviates from the mean of a distribution, measured in units of standard deviation. They provide a means of comparing different data points across varied distributions by standardizing results.**Understanding Z-scores: How do they work?**
  • A z-score expresses the number of standard deviations a sample mean is from the population mean.
  • The formula for calculating a z-score is:\[z = \frac{X - \mu}{SE}\]where \( X \) is the sample mean, \( \mu \) is the population mean, and \( SE \) is the standard error.
  • Once calculated, z-scores can be used to determine probabilities by referencing a standard normal distribution table.
By using z-scores, you can determine probabilities and make informed decisions regarding the likelihood of sample outcomes. For example, in the exercise, the probability that the sample mean age falls within ±2 years of the population mean was found using z-scores.
This ability to translate raw data into probabilities makes z-scores an invaluable tool for statistical analysis.

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Most popular questions from this chapter

The College Board reported the following mean scores for the three parts of the Scholastic Aptitude Test (SAT) (The World Almanac, 2009): $$\begin{array}{ll} \text { Critical Reading } & 502 \\ \text { Mathematics } & 515 \\ \text { Writing } & 494 \end{array}$$ Assume that the population standard deviation on each part of the test is \(\sigma=100\) a. What is the probability that a random sample of 90 test takers will provide a sample mean test score within 10 points of the population mean of 502 on the Critical Reading part of the test? b. What is the probability that a random sample of 90 test takers will provide a sample mean test score within 10 points of the population mean of 515 on the Mathematics part of the test? Compare this probability to the value computed in part (a). c. What is the probability that a random sample of 100 test takers will provide a sample mean test score within 10 of the population mean of 494 on the writing part of the test? Comment on the differences between this probability and the values computed in parts (a) and (b).

After deducting grants based on need, the average cost to attend the University of Southern California (USC) is \(\$ 27,175\) (U.S. News \& World Report, America 's Best Colleges, \(2009 \text { ed. }) .\) Assume the population standard deviation is \(\$ 7400 .\) Suppose that a random sample of 60 USC students will be taken from this population. a. What is the value of the standard error of the mean? b. What is the probability that the sample mean will be more than \(\$ 27,175 ?\) c. What is the probability that the sample mean will be within \(\$ 1000\) of the population mean? d. How would the probability in part (c) change if the sample size were increased to \(100 ?\)

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