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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 \(N=500\).

Short Answer

Expert verified
a) 1.414, b) 1.414, c) 1.407, d) 1.342

Step by step solution

01

Understanding the Standard Error

The standard error (SE) of the mean is calculated using the formula \( SE = \frac{\sigma}{\sqrt{n}} \), where \( \sigma \) is the population standard deviation and \( n \) is the sample size. This formula applies when the population size is infinite or large enough to ignore the finite population correction factor.
02

Compute Standard Error for Infinite Population

For an infinite population size, the formula for standard error is directly applied: \( SE = \frac{10}{\sqrt{50}} = \frac{10}{7.071} \approx 1.414 \). Thus, the standard error is approximately 1.414.
03

Applying Finite Population Correction Factor

When the population size is finite and small relative to the sample size, we use the finite population correction (FPC) factor: \( FPC = \sqrt{\frac{N-n}{N-1}} \). This adjustment is used to refine the standard error estimation.
04

Compute Standard Error for Population Size 50,000

For \(N = 50,000\), the correction factor is \(FPC = \sqrt{\frac{50000 - 50}{49999}} = \sqrt{0.999} \approx 0.9995\). Thus, the adjusted standard error is \(SE \approx 1.414 \times 0.9995 = 1.414\). The effect of the correction is negligible for such a large population.
05

Compute Standard Error for Population Size 5000

For \(N = 5000\), the correction factor is \(FPC = \sqrt{\frac{5000 - 50}{4999}} = \sqrt{0.990} \approx 0.995 \). Thus, the adjusted standard error is \(SE \approx 1.414 \times 0.995 = 1.407\). The corrected standard error is approximately 1.407.
06

Compute Standard Error for Population Size 500

For \(N = 500\), the correction factor is \(FPC = \sqrt{\frac{500 - 50}{499}} = \sqrt{0.902} \approx 0.949 \). Thus, the adjusted standard error is \(SE \approx 1.414 \times 0.949 = 1.342\). The corrected standard error is approximately 1.342.

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Key Concepts

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

Finite Population Correction
When working with samples that are quite large compared to a finite population, we apply the **Finite Population Correction (FPC)** factor. This is crucial to obtain a more accurate measure of the standard error of the mean (SE). The traditional calculation of SE assumes an infinite population:
  • Formula for Standard Error: \( SE = \frac{\sigma}{\sqrt{n}} \)
  • Where \( n \) is the sample size and \( \sigma \) is the population standard deviation.
When a population is finite, say you're drawing a large sample from a smaller group, the potential overlap needs correction to ensure accuracy.
  • Finite Population Correction Factor: \( FPC = \sqrt{\frac{N-n}{N-1}} \)
  • Where \( N \) is the population size.
By multiplying SE with the FPC, the corrected standard error refines our understanding. This adjustment becomes significant when the sample size is 5% or more of the population size. For example, in our original exercise, the effect of FPC is small when \( N = 50,000 \), but more noticeable when \( N = 500 \). Understanding when and how much to correct helps in achieving precise statistical estimates.
Random Sampling
**Random sampling** is the cornerstone of statistical accuracy, ensuring that every member of a population has an equal chance of being chosen. It forms the backbone of obtaining reliable and unbiased data.
  • Ensures every individual has equal selection chances.
  • Reduces bias, leading to more representative samples.
  • Improves the reliability of conclusions drawn from data.
In conducting random sampling, researchers employ techniques such as simple random sampling where each subject is chosen randomly, without any order or system. This process minimizes selection bias, where the randomness highlights the random variation that naturally occurs. For the exercise, simple random sampling of size 50 ensures that each possible sample combination of the population (say 50,000 or 500) has equal opportunity to be selected. This principle supports the unbiased calculation of the standard error with or without the finite population correction.
Population Standard Deviation
The **Population Standard Deviation (\( \sigma \))** quantifies the variability or spread of a set of data in a population. A critical element in determining the standard error of the mean, standard deviation reflects how much the elements of the population differ from the mean.
  • \( \sigma \) is crucial for estimating SE since SE = \( \frac{\sigma}{\sqrt{n}} \).
  • Represents the root mean square deviation of each data point from the mean.
  • Key for determining the degree of spread in population data.
This measure provides insight into the 'normal' deviation within a data set and is essential when calculating standard error, as seen in the term \( SE = \frac{\sigma}{\sqrt{n}} \). For our specific problem, where \( \sigma = 10 \), it assists in determining how reliable the sample mean is as an estimate of the population mean. Changes in \( \sigma \) affect the size of the SE, emphasizing the accuracy and reliability of statistical conclusions.

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Most popular questions from this chapter

The average price of a gallon of unleaded regular gasoline was reported to be \(\$ 2.34\) in northern Kentucky (The Cincinnati Enquirer, January 21,2006 ). Use this price as the population mean, and assume the population standard deviation is \(\$ .20\) a. What is the probability that the mean price for a sample of 30 service stations is within \(\$ .03\) of the population mean? b. What is the probability that the mean price for a sample of 50 service stations is within \(\$ .03\) of the population mean? c. What is the probability that the mean price for a sample of 100 service stations is within \(\$ .03\) of the population mean? d. Which, if any, of the sample sizes in parts (a), (b), and (c) would you recommend to have at least a .95 probability that the sample mean is within \(\$ .03\) of the population mean?

A population proportion is \(.40 .\) A simple random sample of size 200 will be taken and the sample proportion \(\bar{p}\) will be used to estimate the population proportion. a. What is the probability that the sample proportion will be within ±.03 of the population proportion? b. What is the probability that the sample proportion will be within ±.05 of the population proportion?

The Grocery Manufacturers of America reported that \(76 \%\) of consumers read the ingredients listed on a product's label. Assume the population proportion is \(p=.76\) and a sample of 400 consumers is selected from the population.a. Show the sampling distribution of the sample proportion \(\bar{p}\) a. where \(\bar{p}\) is the proportion of the sampled consumers who read the ingredients listed on a product's label. b. What is the probability that the sample proportion will be within ±.03 of the population proportion? c. Answer part (b) for a sample of 750 consumers.

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?

A simple random sample of 5 months of sales data provided the following information: \\[ \begin{array}{lrrrrr} \text {Month:} & 1 & 2 & 3 & 4 & 5 \\ \text {Units Sold:} & 94 & 100 & 85 & 94 & 92 \end{array} \\] a. Develop a point estimate of the population mean number of units sold per month. b. Develop a point estimate of the population standard deviation.

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