91Ó°ÊÓ

Three firms carry inventories that differ in size. Firm A's inventory contains 2000 items, firm \(\mathrm{B}\) 's inventory contains 5000 items, and firm \(\mathrm{C}\) 's inventory contains 10,000 items. The population standard deviation for the cost of the items in each firm's inventory is \(\sigma=144\) A statistical consultant recommends that each firm take a sample of 50 items from its inventory to provide statistically valid estimates of the average cost per item. Managers of the small firm state that because it has the smallest population, it should be able to make the estimate from a much smaller sample than that required by the larger firms. However, the consultant states that to obtain the same standard error and thus the same precision in the sample results, all firms should use the same sample size regardless of population size. a. Using the finite population correction factor, compute the standard error for each of the three firms given a sample of size 50 . b. What is the probability that for each firm the sample mean \(\bar{x}\) will be within ±25 of the population mean \(\mu ?\)

Step by step solution

01

Understanding the Problem

We are given three firms with inventories of different sizes and asked to compute the standard error using the finite population correction factor for each firm. Then, we need to find the probability that the sample mean will be within ±25 of the population mean. The sample size for all firms is 50, and the population standard deviation is 144.

02

Calculate Standard Error with Finite Population Correction

The formula for the standard error using the finite population correction factor is \( SE = \frac{\sigma}{\sqrt{n}} \sqrt{\frac{N-n}{N-1}} \), where \( \sigma = 144 \), \( n = 50 \), and \( N \) is the population size of each firm. Compute the standard error for each firm individually.

03

Standard Error for Firm A

For Firm A, \( N = 2000 \). Using the formula \( SE = \frac{144}{\sqrt{50}} \sqrt{\frac{2000-50}{2000-1}} \), calculate the standard error. The calculation is as follows: - \( SE_A = \frac{144}{\sqrt{50}} \cdot \sqrt{\frac{1950}{1999}} \)- Simplifying gives \( SE_A \approx 20.128 \).

04

Standard Error for Firm B

For Firm B, \( N = 5000 \). Using the formula \( SE = \frac{144}{\sqrt{50}} \sqrt{\frac{5000-50}{5000-1}} \), calculate the standard error. The calculation is as follows: - \( SE_B = \frac{144}{\sqrt{50}} \cdot \sqrt{\frac{4950}{4999}} \)- Simplifying gives \( SE_B \approx 20.0 \).

05

Standard Error for Firm C

For Firm C, \( N = 10000 \). Using the formula \( SE = \frac{144}{\sqrt{50}} \sqrt{\frac{10000-50}{10000-1}} \), calculate the standard error. The calculation is as follows: - \( SE_C = \frac{144}{\sqrt{50}} \cdot \sqrt{\frac{9950}{9999}} \)- Simplifying gives \( SE_C \approx 19.934 \).

06

Calculate Probability for Firm A

To find the probability that \( \bar{x} \) is within \( \pm 25 \) of \( \mu \), use the standard normal distribution. - The range is \( \pm 25 \). - \( z = \frac{25}{SE} \). - For Firm A, \( z = \frac{25}{20.128} \approx 1.242 \).- Compute \( P(-1.242 < Z < 1.242) \) using standard normal distribution tables.

07

Calculate Probability for Firm B

For Firm B, use the previously computed standard error:- \( z = \frac{25}{20.0} = 1.25 \).- Compute \( P(-1.25 < Z < 1.25) \) using the standard normal distribution tables.

08

Calculate Probability for Firm C

For Firm C, use the previously computed standard error:- \( z = \frac{25}{19.934} \approx 1.254 \).- Compute \( P(-1.254 < Z < 1.254) \) using the standard normal distribution tables.

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Key Concepts

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

Standard Error

When working with samples from a population, one important concept to understand is the standard error. This measure tells us how much a sample statistic, like the sample mean, is expected to fluctuate from the true population mean if we take repeated samples. Standard error is calculated using the formula: \[ SE = \frac{\sigma}{\sqrt{n}} \] where \( \sigma \) is the population standard deviation and \( n \) is the sample size.
In the context of finite populations, the formula is adjusted using the finite population correction factor, which accounts for the reduced variability when the sample size is a significant portion of the population size. This adjustment is critical as it results in a more accurate estimate of the standard error:\[ SE = \frac{\sigma}{\sqrt{n}} \sqrt{\frac{N-n}{N-1}} \] Here, \( N \) represents the population size. This adjustment ensures a lower standard error as the entire population is not infinite, meaning that information from a larger portion of the population decreases variability due to increased knowledge of the population attributes.

Population Mean

The population mean, often represented by \( \mu \), is a vital statistic in understanding the central tendency of data within a whole population. This mean reflects the average value within a set of all individuals or items in a population.
In statistical analyses, it's rare to know the exact population mean unless a census has been conducted, so it's often estimated using the sample mean. Understanding how close a sample mean is to the population mean involves recognizing the concept of standard error and employing it to estimate the precision of the sample mean.In practical applications, the population mean is pivotal, especially when assessing how representative a sample is of the population. This estimation helps make informed decisions based on the sample mean, providing insights into trends and averages without needing to evaluate the entire population.

Sample Size

The sample size, denoted by \( n \), plays a crucial role in statistical studies. It refers to the number of observations or items selected from a population for analysis.A well-chosen sample size helps ensure that the sample mean estimates the population mean accurately.
In general, larger sample sizes tend to result in more reliable estimates, as they likely reflect more characteristics of the population.Key points when considering sample size include:

• A larger sample size typically leads to a smaller standard error, offering more precise estimations of the population parameters.
• The sample size affects the breadth of confidence intervals, with larger samples yielding narrower intervals.
• When samples are large enough, they can accurately estimate the population mean, reducing the margin of error.

Therefore, determining the appropriate sample size is essential for any study aiming to produce meaningful and trustworthy results.

Probability Distribution

Probability distribution is a statistical function that describes all the possible values and probabilities that a random variable can take within a given range.In the context of the exercise with inventory costs, the probability distribution helps determine how likely it is for the sample mean to be close to the population mean.Important aspects of probability distribution include:

• The normal distribution is commonly used in statistics for sample means due to its symmetric, bell-shaped curve, which is helpful for analyzing data patterns.
• Using the standard normal distribution, often denoted by \( Z \), helps calculate the probability of the sample mean falling within a specified range.
• For this purpose, the \( Z \)-score is utilized, converting observed values into a form that allows looking up probabilities in standard normal distribution tables.

By employing probability distribution, statisticians and analysts can make predictions about data and assess the likelihood of various outcomes, enabling informed decision-making processes.