/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 A company maintains three office... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A company maintains three offices in a region, each staffed by two employees. Information concerning yearly salaries (1000's of dollars) is as follows: $$ \begin{array}{lcccccc} \text { Office } & 1 & 1 & 2 & 2 & 3 & 3 \\ \text { Employee } & 1 & 2 & 3 & 4 & 5 & 6 \\ \text { Salary } & 29.7 & 33.6 & 30.2 & 33.6 & 25.8 & 29.7 \end{array} $$ a. Suppose two of these employees are randomly selected from among the six (without replacement). Determine the sampling distribution of the sample mean salary \(\bar{X}\). b. Suppose one of the three offices is randomly selected. Let \(X_{1}\) and \(X_{2}\) denote the salaries of the two employees. Determine the sampling distribution of \(X\). c. How does \(E(X)\) from parts (a) and (b) compare to the population mean salary \(\mu\) ?

Short Answer

Expert verified
Both sampling methods yield the same mean as the population mean, 30.43.

Step by step solution

01

Identify Population Salaries

Here, the salaries of all employees are provided as \[ 29.7, 33.6, 30.2, 33.6, 25.8, 29.7 \]. This defines our population.
02

Determine Population Mean

Calculate the population mean \( \mu \) of the given salaries. \[ \mu = \frac{29.7 + 33.6 + 30.2 + 33.6 + 25.8 + 29.7}{6} = 30.4333\ldots \approx 30.43 \]
03

Determine All Possible Samples for Part a

Two employees are selected without replacement from six. List all combinations: (1,2), (1,3), (1,4), (1,5), (1,6), (2,3), (2,4), (2,5), (2,6), (3,4), (3,5), (3,6), (4,5), (4,6), (5,6).
04

Calculate Sample Means for Part a

Calculate the sample mean salary \( \bar{X} \) for each combination: \[ \begin{align*} \bar{X}_{12} &= \frac{29.7 + 33.6}{2} = 31.65, \ \bar{X}_{13} &= \frac{29.7 + 30.2}{2} = 29.95, \ \bar{X}_{14} &= \frac{29.7 + 33.6}{2} = 31.65, \ \bar{X}_{15} &= \frac{29.7 + 25.8}{2} = 27.75, \ \bar{X}_{16} &= \frac{29.7 + 29.7}{2} = 29.7, \ \bar{X}_{23} &= \frac{33.6 + 30.2}{2} = 31.9, \ \bar{X}_{24} &= \frac{33.6 + 33.6}{2} = 33.6, \ \bar{X}_{25} &= \frac{33.6 + 25.8}{2} = 29.7, \ \bar{X}_{26} &= \frac{33.6 + 29.7}{2} = 31.65, \ \bar{X}_{34} &= \frac{30.2 + 33.6}{2} = 31.9, \ \bar{X}_{35} &= \frac{30.2 + 25.8}{2} = 28.0, \ \bar{X}_{36} &= \frac{30.2 + 29.7}{2} = 29.95, \ \bar{X}_{45} &= \frac{33.6 + 25.8}{2} = 29.7, \ \bar{X}_{46} &= \frac{33.6 + 29.7}{2} = 31.65, \ \bar{X}_{56} &= \frac{25.8 + 29.7}{2} = 27.75 \ \end{align*} \]
05

Collect Distinct Sample Means for Part a

Distinct sample means are \[ 27.75, 28.0, 29.7, 29.95, 31.65, 31.9, 33.6 \].
06

Analyze Part b - Selection by Office

A single office is chosen randomly, containing a pair: Office 1: (29.7, 33.6); Office 2: (30.2, 33.6); Office 3: (25.8, 29.7). Calculate \( X \), the average of each pair: \[ \begin{align*} X_1 &= \frac{29.7 + 33.6}{2} = 31.65, \ X_2 &= \frac{30.2 + 33.6}{2} = 31.9, \ X_3 &= \frac{25.8 + 29.7}{2} = 27.75 \end{align*} \]
07

Compare Expectations - Parts a, b, and Population

For part (a), the average of all sample means is \( E(X) = 30.43 \). For part (b), the mean of \( X \) is \[ \frac{31.65 + 31.9 + 27.75}{3} = 30.43 \]. In both cases, this matches the population mean \( \mu = 30.43 \).
08

Conclusion

Both parts (a) and (b) have sample means that equal the population mean salary, demonstrating that the sampling method is unbiased.

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.

Sample Mean
Understanding the concept of the sample mean is fundamental when working with data. The sample mean is the average value of a set of observations taken from a larger population. It provides a way to make estimations about the overall population, without having to analyze every single data point. For example, in a population of employee salaries, if we take a subset of these salaries, their average will be the sample mean. Mathematically, it is represented as \( \bar{X} = \frac{1}{n} \sum_{i=1}^{n}X_i \), where \(n\) is the number of observations in the sample, and \(X_i\) are the observed values. The sample mean helps to simplify large sets of data by synthesizing them into a single representative number. In statistics, this is frequently used to estimate the central tendency of a population, especially when dealing with a very large dataset. As seen in the original problem, calculating the sample mean for combinations of employees helps to understand average salaries across different selected groups.
Population Mean
The population mean is a key concept in statistics, representing the average of all the values in an entire population. It is an essential measurement often denoted by the Greek letter \(\mu\). Knowing the population mean is akin to understanding the central value around which all other data points gather. Calculating the population mean involves summing all the values in the dataset and dividing this sum by the number of values. For example, given the salaries \(29.7, 33.6, 30.2, 33.6, 25.8, 29.7\), the population mean \(\mu\) is calculated as \( \mu = \frac{29.7 + 33.6 + 30.2 + 33.6 + 25.8 + 29.7}{6} = 30.43 \). This value provides a benchmark against which sample distributions can be compared.The population mean is crucial for comparisons in hypothesis testing, quality control, and statistical control methods, as seen in the problem where it is used to compare with sample means to check for unbiased estimation.
Unbiased Estimator
An unbiased estimator is a statistical tool that provides estimates close to the true population parameter. In simpler terms, it means the average value of estimates obtained from multiple samples will equal the true population parameter. The goal is to ensure that the method of sampling does not systematically overestimate or underestimate the true parameter value. In the context of the exercise, both the average of all sample means and the mean of samples calculated based on office selection were equivalent to the population mean of 30.43. This demonstrates that the sample mean is an unbiased estimator of the population mean because, on average, it should match the population mean when taken over numerous samples. Choosing unbiased estimators is critical for reliable statistical analysis and decision-making, as it maintains accuracy and fairness in predicting population parameters.
Probability Distribution
Probability distribution describes how the values of a random variable are distributed. It outlines the likelihood any specific outcome will occur and helps in understanding the properties of data.In the original exercise, the sampling distribution of the sample mean pertains to the distribution of averages calculated from different samples. By understanding this distribution, one can ascertain how the sample mean varies and how often it is close to the population mean. For instance, possible sample means in the exercise, such as \( 27.75, 28.0, 29.7, 29.95, 31.65, 31.9, \) and \( 33.6 \), give insight into the range and frequency of sample outcomes.Studying the probability distribution is crucial for predicting trends, defining expectations, and making informed decisions based on data patterns. It serves as a foundation for many inferential statistical methods used in various scientific and practical applications.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

One piece of PVC pipe is to be inserted inside another piece. The length of the first piece is normally distributed with mean value \(20 \mathrm{in}\). and standard deviation \(.5\) in. The length of the second piece is a normal rv with mean and standard deviation 15 and \(.4\) in., respectively. The amount of overlap is normally distributed with mean value \(1 \mathrm{in}\). and standard deviation \(.1 \mathrm{in}\). Assuming that the lengths and amount of overlap are independent of each other, what is the probability that the total length after insertion is between \(34.5\) and 35 in.?

Show directly from the pdf that the mean of a \(t_{1}\) (Cauchy) random variable does not exist.

A student has a class that is supposed to end at 9:00 a.m. and another that is supposed to begin at 9:10 a.m. Suppose the actual ending time of the 9 a.m. class is a normally distributed rv \(X_{1}\) with mean 9:02 and standard deviation 1.5 min and that the starting time of the next class is also a normally distributed rv \(X_{2}\) with mean 9:10 and standard deviation \(1 \mathrm{~min}\). Suppose also that the time necessary to get from one classroom to the other is a normally distributed rv \(X_{3}\) with mean \(6 \mathrm{~min}\) and standard deviation \(1 \mathrm{~min}\). What is the probability that the student makes it to the second class before the lecture starts? (Assume independence of \(X_{1}, X_{2}\), and \(X_{3}\), which is reasonable if the student pays no attention to the finishing time of the first class.)

A professor has three errands to take care of in the Administration Building. Let \(X_{i}=\) the time that it takes for the \(i\) th errand \((i=1,2,3)\), and let \(X_{4}=\) the total time in minutes that she spends walking to and from the building and between each errand. Suppose the \(X_{i}\) 's are independent, normally distributed, with the following means and standard deviations: \(\mu_{1}=15, \quad \sigma_{1}=4\), \(\mu_{2}=5, \quad \sigma_{2}=1, \quad \mu_{3}=8, \quad \sigma_{3}=2, \quad \mu_{4}=12\), \(\sigma_{4}=3 .\) She plans to leave her office at precisely \(10=00\) a.m. and wishes to post a note on her door that reads, "I will return by \(t\) a.m." What time \(t\) should she write down if she wants the probability of her arriving after \(t\) to be \(.01\) ?

A large university has 500 single employees who are covered by its dental plan. Suppose the number of claims filed during the next year by such an employee is a Poisson rv with mean value \(2.3\). Assuming that the number of claims filed by any such employee is independent of the number filed by any other employee, what is the approximate probability that the total number of claims filed is at least 1200 ?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.