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Chapter 8: Problem 11

Determine \(\mu_{\bar{x}}\) and \(\sigma_{\bar{x}}\) from the given parameters of the population and the sample size. \(\mu=52, \sigma=10, n=21\)

Short Answer

Expert verified
\(\mu_{\bar{x}} = 52\), \(\sigma_{\bar{x}} \approx 2.18\)

Step by step solution

01

Identify Given Parameters

First, identify the given parameters from the problem. Here, we have the population mean \(\mu = 52\), the population standard deviation \(\sigma = 10\), and the sample size \(\ n = 21\).
02

Determine the Mean of the Sample Mean Distribution

The mean of the sample mean distribution \(\mu_{\bar{x}}\) is equal to the mean of the population \(\mu\). Therefore, \(\mu_{\bar{x}} = 52\).
03

Calculate the Standard Error of the Mean

The standard error of the mean \(\sigma_{\bar{x}}\) can be calculated using the formula \[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \]. Plugging in the given values: \[ \sigma_{\bar{x}} = \frac{10}{\sqrt{21}} \approx 2.18 \]

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Key Concepts

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

Population Mean
The population mean, denoted by \(\mu\), represents the average value of a particular characteristic in a given population. It is a vital statistic because it offers insights into the central tendency of the population data.

For example, in our exercise, the population mean \(\mu\) is 52. This implies that if we were to calculate the average of all individuals in the population, the result would be 52. Understanding the population mean is crucial because it acts as a benchmark for various statistical analyses.

Whenever we take samples from this population, we would expect the average value of each sample to be around 52, although not necessarily exactly 52 due to random variations in sampling.
Standard Deviation
The standard deviation, represented by \(\sigma\), measures the dispersion or spread of a set of data points in a population. In simpler terms, it tells us how much individual data points in the population deviate from the population mean.

In our exercise, the standard deviation \(\sigma\) is 10. This means that on average, the data points in the population differ from the mean by 10 units. A smaller standard deviation would indicate that the data points are closely clustered around the mean, while a larger standard deviation would mean they are more spread out.

Understanding standard deviation is crucial because it helps in comparing the variability or consistency within different datasets. It also lays the foundation for calculating the standard error of the mean, a concept we will discuss next.
Standard Error of the Mean
The standard error of the mean, denoted as \(\sigma_{\bar{x}}\), represents the standard deviation of the sampling distribution of the sample mean. It provides an idea of how much the sample mean is expected to vary from the population mean when multiple samples are taken from the same population.

In our exercise, to calculate the standard error of the mean, we use the formula \[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{\}} \].

Given the population standard deviation \(\sigma = 10\) and the sample size \(\ n = 21\), we can plug these values into the formula:
\[ \sigma_{\bar{x}} = \frac{10}{\sqrt{21}} \approx 2.18 \]

This calculation tells us that the sample mean will be around 2.18 units away from the population mean on average. The standard error of the mean helps in understanding the preciseness of the sample mean as an estimate of the population mean. Smaller values indicate more reliable estimates.

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Most popular questions from this chapter

The Food and Drug Administration sets Food Defect Action Levels (FDALs) for some of the various foreign substances that inevitably end up in the food we eat and liquids we drink. For example, the FDAL for insect filth in peanut butter is 3 insect fragments (larvae, eggs, body parts, and so on ) per 10 grams. A random sample of 50 ten-gram portions of peanut butter is obtained and results in a sample mean of \(\bar{x}=3.6\) insect fragments per ten-gram portion. (a) Why is the sampling distribution of \(\bar{x}\) approximately normal? (b) What is the mean and standard deviation of the sampling distribution of \(\bar{x}\) assuming that \(\mu=3\) and \(\sigma=\sqrt{3} ?\) (c) What is the probability that a simple random sample of 50 ten-gram portions results in a mean of at least 3.6 insect fragments? Is this result unusual? What might we conclude?

The S\&P 500 is a collection of 500 stocks of publicly traded companies. Using data obtained from Yahoo! Finance, the monthly rates of return of the S\&P500 since 1950 are normally distributed. The mean rate of return is \(0.007233(0.7233 \%),\) and the standard deviation for rate of return is \(0.04135(4.135 \%)\). (a) What is the probability that a randomly selected month has a positive rate of return? That is, what is \(P(x>0) ?\) (b) Treating the next 12 months as a simple random sample, what is the probability that the mean monthly rate of return will be positive? That is, with \(n=12,\) what is \(P(\bar{x}>0) ?\) (c) Treating the next 24 months as a simple random sample, what is the probability that the mean monthly rate of return will be positive? (d) Treating the next 36 months as a simple random sample, what is the probability that the mean monthly rate of return will be positive? (e) Use the results of parts (b)-(d) to describe the likelihood of earning a positive rate of return on stocks as the investment time horizon increases.

Describe the sampling distribution of \(\hat{p}\). Assume that the size of the population is 25,000 for each problem. $$ n=1000, p=0.103 $$

State the Central Limit Theorem

What happens to the standard deviation of \(\hat{p}\) as the sample size increases? If the sample size is increased by a factor of 4 what happens to the standard deviation of \(\hat{p} ?\)
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