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Chapter 8: Problem 11

Determine \(\mu_{x}\) and \(\sigma_{x}\) from the given parameters of the population and the sample size. $$\mu=80, \sigma=14, n=49$$

Short Answer

Expert verified
\backslashmu_{x} = 80, \backslashsigma_{x} = 2.

Step by step solution

01

Understand the Problem

Given the population parameters \(\backslash\text{mean} = 80, \backslash\text{standard deviation} = 14\), and the sample size \(n = 49\).\ We need to find the mean \(\backslashmu_{x}\) and standard deviation \(\backslashsigma_{x}\) of the sample mean.\
02

Find the Mean of the Sample Mean (\backslashmu_{x})

In the case of the mean of the sample mean, it remains the same as the population mean: \[\backslashmu_{x} = \backslashmu = 80\]
03

Find the Standard Deviation of the Sample Mean (\backslashsigma_{x})

The formula for the standard deviation of the sample mean is \(\backslashsigma_{x} = \frac{\backslashsigma}{\backslashsqrt{n}} \).\ Given \(\backslashsigma = 14\) and \(n = 49\), apply the formula:\[\backslashsigma_{x} = \frac{14}{\backslashsqrt{49}} = \frac{14}{7} = 2\]
04

Conclusion

Therefore, the values are \(\backslashmu_{x} = 80\) and \(\backlashsigma_{x} = 2\).

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Key Concepts

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

Sample Mean
The sample mean is an important concept in statistics. It is the average of a set of data points collected from a sample. The sample mean is denoted by \( \bar{x} \). To calculate the sample mean, sum all the values in the sample and then divide by the number of values in the sample. The formula is given by: \[ \bar{x} = \frac{1}{n} \times \text{Sum of all values in the sample} \]
Since a sample is just a subset of the population, the sample mean is an estimate of the population mean. As you gather more data, the sample mean gets closer to the population mean. This is because larger samples tend to represent the population better.
Population Mean
The population mean is the average of all possible values in a population. It's a key parameter in statistics, represented by the symbol \( \mu \). To find the population mean, sum all the values in the population and divide by the number of values in the population. The formula is: \[ \mu = \frac{\text{Sum of all values in the population}}{N} \]
Unlike the sample mean, which can vary from sample to sample, the population mean is a fixed value and does not change. It's also known as the expected value or the true mean of the population. For example, in our exercise, the population mean is given as 80.
Standard Deviation of the Sample Mean
The standard deviation of the sample mean is a measure of how much the sample mean varies from sample to sample. It is also known as the standard error of the mean and is denoted by \( \sigma_{x} \). Its formula is given by: \[ \sigma_{x} = \frac{\sigma}{\sqrt{n}} \]
Here, \( \sigma \) is the population standard deviation, and \( n \) is the sample size. This formula shows that as the sample size increases, the standard error decreases, indicating more stable and reliable estimates of the population mean. For our exercise, with \( \sigma = 14 \) and \( n = 49 \), the standard deviation of the sample mean is calculated as: \[ \sigma_{x} = \frac{14}{\sqrt{49}} = \frac{14}{7} = 2 \]
Sample Size
Sample size is denoted by \( n \) and represents the number of observations in a sample. It is a critical factor in statistical analysis because larger sample sizes generally lead to more accurate estimates of population parameters. This is due to the law of large numbers, which tells us that as the sample size increases, the sample mean gets closer to the population mean.
A larger sample size reduces the standard error of the mean, as seen in the formula: \[ \sigma_{x} = \frac{\sigma}{\sqrt{n}} \]
In our exercise, the sample size is 49. A larger sample size improves the precision of our estimations regarding the population parameters.
In summary, an appropriate sample size is integral to achieving reliable and accurate statistical analysis.

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Most popular questions from this chapter

State the Central Limit Theorem.

The quality-control manager of a Long John Silver's restaurant wishes to analyze the length of time a car spends at the drive-through window waiting for an order. According to records obtained from the restaurants, it is determined that the mean time spent at the window is 59.3 seconds with a standard deviation of 13.1 seconds. The distribution of time at the window is skewed right (data based on information provided by Danica Williams, student at Joliet Junior College). (a) To obtain probabilities regarding a sample mean using the normal model, what size sample is required? (b) The quality-control manager wishes to use a new delivery system designed to get cars through the drivethrough system faster. A random sample of 40 cars results in a sample mean time spent at the window of 56.8 seconds. What is the probability of obtaining a sample mean of 56.8 seconds or less assuming the population mean is 59.3 seconds? Do you think that the new system is effective?

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\&P 500 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.

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 \(\mu=3\) and \(\sigma=\sqrt{3}\) (c) Suppose a simple random sample of \(n=50\) ten-gram samples of peanut butter results in a sample mean of 3.6 insect fragments. What is the probability 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?

A nationwide study in 2003 indicated that about \(60 \%\) of college students with cell phones send and receive text messages with their phones. Suppose a simple random sample of \(n=1136\) college students with cell phones is obtained. (Source: promomagazine.com) (a) Describe the sampling distribution of \(\hat{p},\) the sample proportion of college students with cell phones who send or receive text messages with their phones. (b) What is the probability that 665 or fewer college students in the sample send and receive text messages with their cell phones? Is this result unusual? (c) What is the probability that 725 or more college students in the sample send and receive text messages with their cell phone? Is this result unusual?
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