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Chapter 8: Problem 12

Determine \(\mu_{\bar{x}}\) and \(\sigma_{\bar{x}}\) from the given parameters of the population and the sample size. \(\mu=27, \sigma=6, n=15\)

Short Answer

Expert verified
\[ \mu_{\bar{x}} = 27 \] \[ \sigma_{\bar{x}} \approx 1.55 \]

Step by step solution

01

- Understand the Central Tendency and Spread

Given the population mean \( \mu = 27 \), standard deviation \( \sigma = 6 \), and sample size \( \ n = 15 \), determine the mean and the standard deviation of the sample mean.
02

- Calculate the Mean of the Sample Mean

The mean of the sample mean, \( \mu_{\bar{x}} \), is equal to the population mean. So, \[ \mu_{\bar{x}} = 27 \]
03

- Calculate the Standard Deviation of the Sample Mean

The standard deviation of the sample mean, \( \sigma_{\bar{x}} \), is given by \[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \] Substituting the given values: \[\begin{aligned} \sigma_{\bar{x}} = \frac{6}{\sqrt{15}} = \frac{6}{3.87} \approx 1.55 \end{aligned} \]

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Key Concepts

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

Central Tendency
Central tendency describes the average or middle value of a set of data. Common measures of central tendency include the mean, median, and mode.
In this exercise, we are particularly interested in the mean, which is the total sum of all values divided by the number of values.
The population mean (\mu) is a measure of the central tendency for the entire population.
For any sample taken from this population, the mean of the sample mean \(\mu_{\bar{x}}\) is also equal to the population mean. Understanding central tendency helps to summarize a large set of data with a single value that represents the center of the distribution.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. In other words, it tells us how spread out the values are around the mean.
In this case, the population standard deviation (\sigma) is given as 6. To find the standard deviation of the sample mean \(\sigma_{\bar{x}}\), we use the formula:
\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \].
The standard deviation of the sample mean shows how much the sample mean deviates from the population mean. Smaller sample standard deviations indicate that the sample means are closely clustered around the population mean, while larger standard deviations indicate more spread.
Sample Size
Sample size (\(n\)) is the number of observations in a sample. It plays a crucial role in statistical calculations and directly affects the standard deviation of the sample mean.
In this exercise, the sample size is provided as 15. When calculating the standard deviation of the sample mean, the formula involves the square root of the sample size:
\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \].
This means that as the sample size increases, the standard deviation of the sample mean decreases, indicating more precision. Conversely, smaller sample sizes lead to higher variability in the sample mean.
Population Mean
The population mean (\(\mu\)) is the average of all values in a population, and it represents the central value of the entire dataset.
The formula for the population mean is:
\[ \mu = \frac{\sum_{i=1}^{N} X_{i}}{N} \],
where \(\sum\_{i=1}^{N} X\_{i}\)) is the sum of all values in the population, and \(\N\) is the total number of values.
In this exercise, the population mean is given as 27. This value helps in understanding the central tendency of the population. When calculating the sample mean, we see that the mean of the sample mean \(\mu_\bar\{x\}\) equals the population mean, showing that the sample's average accurately represents the population.

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Most popular questions from this chapter

A simple random sample of size \(n=36\) is obtained from a population with \(\mu=64\) and \(\sigma=18\). (a) Describe the sampling distribution of \(\bar{x}\). (b) What is \(P(\bar{x}<62.6) ?\) (c) What is \(P(\bar{x} \geq 68.7) ?\) (d) What is \(P(59.8<\bar{x}<65.9) ?\)

The shape of the distribution of the time required to get an oil change at a 10 -minute oil-change facility is unknown. However, records indicate that the mean time for an oil change is 11.4 minutes, and the standard deviation for oilchange time is 3.2 minutes. (a) To compute probabilities regarding the sample mean using the normal model, what size sample would be required? (b) What is the probability that a random sample of \(n=40\) oil changes results in a sample mean time of less than 10 minutes? (c) Suppose the manager agrees to pay each employee a \(\$ 50\) bonus if they meet a certain goal. On a typical Saturday, the oil-change facility will perform 40 oil changes between 10 A.M. and 12 P.M. Treating this as a random sample, what mean oil-change time would there be a \(10 \%\) chance of being at or below? This will be the goal established by the manager.

Afraid to Fly According to a study conducted by the Gallup organization, the proportion of Americans who are afraid to fly is \(0.10 .\) A random sample of 1100 Americans results in 121 indicating that they are afraid to fly. Explain why this is not necessarily evidence that the proportion of Americans who are afraid to fly has increased since the time of the Gallup study.

Foreign Language According to a study done by Wakefield Research, the proportion of Americans who can order a meal in a foreign language is \(0.47 .\) (a) Suppose a random sample of 200 Americans is asked to disclose whether they can order a meal in a foreign language. Is the response to this question qualitative or quantitative? Explain. (b) Explain why the sample proportion, \(\hat{p},\) is a random variable. What is the source of the variability? (c) Describe the sampling distribution of \(\hat{p},\) the proportion of Americans who can order a meal in a foreign language. Be sure to verify the model requirements. (d) In the sample obtained in part (a), what is the probability the proportion of Americans who can order a meal in a foreign language is greater than \(0.5 ?\) (e) Would it be unusual that, in a survey of 200 Americans, 80 or fewer Americans can order a meal in a foreign language? Why?

Credit Cards According to creditcard.com, \(29 \%\) of adults do not own a credit card. (a) Suppose a random sample of 500 adults is asked, "Do you own a credit card?" Describe the sampling distribution of \(\hat{p}\), the proportion of adults who do not own a credit card. (b) What is the probability that in a random sample of 500 adults more than \(30 \%\) do not own a credit card? (c) What is the probability that in a random sample of 500 adults between \(25 \%\) and \(30 \%\) do not own a credit card? (d) Would it be unusual for a random sample of 500 adults to result in 125 or fewer who do not own a credit card? Why?
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