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The mean of the distribution shown in the following histogram is 162 and the standard deviation is 18 . Consider taking random samples of size \(n=9\) from this distribution and calculating the sample mean, \(\bar{y},\) for each sample. (a) What is the mean of the sampling distribution of \(\bar{Y} ?\) (b) What is the standard deviation of the sampling distribution of \(\bar{Y} ?\)

Short Answer

Expert verified
The mean of the sampling distribution of \(\bar{Y}\) is 162, and the standard deviation of the sampling distribution of \(\bar{Y}\) is 6.

Step by step solution

01

Understanding the Mean of the Sampling Distribution

The mean of the sampling distribution of the sample means, often denoted as \(\mu_{\bar{Y}}\), is equal to the mean of the population from which the samples are taken. In this case, since the population mean is given as 162, the mean of the sampling distribution will also be 162.
02

Calculating the Standard Deviation of the Sampling Distribution

The standard deviation of the sampling distribution of the sample means, also known as the standard error, is the population standard deviation divided by the square root of the sample size (n). The formula is \(\sigma_{\bar{Y}} = \frac{\sigma}{\sqrt{n}}\), where \(\sigma\) is the population standard deviation and n is the sample size. Given that \(\sigma = 18\) and \(n = 9\), plug these values into the formula to calculate the standard deviation of the sampling distribution.
03

Computing the Standard Error

Apply the formula from Step 2: \(\sigma_{\bar{Y}} = \frac{\sigma}{\sqrt{n}} = \frac{18}{\sqrt{9}} = \frac{18}{3} = 6\). This gives us the standard deviation (standard error) of the sampling distribution of \(\bar{Y}\).

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Key Concepts

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

Mean of Sampling Distribution
When delving into the concept of a sampling distribution, one of the most fundamental aspects to understand is the mean of this distribution, often denoted as \( \mu_{\bar{Y}} \). This is essentially the average value we'd expect if we took an infinite number of samples from our population and calculated the mean of each sample. What's particularly important here is the fact that regardless of the sample size, the mean of the sampling distribution \( \mu_{\bar{Y}} \) is equal to the mean of the entire population. This is known as the Central Limit Theorem.

For instance, in our exercise, we're working with a population mean of 162. So, if we're considering the sampling distribution of the sample mean \( \bar{Y} \), the mean of this distribution is also 162. This remains true no matter how many samples we collect, as long as the samples are taken at random from the population.
Standard Deviation of Sampling Distribution
Following our understanding of the mean, let's talk about the standard deviation of the sampling distribution, often expressed as \( \sigma_{\bar{Y}} \). This term may sound complex, but it's essentially a measure of how much the sample means can vary from the population mean. A smaller standard deviation indicates that the sample means are clustered closely around the population mean, whereas a larger standard deviation suggests more variation.

In mathematical terms, to find this standard deviation, we take the standard deviation of the population (denoted as \( \sigma \)) and divide it by the square root of the sample size (n), symbolically given as \( \sigma_{\bar{Y}} = \frac{\sigma}{\sqrt{n}} \). This formula is crucial for understanding and calculating the variability of a sampling distribution. The exercise clearly demonstrates this calculation, with a population standard deviation of 18 and a sample size of 9.
Standard Error
Often in statistics, we come across the term 'standard error', which might sound synonymous with standard deviation but caters specifically to sampling distributions. The standard error measures how far the sample mean of a population is likely to be from the actual population mean—not just for one sample, but on average across all possible samples of the same size.

This is why the standard error is simply the standard deviation of the sampling distribution that we discussed earlier. It is represented by the same formula \( \sigma_{\bar{Y}} = \frac{\sigma}{\sqrt{n}} \). Using our earlier exercise as an anchor for explanation, we calculated the standard error to be 6. This value aids researchers in understanding the degree of uncertainty involved in their sample estimates.
Random Sample Mean
The term 'random sample mean', denoted as \( \bar{Y} \), refers to the average of observations from just one sample out of all potential random samples. This mean represents a single point estimate of our population parameter—the central value that we'd expect to get if we could observe the entire population. It's different from the mean of the sampling distribution, which collates the means from all possible samples.

The random sample mean comes into play in practical scenarios, where we deal with one particular subset of the population at a time. In our exercise scenario, we calculate the mean \( \bar{y} \) for multiple random samples, each of size 9, from the population with a mean of 162. Each of these \( \bar{y} \) values will be a random sample mean and can vary from sample to sample. However, the central limit theorem reassures us that the distribution of these means will approximate a normal distribution centered around the population mean, particularly when dealing with larger sample sizes.

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Most popular questions from this chapter

The serum cholesterol levels of a population of 12 to 14-year-olds follow a normal distribution with mean \(155 \mathrm{mg} / \mathrm{dl}\) and standard deviation \(27 \mathrm{mg} / \mathrm{dl}\) (as in Example 4.1.1). (a) What percentage of the 12 - to 14 -year-olds have serum cholesterol values between 145 and \(165 \mathrm{mg} / \mathrm{dl} ?\) (b) Suppose we were to choose at random from the population a large number of groups of nine 12 - to 14-year-olds each. In what percentage of the groups would the group mean cholesterol value be between 145 and \(165 \mathrm{mg} / \mathrm{dl} ?\) (c) If \(\bar{Y}\) represents the mean cholesterol value of a random sample of nine 12 - to 14 -year-olds from the population, what is \(\operatorname{Pr}\\{145 \leq \bar{Y} \leq 165\\} ?\)

An important indicator of lung function is forced expiratory volume (FEV), which is the volume of air that a person can expire in one second. Dr. Hernandez plans to measure FEV in a random sample of \(n\) young women from a certain population, and to use the sample mean \(\bar{y}\) as an estimate of the population mean. Let \(E\) be the event that Hernandez's sample mean will be within \(\pm 100 \mathrm{ml}\) of the population mean. Assume that the population distribution is normal with mean \(3,000 \mathrm{ml}\) and standard deviation \(400 \mathrm{ml}^{3}\) Find \(\operatorname{Pr}\\{E\\}\) if (a) \(n=15\) (b) \(n=60\) (c) How does \(\operatorname{Pr}\\{E\\}\) depend on the sample size? That is, as \(n\) increases, does \(\operatorname{Pr}\\{E\\}\) increase, decrease, or stay the same?

Professor Smith conducted a class exercise in which students ran a computer program to generate random samples from a population that had a mean of 50 and a standard deviation of \(9 \mathrm{~mm}\). Each of Smith's students took a random sample of size \(n\) and calculated the sample mean. Smith found that about \(68 \%\) of the students had sample means between 48.5 and \(51.5 \mathrm{~mm} .\) What was \(n ?\) (Assume that \(n\) is large enough that the Central Limit Theorem is applicable.)

A certain cross between sweet-pea plants will produce progeny that are either purple flowered or white flowered; is \(p=\frac{9}{16}\). Suppose \(n\) progeny are to be examined, and let \(\hat{P}\) be the sample proportion of purple- flowered plants. It might happen, by chance, that \(\hat{P}\) would be closer to \(\frac{1}{2}\) than to \(\frac{9}{16}\). Find the probability that this misleading event would occur if (a) \(n=1\). (b) \(n=64\). (c) \(n=320\). (Use the normal approximation without the continuity correction.)

A certain assay for serum alanine aminotransferase (ALT) is rather imprecise. The results of repeated assays of a single specimen follow a normal distribution with mean equal to the ALT concentration for that specimen and standard deviation equal to \(4 \mathrm{U} / \mathrm{I}\) (as in Exercise \(4 .\) S. 15 ). Suppose a hospital lab measures many specimens every day, and specimens with reported ALT values of 40 or more are flagged as " unusually high." If a patient's true ALT concentration is \(35 \mathrm{U} / \mathrm{I},\) find the probability that his specimen will be flagged as "unusually high" (a) if the reported value is the result of a single assay. (b) if the reported value is the mean of three independent assays of the same specimen.

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