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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

In the United States, \(44 \%\) of the population has type O blood. Suppose a random sample of 12 persons is taken. Find the probability that 6 of the persons will have type \(\mathrm{O}\) blood (and 6 will not) (a) using the binomial distribution formula. (b) using the normal approximation.

Consider taking a random sample of size 20 from the population of students at a certain college and measuring the diastolic blood pressure each of the 20 students. In the context of this setting, explain what is meant by the sampling distribution of the sample mean.

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 \(\operatorname{Pr}\\{145 \leq \bar{Y} \leq 165\\} ?\)

A fair coin is to be tossed 20 times. Find the probability that 10 of the tosses will fall heads and 10 will fall tails, (a) using the binomial distribution formula. (b) using the normal approximation with the continuity correction.

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