/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 What does the sampling distribut... [FREE SOLUTION] | 91Ó°ÊÓ

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What does the sampling distribution of the sample mean look like if samples are taken from an approximately normal distribution? Use the applet Sampling Distribution of the Mean (at https://college.cengage.com/nextbook/statistics/wackerly 966371/student/html/index.html) to complete the following. The population to be sampled is approximately normally distributed with \(\mu=16.50\) and \(\sigma=6.03\) (these values are given above the population histogram and denoted \(M\) and S, respectively). a. Use the button "Next Obs" to select a single value from the approximately normal population. Click the button four more times to complete a sample of size \(5 .\) What value did you obtain for the mean of this sample? Locate this value on the bottom histogram (the histogram for the values of \(Y\) ). b. Click the button "Reset" to clear the middle graph. Click the button "Next Obs" five more times to obtain another sample of size 5 from the population. What value did you obtain for the mean of this new sample? Is the value that you obtained equal to the value you obtained in part (a)? Why or why not? c. Use the button "1 Sample" eight more times to obtain a total of ten values of the sample mean. Look at the histogram of these ten means. i. What do you observe? ii. How does the mean of these \(10 \bar{y}\) -values compare to the population mean \(\mu\) ? d. Use the button "1 Sample" until you have obtained and plotted 25 realized values for the sample mean \(\bar{Y}\), each based on a sample of size 5 . i. What do you observe about the shape of the histogram of the 25 values of \(\bar{y}_{i}\) $$ i=1,2, \ldots, 25 ? $$ ii. How does the value of the standard deviation of the \(25 \bar{y}\) values compare with the theoretical value for \(\sigma_{Y}\) obtained in Example 5.27 where we showed that, if \(Y\) is computed based on a sample of size \(n,\) then \(V(\bar{Y})=\sigma^{2} / n ?\) e. Click the button "1000 Samples" a few times, observing changes to the histogram as you generate more and more realized values of the sample mean. What do you observe about the shape of the resulting histogram for the simulated sampling distribution of \(Y\) ? f. Click the button "Toggle Normal" to overlay (in green) the normal distribution with the same mean and standard deviation as the set of values of \(Y\) that you previously generated. Does this normal distribution appear to be a good approximation to the sampling distribution of \(Y ?\)

Short Answer

Expert verified
The sampling distribution of the sample mean tends to be approximately normal, even for small sample sizes, due to the Central Limit Theorem. As the number of samples increases, this normality becomes more pronounced.

Step by step solution

01

Select Initial Sample

Begin by using the "Next Obs" button to select a single value from the approximately normal population. Repeat this four times to complete a sample of size 5. Record the mean of this sample.
02

Obtain a New Sample

Reset the middle graph using the "Reset" button, then click "Next Obs" five more times to obtain another sample of size 5. Record the mean of this new sample and compare it to the mean from Step 1.
03

Analyze Ten Sample Means

Use the "1 Sample" button eight more times to obtain ten sample mean values. Observe the histogram of these ten means and compare their mean to the population mean \( \mu = 16.50.\)
04

Examine 25 Sample Means

Use "1 Sample" until 25 sample means are plotted. Observe the shape of their histogram and compare the standard deviation of these means to the theoretical standard deviation.
05

Simulate 1000 Samples

Click "1000 Samples" a few times. Observe how the histogram shape evolves as you simulate more sample means. Note the shape of the distribution.
06

Overlay Normal Distribution

Use "Toggle Normal" to overlay a normal distribution over the histogram. Determine whether this normal distribution approximates the sampling distribution well.

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Key Concepts

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

Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental concept in statistics. It explains why many statistical procedures work, even if the data doesn't look like a normal distribution. According to the CLT, if you take a large number of random samples from any population and calculate their means, the distribution of those means will resemble a normal distribution.

This amazing property holds true even if the original population distribution is not normal. However, the process is faster and more reliable if your samples are from a population that does resemble a normal distribution.
  • Larger sample sizes offer better approximations to normal distribution.
  • This theorem supports the use of normal distribution in creating confidence intervals and hypothesis tests.
Imagine sampling from a population like taking spoonfuls from a pot of soup. Even if one spoonful might miss an ingredient, multiple spoonfuls averaged out provide a good taste of the whole pot. The CLT builds the foundation for understanding how sampling distributions form and behave.
Normal Distribution
Normal distribution, often referred to as the bell curve, is a type of continuous probability distribution for a real-valued random variable. It is symmetric about the mean and describes how data naturally clusters around a central point.

In our example, we're dealing with approximately normal distribution, characterized by a population mean (denoted by \( \mu = 16.50 \)) and a standard deviation (denoted by \( \sigma = 6.03 \)). These parameters define the shape and the spread of the distribution.
  • The mean determines the center of the distribution.
  • The standard deviation measures the spread and indicates how much variation exists from the mean.
A real-world application of this could be heights of people, where most heights cluster around an average height (the mean), and fewer people are extremely tall or short.
Sample Mean
The sample mean is an important statistic calculated from a sample of observations. It represents the average value of the sample and serves as an estimate of the population mean. To compute a sample mean, sum all the observations in your sample and divide by the number of observations:
\[ \bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_{i} \]

In the exercise, you repeatedly select samples and calculate their means, observing how these means distribute around the population mean \( \mu = 16.50 \).
  • Each mean provides a snapshot, or mini-estimate, of the population mean.
  • This helps in understanding the variability and reliability of the sample mean as an estimator.
When compared across multiple samples, the sample means provide insight into how close your estimates are to the true population mean.
Standard Deviation
Standard deviation is a key measure in statistics indicating how spread out the numbers are in a set of data. It quantifies the amount of variation or dispersion and is crucial in describing the spread of data in both population and sample contexts.

In our example, the standard deviation \( \sigma = 6.03 \) quantifies the population spread. For sample means, the standard deviation gets adjusted and is often referred to as the standard error. This adjustment accounts for sample size \( n \):
\[ \sigma_{\bar{Y}} = \frac{\sigma}{\sqrt{n}} \]

Using a large number of samples from the population reduces the standard deviation of the sample mean, improving precision in estimating the population mean. In practical terms:
  • A smaller standard deviation means data points are closer to the mean.
  • A larger standard deviation implies more spread and wider range of data values.
Understanding standard deviation helps in evaluating how consistent sample means are with the true population mean. It also assists in calculating confidence intervals and in hypothesis testing.

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Most popular questions from this chapter

In 2004 Florida was hit by four major hurricanes. In 2005 a survey indicated that, in \(2004,48 \%\) of the households in Florida had no plans for escaping an approaching hurricane. Suppose that a recent random sample of 50 households was selected in Gainesville and that those in 29 of the households indicated that their household had a hurricane escape plan. a. If the 2004 state percentages still apply to recent Gainesville households, use the Normal Approximation to Binomial Distribution applet to find the exact and approximate values of the probability that 29 or more of the households sampled have a hurricane escape plan. b. Refer to part (a). Is the normal approximation close to the exact binomial probability? Explain why.

Resistors to be used in a circuit have average resistance 200 ohms and standard deviation 10 ohms. Suppose 25 of these resistors are randomly selected to be used in a circuit. a. What is the probability that the average resistance for the 25 resistors is between 199 and 202 ohms? b. Find the probability that the total resistance does not exceed 5100 ohms. [Hint: see Example 7.9.]

If \(Y\) is a random variable that has an \(F\) distribution with \(\nu_{1}\) numerator and \(\nu_{2}\) denominator degrees of freedom, show that \(U=1 / Y\) has an \(F\) distribution with \(\nu_{2}\) numerator and \(\nu_{1}\) denominator degrees of freedom.

The downtime per day for a computing facility has mean 4 hours and standard deviation. .8 hour. a. Suppose that we want to compute probabilities about the average daily downtime for a period of 30 days.i. What assumptions must be true to use the result of Theorem 7.4 to obtain a valid approximation for probabilities about the average daily downtime? i. Under the assumptions described in part (i), what is the approximate probability that the average daily downtime for a period of 30 days is between 1 and 5 hours? b. Under the assumptions described in part (a), what is the approximate probability that the total downtime for a period of 30 days is less than 115 hours?

Suppose that \(Z\) has a standard normal distribution and that \(Y\) is an independent \(\chi^{2}\) -distributed random variable with \(\nu\) df. Then, according to Definition 7.2, $$T=\frac{Z}{\sqrt{Y / \nu}}$$ has a \(t\) distribution with \(\nu\) df. \(^{*}\) a. If \(Z\) has a standard normal distribution, give \(E(Z)\) and \(E\left(Z^{2}\right)\). [Hint: For any random variable, \(\left.E\left(Z^{2}\right)=V(Z)+(E(Z))^{2} \cdot\right]\) b. According to the result derived in Exercise \(4.112(\mathrm{a}),\) if \(Y\) has a \(\chi^{2}\) distribution with \(\nu\) df, then $$E\left(Y^{a}\right)=\frac{\Gamma([\nu / 2]+a)}{\Gamma(\nu / 2)} 2^{a}, \quad \text { if } \nu>-2 a$$. Use this result, the result from part (a), and the structure of \(T\) to show the following. [Hint: Recall the independence of \(Z \text { and } Y \)] i. \(E(T)=0,\) if \(\nu>1\) ii. \(V(T)=\nu /(\nu-2),\) if \(\nu>2\)

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