/*! 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 39 In \(2005,\) a study was conduct... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In \(2005,\) a study was conducted in West Texas A\&M University. It showed that the average student debt in the United States was \(\$ 18,367\) with a standard deviation of \$4709. (Source: http://swer.wtamu. edu/sites/default/files/Data/15-26-49-178-1-PB.pdf.) a. Suppose 100 students had been randomly sampled instead of collecting data for all of them. Describe the mean, standard deviation, and shape of the sampling distribution of the sample mean. b. Using this sampling distribution, find the \(z\) -score for a sample mean of \(\$ 20,000\) c. Using parts a and b, find the probability that the sample mean would fall within approximately \(\$ 1000\) of the population mean.

Short Answer

Expert verified
Mean = \$18,367; SE = \$470.9; Shape: Normal. Z = 3.47. P(\$1000 range) ≈ 96.61%.

Step by step solution

01

Define the Mean and Standard Deviation of the Sampling Distribution

For any random sample of size \( n \), the mean of the sampling distribution of the sample mean \( \bar{x} \) is equal to the population mean \( \mu \). In this case, \( \mu = 18,367 \).The standard deviation of the sampling distribution (standard error \( SE \)) is given by the formula: \[ SE = \frac{\sigma}{\sqrt{n}} \]where \( \sigma \) is the population standard deviation.For a sample size of 100, the standard error is:\[ SE = \frac{4709}{\sqrt{100}} = \frac{4709}{10} = 470.9 \]
02

Determine the Shape of the Sampling Distribution

The shape of the sampling distribution of the sample mean is approximately normal if the sample size \( n \) is sufficiently large, typically \( n \geq 30 \). Since 100 students are sampled, the Central Limit Theorem tells us that the distribution will be approximately normal.
03

Calculate the Z-score for a Sample Mean of $20,000

To find the \( z \)-score for a sample mean \( \bar{x} = 20,000 \), use the formula:\[ z = \frac{\bar{x} - \mu}{SE} \]Plugging the numbers in gives:\[ z = \frac{20000 - 18367}{470.9} \approx \frac{1633}{470.9} \approx 3.47 \]
04

Find the Probability Corresponding to the Z-score

To find the probability that the sample mean is within \( \$1000 \) of the population mean, calculate the z-scores for both ends of the interval:- Lower end: \( \mu - 1000 = 17367 \)- Upper end: \( \mu + 1000 = 19367 \)Calculate the \( z \)-scores:\[ z_{lower} = \frac{17367 - 18367}{470.9} = \frac{-1000}{470.9} \approx -2.12 \]\[ z_{upper} = \frac{19367 - 18367}{470.9} = \frac{1000}{470.9} \approx 2.12 \]Using the standard normal distribution table, find the probability between \( z_{lower} \) and \( z_{upper} \). Probability between these z-scores is approximately \( 0.9830 - 0.0169 = 0.9661 \).
05

Interpret the Probability

Given the probability is approximately 0.9661, there is about a 96.61% chance that the sample mean of a randomly selected sample of 100 students falls within \\(1000 of the population mean \\)18,367.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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's crucial for understanding why sample distributions often look like a normal distribution, especially with a large sample size. The CLT states that the sampling distribution of the sample mean will approximate a normal distribution as the sample size becomes larger, regardless of the original population's distribution.

This is incredibly useful because it allows statisticians to make inferences about population means even if the population data itself isn't normally distributed. In the context of the exercise, students are sampled to understand their average debt. Even if student debts have a skewed distribution, by sampling 100 students, the distribution of the sample mean becomes approximately normal. This is why the shape of our sampling distribution is normal, enabling us to apply various statistical techniques confidently.

In practice, the CLT kicks in strongly when the sample size is 30 or more, but more is better. Here, the sample size is 100, well within the range where the CLT is effective, so we can trust the results drawn from our sample distribution.
z-score
The z-score is a measure used to express how far data is from the mean in terms of standard deviations. Calculating the z-score allows you to understand where a particular sample mean lies within the sampling distribution. It's crucial for hypothesis testing and probability calculations.

For a given sample mean, the z-score is determined by taking the difference between the sample mean (\( \bar{x} \)) and the population mean (\( \mu \)), and dividing by the standard error (SE):
  • Formula: \[ z = \frac{\bar{x} - \mu}{SE} \]
In the exercise, the sample mean is \( \\(20,000 \). The population mean \( \mu \) is \( \\)18,367 \) and the standard error is \(470.9\). Plugging these values in calculates a z-score of approximately \(3.47\).

This tells us how many standard deviations the sample mean is away from the population mean. A z-score of 3.47 indicates the sample mean is quite a bit higher than the overall average, something that occurs less frequently if the data is normally distributed.
Standard Error
The Standard Error (SE) is like a "yardstick" for sampling distributions. It measures the spread of the sample means around the population mean, providing insight into how much the sample mean is expected to vary from the true population mean.

To calculate the standard error, divide the population standard deviation (\( \sigma \)) by the square root of the sample size (\( n \)):
  • Formula: \[ SE = \frac{\sigma}{\sqrt{n}} \]
In the original exercise, the population standard deviation is \( \$4709 \) and the sample size is \( 100 \). Therefore, the standard error is calculated as \(\frac{4709}{10} = 470.9\).

The smaller the SE, the less variance there is, meaning the sample mean is a more accurate estimate of the population mean. This accuracy is dependent on both the sample size and the population's standard deviation. A smaller standard deviation and larger sample size both contribute to a smaller standard error, which is generally desired in statistical analysis.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

According to the website http://www.digitalbookworld.com, the average price of a bestselling ebook increased to \(\$ 8.05\) in the week of February 18,2015 from \(\$ 6.89\) in the previous week. Assume the standard deviation of the price of a bestselling ebook is \(\$ 1\) and suppose you have a sample of 20 bestselling ebooks with a sample mean of \(\$ 7.80\) and a standard deviation of \(\$ 0.95\) a. Identify the random variable \(X\) in this study. Indicate whether it is quantitative or categorical. b. Describe the center and variability of the population distribution. What would you predict as the shape of the population distribution? Explain. c. Describe the center and variability of the data distribution. What would you predict as the shape of the data distribution? Explain. d. Describe the center and variability of the sampling distribution of the sample mean for 20 bestselling ebooks. What would you predict as the shape of the sampling distribution? Explain.

An exam consists of 50 multiplechoice questions. Based on how much you studied, for any given question you think you have a probability of \(p=0.70\) of getting the correct answer. Consider the sampling distribution of the sample proportion of the 50 questions on which you get the correct answer. a. Find the mean and standard deviation of the sampling distribution of this proportion. b. What do you expect for the shape of the sampling distribution? Why? c. If truly \(p=0.70\), would it be very surprising if you got correct answers on only \(60 \%\) of the questions? Justify your answer by using the normal distribution to approximate the probability of a sample proportion of 0.60 or less.

Which of the following is not correct? The standard deviation of a statistic describes a. The standard deviation of the sampling distribution of that statistic. b. The standard deviation of the sample data measurements. c. How close that statistic falls to the parameter that it estimates. d. The variability in the values of the statistic for repeated random samples of size \(n\).

According to The American Heart Association in \(2011,\) about 1 in every 3 deaths in the U.S. was from cardiovascular diseases. (Source: Data from www.heart.org, December \(17,2014 .)\) a. For a random sample of 100 deaths in America in 2011 , find the mean and standard deviation of the proportion of those who died from cardiovascular diseases. b. In a sample of 100 deaths observed in a specific hospital in the U.S. in \(2011,\) half of the deaths were from cardiovascular diseases. Would this have been a surprising result if the sample were a random sample of Americans? Answer by finding how many standard deviations that sample result falls from the mean of the sampling distribution of the proportion of 100 deaths from cardiovascular diseases. c. In part b, identify the population distribution, the data distribution, and the sampling distribution of the sample proportion.

Suppose \(x=1\) with probability \(p\), and \(x=0\) with probability \((1-p)\). Then, \(x\) is the special case of a binomial random variable with \(n=1,\) so that \(\sigma=\sqrt{n p(1-p)}=\sqrt{p(1-p)} .\) With \(n\) trials, using the formula \(\sigma / \sqrt{n}\) for a standard deviation of a sample mean, explain why the standard deviation of a sample proportion equals \(\sqrt{p(1-p) / n}\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.