/*! 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 10 Suppose that a random sample of ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose that a random sample of size 64 is to be selected from a population with mean 40 and standard deviation 5. a. What are the mean and standard deviation of the sampling distribution of \(\bar{x}\) ? Describe the shape of the sampling distribution of \(\bar{x}\). b. What is the approximate probability that \(\bar{x}\) will be within 0.5 of the population mean \(\mu\) ? c. What is the approximate probability that \(\bar{x}\) will differ from \(\mu\) by more than \(0.7 ?\)

Short Answer

Expert verified
a. The mean and standard deviation of the sampling distribution of \(\bar{x}\) are 40 and 0.625, respectively. The shape of the sampling distribution of \(\bar{x}\) is approximately normally distributed. b. The approximate probability that \(\bar{x}\) will be within 0.5 of the population mean \(\mu\) is 0.5774. c. The approximate probability that \(\bar{x}\) will differ from \(\mu\) by more than 0.7 is 0.2628.

Step by step solution

01

Find Mean and Standard Deviation of the Sampling Distribution of \(\bar{x}\)#

According to the Central Limit Theorem, the mean of the sampling distribution of \(\bar{x}\) is equal to the population mean (\(\mu\)), and the standard deviation of the sampling distribution of \(\bar{x}\) is equal to the population standard deviation (\(\sigma\)) divided by the square root of the sample size (\(n\)). Mean of the sampling distribution of \(\bar{x}\) (\(\mu_{\bar{x}}\)) = \(\mu\) = 40. Standard deviation of the sampling distribution of \(\bar{x}\) (\(\sigma_{\bar{x}}\)) = \(\frac{\sigma}{\sqrt{n}}\) = \(\frac{5}{\sqrt{64}} = \frac{5}{8} = 0.625\).
02

Describe the Shape of the Sampling Distribution of \(\bar{x}\)#

The Central Limit Theorem states that the shape of the sampling distribution of the sample mean approaches a normal distribution as the sample size increases. Since our sample size is 64, which is large enough, we can assume that the sampling distribution of \(\bar{x}\) will be approximately normally distributed.
03

Calculate the Probability of \(\bar{x}\) Being Within 0.5 of the Population Mean (\(| \bar{x} - \mu | \leq 0.5\))#

To calculate the probability, we will use the z-score, which is defined as the number of standard deviations a value is away from the mean: \(z = \frac{(\bar{x} - \mu)}{\sigma_{\bar{x}}}\). Now, we want to find the probability that the sample mean is within 0.5 of the population mean: \(| \bar{x} - \mu | \leq 0.5\). This can be rewritten as two z-scores: \(z_1 = \frac{(\mu - 0.5 - \mu)}{\sigma_{\bar{x}}}\) and \(z_2 = \frac{(\mu + 0.5 - \mu)}{\sigma_{\bar{x}}}\). Calculating the z-scores: \(z_1 = \frac{-0.5}{0.625} = -0.8\), \(z_2 = \frac{0.5}{0.625} = 0.8\). Now, we can look up these z-scores in the standard normal distribution z-table or use a calculator to find the probability that the z-score is between -0.8 and 0.8, which is approximately \(0.7887 - 0.2113 = 0.5774\).
04

Calculate the Probability of \(\bar{x}\) Differing from \(\mu\) by More than 0.7 (\(|\bar{x} - \mu| > 0.7\))#

To find the probability of the sample mean differing from the population mean by more than 0.7, we need to calculate two z-scores again: \(z_3 = \frac{(\mu - 0.7 - \mu)}{\sigma_{\bar{x}}}\) and \(z_4 = \frac{(\mu + 0.7 - \mu)}{\sigma_{\bar{x}}}\). Calculating the z-scores: \(z_3 = \frac{-0.7}{0.625} = -1.12\), \(z_4 = \frac{0.7}{0.625} = 1.12\). Now, find the probability for z-scores between -1.12 and 1.12 in the standard normal distribution z-table or use a calculator, which is approximately \(0.8686 - 0.1314 = 0.7372\). Finally, to find the probability of the sample mean differing from the population mean by more than 0.7, we need to subtract the probability found above from 1: Probability (\(|\bar{x} - \mu| > 0.7\)) = \(1 - 0.7372 = 0.2628\).

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 statistical principle with profound importance in probability theory. When dealing with the mean of independent and identically distributed random variables, the CLT tells us that as the sample size grows, the distribution of the sample means will approximate a normal distribution (also known as a Gaussian distribution), irrespective of the original distribution of the data. This allows statisticians to make inferences about population parameters using the normal distribution, even when the sample comes from a non-normal distribution. This is key to understanding why, in our exercise, with a large sample size of 64, we can assume that the sampling distribution of \(\bar{x}\) is approximately normally distributed.
standard deviation
Standard deviation is a measure that quantifies the amount of variation or dispersion of a set of data values. A low standard deviation indicates that the values tend to be close to the mean (also called the expected value) of the set, while a high standard deviation indicates that the values are spread out over a wider range. In the exercise, we calculated the standard deviation of the sampling distribution of the sample mean, \(\sigma_{\bar{x}}\), which is the population standard deviation \(\sigma\) divided by the square root of the sample size \(n\). This calculation of the standard deviation of the sampling distribution is crucial because it allows us to measure the spread of all possible sample means about the population mean and, consequently, to calculate probabilities using the z-score.
z-score
A z-score, also known as a standard score, measures the number of standard deviations an individual data point or sample mean is from the mean of a distribution. It is a dimensionless quantity that provides a way to compare different data points from different normal distributions. In our context, calculating the z-score for the sample mean allows us to determine where it falls within the normal distribution of sample means. We can then use this score to calculate probabilities associated with that sample mean. For example, to find the probability that the sample mean is within a certain range of the population mean, as the exercise requested, we calculate corresponding z-scores and use the standard normal distribution to find these probabilities. The z-score transforms the problem into one that can be easily handled by referring to standard normal distribution tables or software calculations.
normal distribution
The normal distribution is a continuous probability distribution that is symmetrical around its mean, representing a bell-shaped curve where most of the observations cluster around the central peak and probabilities for values further from the mean taper off equally in both directions. It is paramount in many statistical analyses because of its unique properties and the CLT, which allows us to assume a normal distribution for the sampling distribution of the sample mean, given a large enough sample size. In the exercise, the normal distribution is the basis for comparing the sample mean to the population mean and calculating the associated probabilities of divergence between the two. The normal distribution is the reference point for z-scores, allowing us to determine the likelihood of observing sample means within specific ranges.

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

Acrylic bone cement is sometimes used in hip and knee replacements to secure an artificial joint in place. The force required to break an acrylic bone cement bond was measured for six specimens, and the resulting mean and standard deviation were 306.09 Newtons and 41.97 Newtons, respectively. Assuming that it is reasonable to believe that breaking force has a distribution that is approximately normal, use a confidence interval to estimate the mean breaking force for acrylic bone cement.

The formula used to calculate a confidence interval for the mean of a normal population is $$ \bar{x} \pm(t \text { critical value }) \frac{s}{\sqrt{n}} $$ What is the appropriate \(t\) critical value for each of the following confidence levels and sample sizes? a. \(90 \%\) confidence, \(n=12\) b. \(90 \%\) confidence, \(n=25\) c. \(95 \%\) confidence, \(n=10\)

The Economist collects data each year on the price of a Big Mac in various countries around the world. A sample of McDonald's restaurants in Europe in July 2016 resulted in the following Big Mac prices (after conversion to U.S. dollars): \(\begin{array}{llll}4.44 & 3.15 & 2.42 & 3.96\end{array}\) \(\begin{array}{llll}4.51 & 4.17 & 3.69 & 4.62\end{array}\) \(\begin{array}{lll}3.80 & 3.36 & 3.85\end{array}\) The mean price of a Big Mac in the U.S. in July 2016 was \$5.04. For purposes of this exercise, you can assume it is reasonable to regard the sample as representative of European McDonald's restaurants. Does the sample provide convincing evidence that the mean July 2016 price of a Big Mac in Europe is less than the reported U.S. price? Test the relevant hypotheses using \(\alpha=0.05 .\) (Hint: See Example 12.12.)

Explain the difference between \(\bar{x}\) and \(\mu_{\vec{x}}\).

A study of fast-food intake is described in the paper "What People Buy From Fast-Food Restaurants" (Obesity [2009]:1369-1374). Adult customers at three hamburger chains (McDonald's, Burger King, and Wendy's) in New York City were approached as they entered the restaurant at lunchtime and asked to provide their receipt when exiting. The receipts were then used to determine what was purchased and the number of calories consumed was determined. In all, 3857 people participated in the study. The sample mean number of calories consumed was 857 and the sample standard deviation was 677 . a. The sample standard deviation is quite large. What does this tell you about number of calories consumed in a hamburgerchain lunchtime fast-food purchase in New York City? b. Given the values of the sample mean and standard deviation and the fact that the number of calories consumed can't be negative, explain why it is not reasonable to assume that the distribution of calories consumed is normal. c. Based on a recommended daily intake of 2000 calories, the online Healthy Dining Finder (www.healthydiningfinder .com) recommends a target of 750 calories for lunch. Assuming that it is reasonable to regard the sample of 3857 fast-food purchases as representative of all hamburger-chain lunchtime purchases in New York City, carry out a hypothesis test to determine if the sample provides convincing evidence that the mean number of calories in a New York City hamburger-chain lunchtime purchase is greater than the lunch recommendation of 750 calories. Use \(\alpha=0.01\) d. Would it be reasonable to generalize the conclusion of the test in Part (c) to the lunchtime fast-food purchases of all adult Americans? Explain why or why not. e. Explain why it is better to use the customer receipt to determine what was ordered rather than just asking a customer leaving the restaurant what he or she purchased. f. Do you think that asking a customer to provide his or her receipt before they ordered could have introduced a bias? Explain.

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.