/*! 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 52 In Exercises 3.51 to 3.56 , info... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In Exercises 3.51 to 3.56 , information about a sample is given. Assuming that the sampling distribution is symmetric and bell-shaped, use the information to give a \(95 \%\) confidence interval, and indicate the parameter being estimated. $$ \bar{x}=55 \text { and the standard error is } 1.5 . $$

Short Answer

Expert verified
The \(95\%\) confidence interval lies between \(55 - 1.96 * 1.5\) and \(55 + 1.96 * 1.5\), and the parameter being estimated is the population mean.

Step by step solution

01

Identify the parameter

Here, we are trying to estimate the population mean \(\mu\). The parameter being estimated in this scenario is therefore the population mean, which we are getting via the sample mean.
02

Apply the formula for confidence interval

The formula to find the \(95\%\) Confidence Interval for a normal distribution is: \(\bar{x} ± Z*(SE)\). Here, \(\bar{x}\) is the sample mean, \(Z\) is the Z-value corresponding to the desired confidence level, and \(SE\) is the standard error.
03

Substitute the given values into the formula

The given values are \(\bar{x} = 55\), \(Z = 1.96\) (Z-value for a 95% confidence level) and \(SE = 1.5\). Substituting these values into the formula gives: \(55 ± 1.96 * 1.5\)
04

Calculation

The calculation yields: \(55 - 1.96 * 1.5\) and \(55 + 1.96 * 1.5\), which gives you the lower and upper limits of the confidence interval respectively.
05

Result interpretation

The resulting values represent the range within which we can be \(95\%\) certain that the population mean lies.

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.

Sample Mean
The sample mean, often denoted as \( \bar{x} \), is the average value of a numerical dataset. Simply put, it is the sum of all sample observations divided by the number of observations in the sample.
It's a useful measure that suggests the central location of data points from a sample.The sample mean is used as an estimator for the population mean, \( \mu \). While the sample mean itself is derived solely from the sample data, it serves as a key component in statistical methods like confidence intervals and hypothesis testing.
By employing a sample mean from a data subset, statisticians can make inferences about the entire population.Here are some points to consider about the sample mean:
  • It is sensitive to outliers, meaning extreme values can significantly shift the mean.
  • It is easy to compute and provides a straightforward estimate of the center of the data's distribution.
  • Helps in estimating the population mean when the sample is random and representative of the population intended to study.
Population Mean
The population mean, denoted by the Greek letter \( \mu \), is a pivotal concept in statistics. It represents the average of all possible values in a given population. Unlike the sample mean, which is calculated from a subset, the population mean accounts for every single data point in the entire set.
This comprehensive characteristic of the population mean makes it often impractical to calculate directly, especially when dealing with large populations.In statistics, we use estimators like the sample mean to make educated guesses about this true population parameter.
  • It reflects the actual center of the entire data set without the influence of sampling variability.
  • Whenever we state a population mean, it implies we are including every data point in the population.
  • It is usually known only when an entire population can be measured directly.
  • Using a confidence interval, we can estimate the range within which the population mean is likely to fall with a given confidence level, like 95%.
Standard Error
The standard error (SE) plays a crucial role in statistics. It measures the precision of a sample mean as an estimate of the population mean. Essentially, it indicates how much variability one can expect in the sample mean from sample to sample due to random sampling.
A smaller standard error suggests a more accurate estimate of the population mean, indicating that the sample mean is close to the true mean.The formula for standard error depends on the variability of the population and the size of the sample:\[SE = \frac{\sigma}{\sqrt{n}}\]where \( \sigma \) is the standard deviation of the population, and \( n \) is the sample size.Key points about the standard error:
  • Larger samples yield a smaller SE, because they better represent the population.
  • For a fixed sample size, greater population variability leads to a larger SE.
  • It is instrumental in constructing confidence intervals and conducting hypothesis testing.
  • The SE decreases as the sample size increases, provided that the population variance remains constant.
By understanding the standard error, we gain insights into the reliability of sample statistics as estimates of population parameters.

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

A Sampling Distribution for Performers in the Rock and Roll Hall of Fame Exercise 3.38 tells us that 206 of the 303 inductees to the Rock and Roll Hall of Fame have been performers. The data are given in RockandRoll. Using all inductees as your population: (a) Use StatKey or other technology to take many random samples of size \(n=10\) and compute the sample proportion that are performers. What is the standard error of the sample proportions? What is the value of the sample proportion farthest from the population proportion of \(p=0.68 ?\) How far away is it? (b) Repeat part (a) using samples of size \(n=20\). (c) Repeat part (a) using samples of size \(n=50\). (d) Use your answers to parts (a), (b), and (c) to comment on the effect of increasing the sample size on the accuracy of using a sample proportion to estimate the population proportion.

Average Salary of NFL Players The dataset NFLContracts2015 contains the yearly salary (in millions of dollars) from the contracts of all players on a National Football League (NFL) roster at the start of the 2015 season. \({ }^{19}\) (a) Use StatKey or other technology to select a random sample of 5 NFL contract YearlySalary values. Indicate which players you've selected and compute the sample mean. (b) Repeat part (a) by taking a second sample of 5 values, again indicating which players you selected and computing the sample mean. (c) Find the mean for the entire population of players. Include notation for this mean. Comment on the accuracy of using the sample means found in parts (a) and (b) to estimate the population mean.

Effect of Overeating for One Month: Average Long-Term Weight Gain Overeating for just four weeks can increase fat mass and weight over two years later, a Swedish study shows \(^{35}\) Researchers recruited 18 healthy and normal-weight people with an average age of \(26 .\) For a four-week period, participants increased calorie intake by \(70 \%\) (mostly by eating fast food) and limited daily activity to a maximum of 5000 steps per day (considered sedentary). Not surprisingly, weight and body fat of the participants went up significantly during the study and then decreased after the study ended. Participants are believed to have returned to the diet and lifestyle they had before the experiment. However, two and a half years after the experiment, the mean weight gain for participants was 6.8 lbs with a standard error of 1.2 lbs. A control group that did not binge had no change in weight. (a) What is the relevant parameter? (b) How could we find the actual exact value of the parameter? (c) Give a \(95 \%\) confidence interval for the parameter and interpret it. (d) Give the margin of error and interpret it.

Automobile Depreciation For a random sample of 20 automobile models, we record the value of the model as a new car and the value after the car has been purchased and driven 10 miles. \({ }^{47}\) The difference between these two values is a measure of the depreciation on the car just by driving it off the lot. Depreciation values from our sample of 20 automobile models can be found in the dataset CarDepreciation. (a) Find the mean and standard deviation of the Depreciation amounts in CarDepreciation. (b) Use StatKey or other technology to create a bootstrap distribution of the sample mean of depreciations. Describe the shape, center, and spread of this distribution. (c) Use the standard error obtained in your bootstrap distribution to find and interpret a \(95 \%\) confidence interval for the mean amount a new car depreciates by driving it off the lot.

3.62 Employer-Based Health Insurance A report from a Gallup poll \(^{29}\) in 2011 started by saying, "Forty-five percent of American adults reported getting their health insurance from an employer...." Later in the article we find information on the sampling method, "a random sample of 147,291 adults, aged 18 and over, living in the US," and a sentence about the accuracy of the results, "the maximum margin of sampling error is ±1 percentage point." (a) What is the population? What is the sample? What is the population parameter of interest? What is the relevant statistic? (b) Use the margin of error \(^{30}\) to give an interval showing plausible values for the parameter of interest. Interpret it in terms of getting health insurance from an employer.

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.