/*! 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 15 The standard deviation for a pop... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The standard deviation for a population is \(\sigma=6.30\). A random sample selected from this population gave a mean equal to \(81.90\). The population is known to be normally distributed. a. Make a \(99 \%\) confidence interval for \(\mu\) assuming \(n=16\). b. Construct a \(99 \%\) confidence interval for \(\mu\) assuming \(n=20\). c. Determine a \(99 \%\) confidence interval for \(\mu\) assuming \(n=25\). d. Does the width of the confidence intervals constructed in parts a through c decrease as the sample size increases? Explain.

Short Answer

Expert verified
The 99% confidence interval for population mean, assuming \(n=16\) is (78.755, 85.045). Assuming \(n=20\), the confidence interval is (79.003, 84.797). Assuming \(n=25\), the confidence interval is (79.325, 84.475). As the sample size increases, the width of the confidence interval decreases implying that increasing sample size leads to better precision of the estimation of the population parameter.

Step by step solution

01

Calculation For n=16

Substitute the given values into the formula: \(\mu = 81.90 ± 2.576 * \frac{6.30}{\sqrt{16}}\). So the 99% confidence interval for the population mean is (78.755, 85.045)
02

Calculation For n=20

Substitute the given values into the formula: \(\mu = 81.90 ± 2.576 * \frac{6.30}{\sqrt{20}}\). So the 99% confidence interval for the population mean is (79.003, 84.797).
03

Calculation For n=25

Substitute the given values into the formula: \(\mu = 81.90 ± 2.576 * \frac{6.30}{\sqrt{25}}\). So the 99% confidence interval for the population mean is (79.325, 84.475).
04

Compare Confidence Interval Widths

Upon comparing the results from step 1 to 3, it is evident that as the sample size increases, the width of the confidence interval decreases. This is because as sample size increases, the standard error of the mean decreases, resulting in a tighter confidence interval. Hence, more precision is achieved in estimating the population parameter.

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 Size
The concept of sample size is integral when conducting statistical analyses, especially in determining confidence intervals. Sample size refers to the number of individual observations or data points collected from a larger population.
It is symbolized by "n" in statistical formulas. In general, a larger sample size provides more reliable and accurate estimates of the population parameters.
Understanding Sample Size Effects:
  • As the sample size increases, the estimate of the population parameter becomes more precise.
  • A larger sample size reduces the standard error, which is the measure of variability between the sample mean and the true population mean.
In the context of our exercise, increasing the sample size from 16 to 20, and then to 25, showed a clear reduction in the width of the confidence intervals. This illustrates that a larger sample size results in more confidence in the interval estimate of a population mean.
Standard Deviation
Standard deviation is a measure of how much individual data points deviate from the mean of the dataset. It is denoted by the symbol \(\sigma\).
In a population, it represents how spread out the numbers are. In our example, the population standard deviation is given as \(\sigma = 6.30\).
Key Features of Standard Deviation:
  • A higher standard deviation implies more variability and less predictability in the data.
  • It influences the width of the confidence intervals, wider for larger standard deviations.
Standard deviation is crucial in calculating the confidence interval, as it helps determine the standard error. The standard error, in turn, is used to estimate how much a sample mean is expected to vary from the true population mean. Hence, the knowledge of standard deviation, together with sample size, helps in creating more accurate and reliable confidence intervals.
Population Mean
The population mean, often represented by the Greek letter \(\mu\), is the average of all values in the population. It is a central measure in statistics and represents the expected value.
Estimating Population Mean:
  • Directly measuring the population mean is often impractical, especially if the population is large.
  • Instead, we use sample data to make estimates about the population mean.
In the exercise, we are tasked to construct confidence intervals for the population mean based on different sample sizes. Confidence intervals give us a range where we believe, with a certain level of confidence, the true population mean lies. For instance, a mean of 81.90 from our sample indicates our best estimate of the population mean, but not with certainty. Hence, using confidence intervals allows us to have a statistical assurance on the reliability of our estimate.
Normal Distribution
Normal distribution is a fundamental concept in statistics. It describes how data is symmetrically distributed around the mean, forming a bell-shaped curve.
The properties of normal distribution make it extremely useful for statistical methods, including confidence intervals.
Characteristics of Normal Distribution:
  • Most of the data points are clustered around the mean, with fewer and fewer as you move away.
  • It is fully defined by the mean and the standard deviation.
In the provided exercise, the underlying population is assumed to be normally distributed. This assumption allows the use of specific statistical properties to calculate confidence intervals effectively. For normally distributed data, about 68% of data lies within one standard deviation from the mean, 95% lies within two, and 99.7% lies within three. Hence, normal distribution provides a strong foundation for making estimations about population parameters when little is known about the actual population.

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

Fifteen randomly selected ripe Macintosh apples had the following weights (in ounces). \(\begin{array}{llllllll}8.9 & 6.8 & 7.2 & 8.3 & 8.1 & 7.9 & 7.1 & 8.0 \\ 8.5 & 6.7 & 7.0 & 7.4 & 7.7 & 6.2 & 9.2 & \end{array}\) Assume that the weight of a ripe Macintosh apple is normally distributed. Construct a \(95 \%\) confidence interval for the average weight of all ripe Macintosh apples.

A Centers for Disease Control and Prevention survey about cell phone use noted that \(14.7 \%\) of U.S. households are wireless-only, which means that the household members use only cell phones and do not have a landline. Suppose that this percentage is based on a random sample of 855 U.S. households (Source: http://www.cdc.gov/nchs/data/nhsr/nhsr014.pdf). a. Construct a \(95 \%\) confidence interval for the proportion of all U.S. households that are wireless-only. b. Explain why we need to construct a confidence interval. Why can we not simply say that \(14.7 \%\) of all U.S. households are wireless-only?

a. Find the value of \(t\) from the \(t\) distribution table for a sample size of 22 and a confidence level of \(95 \%\) b. Find the value of \(t\) from the \(t\) distribution table for 60 degrees of freedom and a \(90 \%\) confidence level. c. Find the value of \(t\) from the \(t\) distribution table for a sample size of 24 and a confidence level of \(99 \%\)

A jumbo mortgage is a mortgage with a loan amount above the industry-standard definition of conyentional conforming loan limits. As of January 2009, approximately \(2.57 \%\) of people who took out a jumbo mortgage during the previous 12 months were at least 60 days late on their payments. Suppose that this percentage is based on a random sample of 1430 people who took out a jumbo mortgage during the previous 12 months. a. Construct a \(95 \%\) confidence interval for the proportion of all people who took out a jumbo mortgage during the previous 12 months and were at least 60 days late on their payments. b. Suppose the confidence interval obtained in part a is too wide. How can the width of this interval be reduced? Discuss all possible alternatives. Which alternative is the best?

A drug that provides relief from headaches was tried on 18 randomly selected patients. The experiment showed that the mean time to get relief from headaches for these patients after taking this drug was 24 minutes with a standard deviation of \(4.5\) minutes. Assuming that the time taken to get relief from a headache after taking this drug is (approximately) normally distributed, determine a \(95 \%\) confidence interval for the mean relief time for this drug for all patients.

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.