/*! 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 24 Finding Sample Size Instead of u... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 24

Finding Sample Size Instead of using Table \(7-2\) for determining the sample size required to estimate a population standard deviation \(\sigma,\) the following formula can be used $$n=\frac{1}{2}\left(\frac{z_{\alpha / 2}}{d}\right)^{2}$$ where \(z_{\alpha / 2}\) corresponds to the confidence level and \(d\) is the decimal form of the percentage error. For example, to be \(95 \%\) confident that \(s\) is within \(15 \%\) of the value of \(\sigma,\) use \(z_{\alpha / 2}=1.96\) and \(d=0.15\) to get a sample size of \(n=86 .\) Find the sample size required to estimate \(\sigma,\) assuming that we want \(98 \%\) confidence that \(s\) is within \(15 \%\) of \(\sigma .\)

Short Answer

Expert verified
The required sample size is 121.

Step by step solution

01

Identify the given values

Given: the confidence level is 98%, the percentage error is 15%, and the confidence level corresponds to a particular value of standard normal distribution.Convert the percentage error to a decimal form: 15% = 0.15.Find the value of \(z_{\frac{\beta}{2}}\) for a 98% confidence interval, which is 2.33.
02

Substitute values into the formula

Place the values from step 1 into the formula: $$n=\frac{1}{2}\times\frac{(z_{\frac{\beta}{2}})}{(d)}^{2}.$$ Here, \(z_{\frac{\beta}{2}}=2.33\) and \(d=0.15\).
03

Calculate the square of the z-value over the decimal error

Compute \({(z_{\frac{\beta}{2}}) /d}^{2}\) which is \(\frac{2.33}{0.15}^{2}\) and obtain \((15.53)^2=240.88\).
04

Finish the calculation

Divide the result by 2: $$n=\frac{1}{2}\times240.88=120.44$$ The sample size, rounded to the nearest integer, is 121.

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 91影视!

Key Concepts

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

confidence interval
A confidence interval is a range of values within which we expect a population parameter to lie, given a certain level of confidence. For instance, if we have a 98% confidence interval, we are saying that we are 98% sure that the true value of the population parameter falls within this range.

The width of the confidence interval depends on several factors:
  • The size of the sample
  • The variability in the data (measured by the standard deviation)
  • The confidence level we choose (e.g., 95%, 98%)
For a higher confidence level, the interval will be wider because higher confidence means more certainty, requiring a broader range of values.
population standard deviation
The population standard deviation, denoted as \(\sigma\), is a measure of the variability or spread of a set of values in a population. It tells us how much the values in the population deviate from the mean of the population.

If the standard deviation is small, the values are close to the mean, showing less variability. But if the standard deviation is large, the values are spread out over a wider range, indicating more variability.

To estimate the population standard deviation accurately, we often need to determine an appropriate sample size. Using the formula from the exercise, we can calculate this by combining the desired confidence level and the allowable error margin. Doing these calculations helps ensure our sample represents the population accurately.
z-value
A z-value, also known as a z-score, represents the number of standard deviations a value is from the mean. In the context of confidence intervals, the z-value corresponds to the point on the standard normal distribution for a given confidence level.

For example, a z-value of 1.96 is used for a 95% confidence level, meaning we are 95% confident that the sample mean lies within 1.96 standard deviations of the population mean.
In our exercise, we used a z-value of 2.33 for a 98% confidence level, which can be found using standard z-tables or statistical software.
  • First, recognize the confidence level needed (e.g., 98%)
  • Find the corresponding z-value (for 98%, z-value is 2.33)
  • Use the z-value in the formula to estimate the sample size
These steps ensure the calculated sample size provides an accurate estimate within the specified confidence level.

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

Construct the confidence interval estimate of the mean. In a test of weight loss programs, 40 adults used the Atkins weight loss program. After 12 months, their mean weight loss was found to be 2.1 lb, with a standard deviation of 4.8 ib. Construct a \(90 \%\) confidence interval estimate of the mean weight loss for all such subjects. Does the Atkins program appear to be effective? Does it appear to be practical?

Use the given data to find the minimum sample size required to estimate a population proportion or percentage. You plan to develop a new iOS social gaming app that you believe will surpass the success of Angry Birds and Facebook combined. In forecasting revenue, you need to estimate the percentage of all smartphone and tablet devices that use the iOS operating system versus Android and other operating systems. How many smartphones and tablets must be surveyed in order to be \(99 \%\) confident that your estimate is in error by no more than two percentage points? a. Assume that nothing is known about the percentage of portable devices using the iOS operating system. b. Assume that a recent survey suggests that about \(43 \%\) of smartphone and tablets are using the iOS operating system (based on data from NetMarkesShare). c. Does the additional survey information from part (b) have much of an effect on the sample size that is required?

Use the given data to find the minimum sample size required to estimate a population proportion or percentage. An epidemiologist plans to conduct a survey to estimate the percentage of women who give birth. How many women must be surveyed in order to be \(99 \%\) confident that the estimated percentage is in error by no more than two percentage points? a. Assume that nothing is known about the percentage to be estimated. b. Assume that a prior study conducted by the U.S. Census Bureau showed that \(82 \%\) of women give birth. c. What is wrong with surveying randomly selected adult women?

Construct the confidence interval estimate of the mean. Listed below are the amounts of net worth (in millions of dollars) of these ten wealthiest celebrities: Tom Cruise, Will Smith, Robert De Niro, Drew Carey, George Clooney, John Travolta, Samuel L. Jackson, Larry King. Demi Moore, and Bruce Willis. Construct a \(98 \%\) confidence interval. What does the result tell us about the population of all celebrities? Do the data appear to be from a normally distributed population as required? $$\begin{array}{rrrrrrrrrr} 250 & 200 & 185 & 165 & 160 & 160 & 150 & 150 & 150 & 150 \end{array}$$

Use the data and confidence level to construct a confidence interval estimate of \(p,\) then address the given question. Before its clinical trials were discontinued, the Genetics \& IVF Institute conducted a clinical trial of the XSORT method designed to increase the probability of conceiving a girl and, among the 945 babies born to parents using the XSORT method, there were 879 girls. The YSORT method was designed to increase the probability of conceiving a boy and, among the 291 babies born to parents using the YSORT method, there were 239 boys. Construct the two \(95 \%\) confidence interval estimates of the percentages of success. Compare the results. What do you conclude?
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Decision Maths

Read Explanation

Discrete Mathematics

Read Explanation

Applied Mathematics

Read Explanation

Calculus

Read Explanation

Mechanics Maths

Read Explanation
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.