/*! 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 13 A market researcher wants to eva... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A market researcher wants to evaluate car insurance savings at a competing company. Based on past studies he is assuming that the standard deviation of savings is $$\$ 100$$. He wants to collect data such that he can get a margin of error of no more than $$\$ 10$$ at a \(95 \%\) confidence level. How large of a sample should he collect?

Short Answer

Expert verified
A sample size of 385 is required.

Step by step solution

01

Identify the Known Values

We know the standard deviation of the savings is \( \sigma = 100 \) dollars, the margin of error is \( E = 10 \) dollars, and the confidence level is \( 95\% \). These values will be used to determine the sample size.
02

Determine the Z-Score

For a \( 95\% \) confidence level, the Z-score, which is the number of standard deviations a data point is from the mean in a standard normal distribution, is approximately \( 1.96 \).
03

Use the Margin of Error Formula

The margin of error in a normal distribution is calculated by the formula \[ E = Z \frac{\sigma}{\sqrt{n}} \]. We need to solve for \( n \) (the sample size) using this equation.
04

Rearrange the Formula to Solve for Sample Size

Rearrange the formula from Step 3 to isolate \( n \):\[ n = \left( \frac{Z \sigma}{E} \right)^2 \].
05

Plug Values into the Formula

Substitute the known values into the formula from Step 4: \[ n = \left( \frac{1.96 \times 100}{10} \right)^2 \].
06

Calculate the Sample Size

After performing the calculation:\[ n = \left( \frac{196}{10} \right)^2 = \left( 19.6 \right)^2 = 384.16 \].Since the sample size must be a whole number, round up to the nearest whole number, so \( n = 385 \).

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.

Margin of Error
The margin of error (MOE) is a crucial concept in statistics and research studies. It helps us understand the range within which we can expect our sample estimate to fall, relative to the true population parameter. In other words, it tells us how much a sample statistic, like a mean or proportion, might differ from the actual population value.
Imagine you’re making an educated guess about something; the margin of error represents how far off this guess might be. It is determined by several factors: the confidence level, the standard deviation, and the sample size.
  • **Confidence Level**: The higher the confidence level, the larger the margin of error.
  • **Standard Deviation**: Greater variability in the data results in a larger margin of error.
  • **Sample Size**: A larger sample size reduces the margin of error.
To calculate the margin of error, you use the formula: \[ E = Z \frac{\sigma}{\sqrt{n}} \] where
  • \(E\) is the margin of error,
  • \(Z\) is the Z-score associated with your desired confidence level,
  • \(\sigma\) is the standard deviation,
  • \(n\) is the sample size.
Confidence Interval
A confidence interval provides a range of values within which we can be fairly certain the population parameter lies. It's like saying, "I’m confident that the true value is between X and Y." The width of this interval is influenced by the margin of error.
When researchers report results using confidence intervals, they aren't giving a precise number. Instead, they provide a range that reflects their certainty level. The most common confidence levels are 90%, 95%, and 99%. For example, a 95% confidence interval suggests that if the same study were repeated 100 times, the interval would contain the actual parameter 95 times.
To construct a confidence interval, use this formula: \[ \text{Confidence Interval} = \text{Sample Statistic} \pm E \] where
  • \(E\) is the margin of error.
Confidence intervals convey the reliability of an estimate, with narrower intervals indicating more precise estimates given a fixed confidence level.
Standard Deviation
Standard deviation is a measure of how spread out numbers are in a data set. It indicates the average distance of each data point from the mean, allowing researchers to understand the variability or consistency of the data.
A smaller standard deviation means data points are closer to the mean, indicating less variability, whereas a larger standard deviation shows more spread, suggesting greater variability.
  • In the field of statistics, standard deviation is symbolized by the Greek letter \(\sigma\).
  • It plays a critical role in the calculation of both the confidence interval and margin of error.
  • The standard deviation affects how confident we are in our estimates: more variability typically requires a larger sample to obtain a given confidence level.
To calculate standard deviation, first find the variance by averaging the squared differences between each data point and the mean, then take the square root of that average. The formula for standard deviation is:\[\sigma = \sqrt{\frac{1}{N} \sum_{i=1}^{N} (x_i - \bar{x})^2}\]where
  • \(N\) is the number of data points,
  • \(x_i\) is each individual data point, and
  • \(\bar{x}\) is the mean of the data points.
Z-score
The Z-score is a statistical measurement that describes a value's relationship to the mean of a group of values. It is measured in terms of standard deviations from the mean.
Essentially, a Z-score tells us how many standard deviations away a particular data point is from the average of the data set. A positive Z-score indicates the data point is above the mean, while a negative Z-score shows it's below the mean.
For calculating sample size in the context of a specific confidence level, the Z-score provides the number of standard deviations we must look away from the mean in a standard normal distribution. Commonly used Z-scores include:
  • 1.645 for 90% confidence level
  • 1.96 for 95% confidence level
  • 2.576 for 99% confidence level
In the example of finding a sample size with a 95% confidence level, the Z-score of 1.96 signifies the number of standard deviations required to encompass 95% of the data under a normal distribution curve. The Z-score is integral in the formula for calculating margin of error and determining required sample size:\[ E = Z \frac{\sigma}{\sqrt{n}} \] By understanding the Z-score, researchers can accurately interpret data and ensure their study results are reliable and valid.

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

We considered the change in the number of days exceeding \(90^{\circ} \mathrm{F}\) from 1948 and 2018 at 197 randomly sampled locations from the NOAA database in Exercise \(7.19 .\) The mean and standard deviation of the reported differences are 2.9 days and 17.2 days. (a) Calculate a \(90 \%\) confidence interval for the average difference between number of days exceeding \(90^{\circ} \mathrm{F}\) between 1948 and 2018 . We've already checked the conditions for you. (b) Interpret the interval in context. (c) Does the confidence interval provide convincing evidence that there were more days exceeding \(90^{\circ} \mathrm{F}\) in 2018 than in 1948 at NOAA stations? Explain.

Georgianna claims that in a small city renowned for its music school, the average child takes less than 5 years of piano lessons. We have a random sample of 20 children from the city, with a mean of 4.6 years of piano lessons and a standard deviation of 2.2 years. (a) Evaluate Georgianna's claim (or that the opposite might be true) using a hypothesis test. (b) Construct a \(95 \%\) confidence interval for the number of years students in this city take piano lessons, and interpret it in context of the data. (c) Do your results from the hypothesis test and the confidence interval agree? Explain your reasoning.

The standard deviation of SAT scores for students at a particular Ivy League college is 250 points. Two statistics students, Raina and Luke, want to estimate the average SAT score of students at this college as part of a class project. They want their margin of error to be no more than 25 points. (a) Raina wants to use a \(90 \%\) confidence interval. How large a sample should she collect? (b) Luke wants to use a \(99 \%\) confidence interval. Without calculating the actual sample size, determine whether his sample should be larger or smaller than Raina's, and explain your reasoning. (c) Calculate the minimum required sample size for Luke.

A large farm wants to try out a new type of fertilizer to evaluate whether it will improve the farm's corn production. The land is broken into plots that produce an average of 1,215 pounds of corn with a standard deviation of 94 pounds per plot. The owner is interested in detecting any average difference of at least 40 pounds per plot. How many plots of land would be needed for the experiment if the desired power level is \(90 \%\) ? Assume each plot of land gets treated with either the current fertilizer or the new fertilizer.

Caffeine is the world's most widely used stimulant, with approximately \(80 \%\) consumed in the form of coffee. Participants in a study investigating the relationship between coffee consumption and exercise were asked to report the number of hours they spent per week on moderate (e.g., brisk walking) and vigorous (e.g., strenuous sports and jogging) exercise. Based on these data the researchers estimated the total hours of metabolic equivalent tasks (MET) per week, a value always greater than \(0 .\) The table below gives summary statistics of MET for women in this study based on the amount of coffee consumed. \(^{36}\) (a) Write the hypotheses for evaluating if the average physical activity level varies among the different levels of coffee consumption. (b) Check conditions and describe any assumptions you must make to proceed with the test. (c) Below is part of the output associated with this test. Fill in the empty cells. (d) What is the conclusion of the test?

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.