The range for a set of data is estimated to be 36 a. What is the planning value for the population standard deviation? b. At \(95 \%\) confidence, how large a sample would provide a margin of error of \(3 ?\) c. At \(95 \%\) confidence, how large a sample would provide a margin of error of \(2 ?\)

Short Answer

Expert verified
a) The planning value for the population standard deviation is 9. b) At \(95\%\) confidence, a sample size of 34 would provide a margin of error of \(3\). c) At \(95\%\) confidence, a sample size of 76 would provide a margin of error of \(2\).

Step by step solution

01

Calculate the planning value for the population standard deviation

To find the planning value for the population standard deviation, we can use the formula: \(s = \frac{Range}{IQRScalingFactor}\) For this problem, it is given that the range of the data is 36. The IQRScalingFactor is usually taken as 4 to approximate the standard deviation. Using these values, we can calculate the planning value for the population standard deviation: \(s = \frac{36}{4} = 9\)
02

Formula for Margin of Error

To find the necessary sample size, we need to use the formula for the margin of error: \(Margin\ of\ Error = Z_{\frac{1-\alpha}{2}}*\frac{s}{\sqrt{n}}\) \(\alpha\) is the significance level, which is equal to \(1\) minus the confidence level. Since we are considering a \(95\%\) confidence level, \(\alpha = 1-0.95 = 0.05\). For a \(95\%\) confidence interval, the critical Z value, \(Z_{\frac{1-\alpha}{2}}= 1.96\), which can be found from a standard normal table.
03

Calculate sample size for a margin of error of 3

We will now rearrange the formula to find the sample size 'n' for a margin of error of 3: \(n = (\frac{Z_{\frac{1-\alpha}{2}}*s}{Margin\ of\ Error})^2\) Plugging in the values: \(n = (\frac{1.96*9}{3})^2\) \(n \approx 33.64\) Since we need a whole number for sample size, we will round up to the nearest whole number; \(n = 34\)
04

Calculate sample size for a margin of error of 2

Similarly, we will now find the sample size 'n' for a margin of error of 2: \(n = (\frac{Z_{\frac{1-\alpha}{2}}*s}{Margin\ of\ Error})^2\) Plugging values: \(n = (\frac{1.96*9}{2})^2\) \(n \approx 75.6\) Again, rounding up to the nearest whole number; \(n = 76\) In conclusion: a) The planning value for the population standard deviation is 9. b) At \(95\%\) confidence, a sample size of 34 would provide a margin of error of \(3\). c) At \(95\%\) confidence, a sample size of 76 would provide a margin of error of \(2\).

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 Vaia!

Key Concepts

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

Range of Data
The range of data is a measure of dispersion or variability that indicates the difference between the highest and lowest values in a dataset. It is calculated simply by subtracting the smallest value from the largest value. Understanding the range is fundamental in statistics because it provides a quick sense of the spread of values in the dataset.

For example, in a set of test scores where the highest score is 95 and the lowest is 55, the range would be 40 points (95 - 55 = 40). However, the range doesn't take into account how the values are distributed between the highest and lowest points. This is where other measures, like standard deviation, come into play as they consider the variation of each data point from the mean.

When estimating the population standard deviation from the range, as shown in the exercise solution, the range is divided by an interquartile range scaling factor, often 4, to approximate the standard deviation. This method assumes a bell-shaped distribution (normal distribution) of the data.
Margin of Error
In statistics, the margin of error measures the extent of the possible discrepancy between the sample estimate and the actual population parameter. It is an expression of the amount of random sampling error in the results of a survey or experiment. The margin of error is critically important in understanding how reliable the findings are from sample data.

The formula for margin of error is essentially a function of three key elements: the standard deviation of the population (\( s \)), the size of the sample (\( n \)), and the level of confidence one wants to have in the estimate. A higher confidence level or a lower margin of error requires a larger sample size. Conversely, if a researcher is willing to accept a higher margin of error, the required sample size can be smaller.

Understanding the relationship between the margin of error and sample size is crucial, particularly in planning studies and surveys to ensure sufficient precision of the results without unnecessarily oversampling.
Sample Size Calculation
Calculating the correct sample size is essential for obtaining statistically significant results. It is a balance between precision, confidence, and the resources available. The formula for calculating sample size takes into account the desired margin of error, the standard deviation of the population, and the z-value corresponding to the desired level of confidence.

To obtain the sample size, statisticians rearrange the margin of error formula, isolating the sample size (\( n \)). This produces a formula that predicts how large a sample is needed to estimate the population parameter with a specified margin of error and confidence level. In practical terms, a larger sample will produce a more precise estimate but will also require more resources to collect.

As observed in the provided exercise, the required sample size increases as the desired margin of error decreases. It's also important to round up the calculated sample size to the next whole number, as you cannot have a fraction of a sample. When planning experiments or surveys, the calculation of sample size is a fundamental step that influences the quality and credibility of the findings.

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

The National Quality Research Center at the University of Michigan provides a quarterly measure of consumer opinions about products and services (The Wall Street Journal, February 18,2003 ). A survey of 10 restaurants in the Fast Food/Pizza group showed a sample mean customer satisfaction index of \(71 .\) Past data indicate that the population standard deviation of the index has been relatively stable with \(\sigma=5\) a. What assumption should the researcher be willing to make if a margin of error is desired? b. Using \(95 \%\) confidence, what is the margin of error? c. What is the margin of error if \(99 \%\) confidence is desired?

For a \(t\) distribution with 16 degrees of freedom, find the area, or probability, in each region. a. To the right of 2.120 b. To the left of 1.337 c. To the left of -1.746 d. To the right of 2.583 e. Between -2.120 and 2.120 f. Between -1.746 and 1.746

The average cost per night of a hotel room in New York City is \(\$ 273\) (SmartMoney, March 2009 . Assume this estimate is based on a sample of 45 hotels and that the sample standard deviation is \(\$ 65\) a. With \(95 \%\) confidence, what is the margin of error? b. What is the \(95 \%\) confidence interval estimate of the population mean? c. Two years ago the average cost of a hotel room in New York City was \(\$ 229 .\) Discuss the change in cost over the two-year period.

Although airline schedules and cost are important factors for business travelers when choosing an airline carrier, a USA Today survey found that business travelers list an airline's frequent flyer program as the most important factor. From a sample of \(n=1993\) business travelers who responded to the survey, 618 listed a frequent flyer program as the most important factor. a. What is the point estimate of the proportion of the population of business travelers who believe a frequent flyer program is the most important factor when choosing an airline carrier? b. Develop a \(95 \%\) confidence interval estimate of the population proportion. c. How large a sample would be required to report the margin of error of .01 at \(95 \%\) confidence? Would you recommend that \(U S A\) Today attempt to provide this degree of precision? Why or why not?

In an effort to estimate the mean amount spent per customer for dinner at a major Atlanta restaurant, data were collected for a sample of 49 customers. Assume a population standard deviation of \$5. a. At \(95 \%\) confidence, what is the margin of error? b. If the sample mean is \(\$ 24.80\), what is the \(95 \%\) confidence interval for the population mean?

See all solutions

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free