/*! 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 57 A study conducted several years ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A study conducted several years ago reported that 21 percent of public accountants changed companies within 3 years. The American Institute of CPAs would like to update the study. They would like to estimate the population proportion of public accountants who changed companies within 3 years with a margin of error of \(3 \%\) and a \(95 \%\) level of confidence. a. To update this study, the files of how many public accountants should be studied? b. How many public accountants should be contacted if no previous estimates of the population proportion are available?

Short Answer

Expert verified
684 accountants if prior proportion is known; 1068 if not.

Step by step solution

01

Identify the Known Values

To find the required sample size, we need to identify certain known values: the previously reported proportion \( p = 0.21 \), the desired margin of error \( E = 0.03 \), and the critical z-value for a \(95\%\) confidence level, which is approximately \( z = 1.96 \).
02

Calculate Sample Size Using Previous Estimate

Use the formula for the sample size: \[ n = \left( \frac{z^2 \cdot p \cdot (1 - p)}{E^2} \right) \]Substitute the known values:\[ n = \left( \frac{1.96^2 \cdot 0.21 \cdot (1 - 0.21)}{0.03^2} \right) \]Calculate the value:\[ n \approx 1.96^2 \times \frac{0.21 \times 0.79}{0.0009} \approx 683.5 \]Round up to the nearest whole number since a sample size must be whole numbers.Thus, the sample size is \( n = 684 \).
03

Determine Sample Size with No Prior Proportion

If no prior estimate is available, assume \( p = 0.5 \) for the most conservative estimate. Use the same formula but with \( p = 0.5 \):\[ n = \left( \frac{1.96^2 \cdot 0.5 \cdot (1 - 0.5)}{0.03^2} \right) \]Calculate:\[ n \approx 1.96^2 \times \frac{0.5 \times 0.5}{0.0009} \approx 1067.1 \]Round up to the nearest whole number, making \( n = 1068 \).

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.

Population Proportion
When we talk about population proportion, we're referring to the fraction or percentage of the whole population that shares a specific characteristic. In the case of our exercise, it's the percentage of public accountants changing companies within a three-year period. Knowing this helps us understand how common a certain behavior is within a large group.

In our study, the population proportion is given as 21%, meaning 21 out of every 100 public accountants were reported to switch jobs in the defined time frame. Suppose we have a prior study or reliable data. In that case, this proportion can be used as a basis for calculating other statistics, like determining the sample size necessary for a reliable new study. This percentage acts like a central piece of information that guides other calculations in our statistical efforts.

If past estimates weren’t available, we would use a default guess for population proportion, which is often 0.5 due to its conservative nature, offering maximum variability and hence a "safe-side" approach to sample size estimation.
Margin of Error
The margin of error is a concept used in statistics to reflect how much we expect our sample's result to deviate from the true population figure. It gives us the range within which we believe the true population proportion lies.

In our exercise, we aim for a 3% margin of error. This means if our sample study finds a proportion of 21% for accountants switching jobs, the actual population proportion is likely between 18% and 24% (subtracting and adding the margin of error to our estimate).

This concept is crucial when planning studies, because it indicates the precision of our estimates. Smaller margins of error require larger sample sizes, ensuring the results closely reflect the overall population.
Confidence Level
A confidence level in statistical terms indicates the probability that the population parameter falls within the margin of error of the sample statistic. It tells us how confident we can be in our estimates.

In this exercise, a 95% confidence level is used. It implies that if we were to take many samples and build intervals, 95% of these intervals would contain the actual population proportion. So, it's about certainty in the results: the higher the confidence level, the more certain we are about the data, but this also demands larger sample sizes.

Confidence level and margin of error are connected. Larger sample sizes can provide higher confidence without increasing the margin of error. This happens because greater numbers can better represent the population.
z-value
The z-value, sometimes known as the z-score, is a measure that describes the number of standard deviations a data point is from the mean. For our purposes in sample size calculation, it links directly to the confidence level.

In a normal distribution, a 95% confidence level corresponds to an approximate z-value of 1.96. This means that, for most data under a normal distribution condition, 95% falls within 1.96 standard deviations from the mean.

In our exercise, using the z-value allows us to compute how wide we need our interval, factoring in both our sample size and margin of error. It's a critical part of ensuring our estimates stay within the desired confidence range. By understanding and applying the appropriate z-value, we ensure the reliability of our statistical conclusions.

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 Weight Control Registry tries to mine secrets of success from people who lost at least 30 pounds and kept it off for at least a year. It reports that out of 2,700 registrants, 459 were on a low-carbohydrate diet (less than 90 grams a day). a. Develop a \(95 \%\) confidence interval for the proportion of people on a low-carbohydrate diet. b. Is it possible that the population percentage is \(18 \% ?\) c. How large a sample is needed to estimate the proportion within \(0.5 \% ?\)

A large on-demand video streaming company is designing a large-scale survey to determine the mean amount of time corporate executives watch on-demand television. A small pilot survey of 10 executives indicated that the mean time per week is 12 hours, with a standard deviation of 3 hours. The estimate of the mean viewing time should be within one-quarter hour. The \(95 \%\) level of confidence is to be used. How many executives should be surveyed?

Bob Nale is the owner of Nale's Quick Fill. Bob would like to estimate the mean number of gallons of gasoline sold to his customers. Assume the number of gallons sold follows the normal distribution with a population standard deviation of 2.30 gallons. From his records, he selects a random sample of 60 sales and finds the mean number of gallons sold is \(8.60 .\) a. What is the point estimate of the population mean? b. Develop a \(99 \%\) confidence interval for the population mean. c. Interpret the meaning of part (b).

As part of their business promotional package, the Milwaukee Chamber of Commerce would like an estimate of the mean cost per month to lease a one- bedroom apartment. The mean cost per month for a random sample of 40 apartments currently available for lease was 884. The standard deviation of the sample was 50. a. Develop a 98% confidence interval for the population mean. b. Would it be reasonable to conclude that the population mean is 950 per month?

A sample of 352 subscribers to Wired magazine shows the mean time spent using the Internet is 13.4 hours per week, with a sample standard deviation of 6.8 hours. Find the \(95 \%\) confidence interval for the mean time Wired subscribers spend on the Internet.

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.