/*! 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 25 The estimate of the population p... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The estimate of the population proportion is to be within plus or minus .05, with a 95 percent level of confidence. The best estimate of the population proportion is .15. How large a sample is required?

Short Answer

Expert verified
A sample size of 192 is required.

Step by step solution

01

Identify the Formula

To find the required sample size for a given confidence level and margin of error, we use the formula for sample size in estimating a population proportion:\[ n = \left( \frac{Z^2 \cdot p \cdot (1-p)}{E^2} \right)\]where \(n\) is the sample size, \(Z\) is the Z-value corresponding to the confidence level, \(p\) is the estimated population proportion, and \(E\) is the margin of error.
02

Find the Z-value

For a 95% confidence level, we look up the Z-value from the standard normal distribution table. The Z-value for 95% confidence is 1.96.
03

Substitute Values into the Formula

We have the estimated population proportion \(p = 0.15\), the margin of error \(E = 0.05\), and the Z-value \(Z = 1.96\). Substitute these values into the formula:\[ n = \left( \frac{1.96^2 \cdot 0.15 \cdot (1-0.15)}{0.05^2} \right)\]
04

Calculate the Numerator

Calculate the influence of the Z-value and the estimated proportion in the numerator:\[ 1.96^2 \cdot 0.15 \cdot 0.85 = 0.477885\]
05

Calculate the Denominator

The margin of error squared is:\[ E^2 = 0.05^2 = 0.0025\]
06

Compute the Sample Size

Divide the numerator by the denominator to find the required sample size:\[ n = \frac{0.477885}{0.0025} = 191.154\]Since the sample size must be a whole number, we round up to the nearest whole number, resulting in a required sample size of 192.

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 Estimation
In the world of statistics, estimating a population proportion means guessing what percentage or fraction of a total population has a certain characteristic. Imagine you're looking at a jar of jellybeans and want to know how many are red without actually counting them all. This is similar in statistics when we want to figure out a trait in a large group.

When we take a sample, we are looking for an estimate, which gives us an idea of what the whole population might look like based on a smaller group. The sample proportion is our best guess, represented by "p" in statistical formulas. In our example, "p" represents the estimated population proportion, which is 0.15 or 15%.

Population proportion estimation is a useful way to make predictions or prepare for decision-making processes, without going through every single individual in a survey or a study.
Confidence Level
The confidence level in statistics helps us understand how sure we are about our estimates. It does this by showing us the likelihood that our sample statistics are correct in relation to the true population values.

A common choice for a confidence level is 95%, meaning that if we were to take 100 different samples, about 95 of them would be expected to contain the true population proportion within their calculated range. This "range" is known as the confidence interval.

In our context, a 95% confidence level uses a Z-value of 1.96, which we'll discuss further. The confidence level increases as the interval gets wider, offering a greater chance the actual proportion is captured, but also requiring a larger sample size to maintain that level of confidence.
Margin of Error
The margin of error is like the safety net that tells us how far off our estimate might be from the actual population proportion. It's an interval range around the sample proportion within which the true population proportion is expected to lie.

In simple terms, if our estimated proportion is off slightly due to sampling variability, the margin of error provides a cushion so we account for that potential error. In the example we worked with, the margin of error is given as 0.05, meaning our estimate could be 5% higher or lower than the actual value.

The smaller the margin of error, the more precise our estimate is, but achieving a smaller margin requires a larger sample size.
Z-value
The Z-value is crucial in calculating sample sizes and comes from the standard normal distribution. It's a factor that relates to the confidence level we choose.

The Z-value describes how many standard deviations away from the mean our estimate would be on a standard normal distribution. For a 95% confidence level, the corresponding Z-value is 1.96, which we use in our formula to find sample size.

Think of the Z-value as a kind of translator that scales our confidence level into a concrete number we can use in equations, ensuring that our statistical estimates hold up with the desired level of confidence. Using it properly ensures we have a sample that's large enough to make sense of the variability and get useful results.

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

A random sample of 25 people employed by the Florida Department of Transportation earned an average wage (including benefits) of \(\$ 65.00\) per hour. The sample standard deviation was \(\$ 6.25\) per hour. a. What is the population mean? What is the best estimate of the population mean? b. Develop a 99 percent confidence interval for the population mean wage (including benefits) for these employees. c. How large a sample is needed to assess the population mean with an allowable error of \(\$ 1.00\) at 95 percent confidence?

The Human Relations Department of Electronics, Inc. would like to include a dental plan as part of the benefits package. The question is: How much does a typical employee and his or her family spend per year on dental expenses? A sample of 45 employees reveals the mean amount spent last year was \(\$ 1,820,\) with a standard deviation of \(\$ 660 .\) a. Construct a 95 percent confidence interval for the population mean. b. The information from part (a) was given to the president of Electronics, Inc. He indicated he could afford \(\$ 1,700\) of dental expenses per employee. Is it possible that the population mean could be \(\$ 1,700 ?\) Justify your answer.

A population is estimated to have a standard deviation of \(10 .\) We want to estimate the population mean within 2, with a 95 percent level of confidence. How large a sample is required?

As part of an annual review of its accounts, a discount brokerage selects a random sample of 36 customers. Their accounts are reviewed for total account valuation. which showed a mean of \(\$ 32,000\), with a sample standard deviation of \(\$ 8,200 .\) What is a 90 percent confidence interval for the mean account valuation of the population of customers?

A sample of 10 observations is selected from a normal population for which the population standard deviation is known to be 5 . The sample mean is 20 . a. Determine the standard error of the mean. b. Explain why we can use formula \((9-1)\) to determine the 95 percent confidence interval even though the sample is less than \(30 .\) c. Determine the 95 percent confidence interval for the population mean.

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.