/*! 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 15 An internet service provider wis... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An internet service provider wishes to estimate, to within one percentage point, the current proportion of all email that is spam, with \(99.9 \%\) confidence. Last year the proportion that was spam was \(71 \%\). Estimate the minimum size sample required.

Short Answer

Expert verified
The minimum sample size required is 22,305 emails.

Step by step solution

01

Understanding the Confidence Interval Formula

The formula to estimate the required sample size for a proportion is \( n = \left( \frac{Z^2 \cdot p \cdot (1-p)}{E^2} \right) \), where \( n \) is the sample size, \( Z \) is the Z-score for the desired confidence level, \( p \) is the estimated proportion, and \( E \) is the margin of error.
02

Identifying Given Values and Required Variables

Based on the problem statement: the estimated proportion of spam emails last year is \( p = 0.71 \) and the margin of error \( E \) is 1 percentage point (\( E = 0.01 \)). The confidence level is \(99.9\%\), corresponding to a Z-score of approximately 3.291.
03

Plugging Values into the Formula

Substitute the given values into the sample size formula: \[ n = \left( \frac{3.291^2 \cdot 0.71 \cdot (1-0.71)}{0.01^2} \right) \]
04

Calculating the Sample Size

First, calculate the numerator: \( 3.291^2 \approx 10.835 \), \( 0.71 \cdot (1-0.71) = 0.71 \cdot 0.29 = 0.2059 \), and then \( 10.835 \cdot 0.2059 \approx 2.2305 \). Now divide by the square of the margin of error: \( \frac{2.2305}{0.0001} = 22305 \).

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.

Confidence Interval
A confidence interval is a range of values used to estimate an unknown population parameter. It gives us a measure of reliability about the estimate. In other words, a confidence interval reflects how certain we are about the estimate's accuracy. Confidence intervals are generally expressed with a specific confidence level, such as 95% or 99.9%.
For instance, if we have a 99.9% confidence interval, it indicates that we are 99.9% confident that the true proportion lies within this interval. In the context of the exercise, a confidence interval helps us to determine the proportion of all emails that are spam, with high reliability. This ensures that the sample we consider will provide a highly accurate estimation of spam emails across the whole population.
  • High confidence levels lead to wider intervals.
  • Wider intervals are more likely to include the true population parameter.
Understanding the concepts of confidence intervals is essential for making informed decisions based on data.
Z-score
The Z-score is a statistical measurement that describes a value's relationship to the mean of a group of values, expressed in terms of standard deviations. In the context of sample size calculation, the Z-score is critical for determining the appropriate sample size necessary for estimating the population parameter with a specific level of confidence.
For a given confidence level, the Z-score helps us determine the margin of error. A higher Z-score corresponds to a higher confidence level, as it covers more of the data distribution.
In the exercise, the chosen confidence level is 99.9%, which relates to a Z-score of approximately 3.291. This means that the sample mean should be within 3.291 standard deviations of the population mean.
  • Z-scores convert confidence levels into actionable data.
  • They are derived from the normal distribution.
Recognizing the significance of Z-scores allows us to assess how many data points are necessary to draw accurate conclusions.
Margin of Error
The margin of error is a statistic that quantifies the amount of random sampling error in an estimate. It tells us how much we can expect the true population parameter to differ from the estimated sample proportion.
For example, a margin of error of 1 percentage point indicates that the true proportion could vary by this amount. In our exercise, the margin of error is set at 1%, which implies our estimates should be that close to the real proportion of spam emails in the entire population. The margin of error is a crucial factor in determining the sample size - smaller margins require larger sample sizes to maintain confidence in the estimate.
  • It reflects the precision of an estimate.
  • Higher requirements for precision yield larger sample sizes.
Understanding the margin of error helps in comprehending the trade-offs between sample size, confidence, and precision.
Proportion Estimation
Proportion estimation involves determining the fraction of a population that exhibits a particular characteristic. This is especially important in surveys and samples, such as estimating the proportion of spam emails from the total.
In the given example, the estimation starts by using last year's data of 71% spam as a basis for the current proportion estimate. This existing data helps guide what the proportion might look like, allowing for a more effective sample size estimate. We apply this initial proportion value in our formula to calculate the sample size.
Accurate proportion estimation ensures reliable predictions and decisions. It requires careful planning, especially when considering large populations.
  • It's essential for forecasting and strategic planning.
  • Informs decisions based on population characteristics.
Understanding how to estimate proportions accurately is fundamental for any data-driven analysis or prediction.

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 number of trips to a grocery store per week was recorded for a randomly selected collection of households, with the results shown in the table. $$ \begin{array}{lllllllll} 2 & 2 & 2 & 1 & 4 & 2 & 3 & 2 & 5 & 4 \\ 2 & 3 & 5 & 0 & 3 & 2 & 3 & 1 & 4 & 3 \\ 3 & 2 & 1 & 6 & 2 & 3 & 3 & 2 & 4 & 4 \end{array} $$ Construct a \(95 \%\) confidence interval for the average number of trips to a grocery store per week of all households.

The administration at a college wishes to estimate, to within two percentage points, the proportion of all its entering freshmen who graduate within four years, with \(90 \%\) confidence. Estimate the minimum size sample required.

A random sample of size 256 is drawn from a population whose distribution, mean, and standard deviation are all unknown. The summary statistics are \(x-=1011\) and \(s=34\). a. Construct a \(90 \%\) confidence interval for the population mean \(\mu\). b. Construct a \(99 \%\) confidence interval for the population mean \(\mu\). c. Comment on why one interval is longer than the other.

A college athletic program wishes to estimate the average increase in the total weight an athlete can lift in three different lifts after following a particular training program for six weeks. Twenty-five randomly selected athletes when placed on the program exhibited a mean gain of \(47.3 \mathrm{lb}\) with standard deviation \(6.4 \mathrm{lb}\). Construct a \(90 \%\) confidence interval for the mean increase in lifting capacity all athletes would experience if placed on the training program. Assume increases among all athletes are normally distributed.

A random sample of size 28 is drawn from a normal population. The summary statistics are \(\bar{x}-68.6\) and \(s=1.28\). a. Construct a \(95 \%\) confidence interval for the population mean \(\mu\). b. Construct a \(99.5 \%\) confidence interval for the population mean \(\mu\). c. Comment on why one interval is longer than the other.

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.