/*! 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 58 Retailers report that the use of... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Retailers report that the use of cents-off coupons is increasing. The Scripps Howard News Service (July 9 , 1991) reported the proportion of all households that use coupons as .77. Suppose that this estimate was based on a random sample of 800 households (i.e., \(n=800\) and \(p=.77\) ). Construct a \(95 \%\) confidence interval for \(\pi\), the true proportion of all households that use coupons.

Short Answer

Expert verified
The 95% confidence interval for the true proportion of all households that use coupons can be calculated using the given sample size and sample proportion. The interval will be found by adding and subtracting the Margin of Error from the sample proportion.

Step by step solution

01

Calculate the Standard Error (SE)

First, calculate the standard error using the formula \( SE = \sqrt{\frac{p(1-p)}{n}} \) where p is the sample proportion and n is the sample size. Therefore, \( SE = \sqrt{\frac{.77(1-.77)}{800}} \).
02

Determine the Z-Score for 95% Confidence

Next, determine the Z-score associated with a 95% confidence interval. For a 95% confidence level, the Z-score is approximately 1.96.
03

Calculate the Margin of Error

Then, calculate the margin of error by multiplying the Z-score by the standard error. This is given by the formula \( E = Z * SE \). Therefore, \( E = 1.96 * SE \).
04

Find the Confidence Interval

Finally, the confidence interval is found by adding and subtracting the margin of error (E) from the sample proportion (p). Therefore, confidence interval \( CI = (p-E , p+E) \).

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.

Standard Error
The standard error (SE) is a crucial concept when estimating the precision of a sample statistic. It's essentially a measure of how much the sample statistic (like the sample proportion) is expected to vary from the true population parameter.

To compute the standard error for a proportion, we rely on the formula: \[SE = \sqrt{\frac{p(1-p)}{n}}\] Here, \(p\) represents the sample proportion, while \(n\) is the sample size. For the given problem, we substitute \(p = 0.77\) and \(n = 800\). This calculation helps us understand the variability of our sample proportion relative to the true population proportion.

The standard error provides a baseline for how much we expect our sample proportion to "jump around" due to sampling variability. Thus, it's an important component in forming a confidence interval, as it incorporates the sample size and sample characteristics into the estimation process.
Sample Proportion
The sample proportion is a simple yet powerful statistic used to estimate a population proportion. It represents the fraction or percentage of the sample exhibiting a particular trait. In our scenario, the sample proportion, \(p\), is given as 0.77, indicating that 77% of the sampled households use coupons.

Understanding the sample proportion is important because it acts as the point estimate for the population proportion—meaning it's our best guess at the moment for the overall proportion of coupon users in all households. Because this estimate comes from a sample, it's accompanied by some uncertainty, which is precisely where the concept of confidence intervals becomes indispensable.
  • The sample proportion serves as the center of our confidence interval.
  • Variability in sample proportions can arise from the randomness of sample selection.
  • The size of the sample affects how close the sample proportion is to the true population proportion.
Z-Score
Z-Score is a statistical measurement that describes a data point's relation to the mean of a group of points. In constructing confidence intervals, especially for proportions, the Z-Score aligns with the desired confidence level.

For a 95% confidence level, typically used if not specified otherwise, the Z-Score is approximately 1.96. This value is derived from the standard normal distribution, reflecting that we want to capture the area within 1.96 standard deviations from the mean (which accounts for 95%).

In the context of confidence intervals:
  • The Z-Score helps adjust the range of the interval based on desired confidence.
  • It multiplies the standard error to calculate the margin of error—essentially stretching or shrinking the interval.
  • Understanding Z-Scores helps comprehend how confident we are that the interval truly captures the population parameter.
Margin of Error
The margin of error (ME) quantifies the uncertainty in our sample estimate. It's an essential part of the confidence interval, providing a range within which we expect the true population parameter to fall.

To calculate the margin of error, we multiply the Z-Score by the standard error:
\[E = Z \times SE \]
In our problem, substituting the known values—Z = 1.96 and SE calculated previously—gives us the margin of error. This outcome tells us how far from the sample proportion our estimate might reasonably fall if we repeated the sampling process.

Understanding the margin of error is key because:
  • It indicates the precision of our estimate.
  • Larger sample sizes tend to decrease the margin of error, providing more accurate estimates.
  • It's a cornerstone for interpreting confidence intervals, which aim to cover the true parameter with a specified level of certainty.

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 manufacturer of college textbooks is interested in estimating the strength of the bindings produced by a particular binding machine. Strength can be measured by recording the force required to pull the pages from the binding. If this force is measured in pounds, how many books should be tested to estimate with \(95 \%\) confidence to within \(0.1 \mathrm{lb}\), the average force required to break the binding? Assume that \(\sigma\) is known to be \(0.8 \mathrm{lb}\).

In 1991, California imposed a "snack tax" (a sales \(\operatorname{tax}\) on snack food) in an attempt to help balance the state budget. A proposed alternative tax was a \(12 \phi\) -per-pack increase in the cigarette tax. In a poll of 602 randomly selected California registered voters, 445 responded that they would have preferred the cigarette tax increase to the snack tax (Reno Gazette-Journal, August 26,1991 ). Estimate the true proportion of California registered voters who preferred the cigarette tax increase; use a \(95 \%\) confidence interval.

One thousand randomly selected adult Americans participated in a survey conducted by the Associated Press (June, 2006). When asked "Do you think it is sometimes justified to lie or do you think lying is never justified?" \(52 \%\) responded that lying was never justified. When asked about lying to avoid hurting someone's feelings, 650 responded that this was often or sometimes OK. a. Construct a \(90 \%\) confidence interval for the proportion of adult Americans who think lying is never justified. b. Construct a \(90 \%\) confidence interval for the proportion of adult American who think that it is often or sometimes OK to lie to avoid hurting someone's feelings. c. Based on the confidence intervals from Parts (a) and (b), comment on the apparent inconsistency in the responses given by the individuals in this sample.

Discuss how each of the following factors affects the width of the confidence interval for \(\pi\) : a. The confidence level b. The sample size c. The value of \(p\)

Recent high-profile legal cases have many people reevaluating the jury system. Many believe that juries in criminal trials should be able to convict on less than a unanimous vote. To assess support for this idea, investigators asked each individual in a random sample of Californians whether they favored allowing conviction by a 10-2 verdict in criminal cases not involving the death penalty. The Associated Press (San Luis Obispo TelegramTribune, September 13,1995 ) reported that \(71 \%\) supported the \(10-2\) verdict. Suppose that the sample size for this survey was \(n=900\). Compute and interpret a \(99 \%\) confidence interval for the proportion of Californians who favor the \(10-2\) verdict.

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.