/*! 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 86 A random sample of \(n=500\) obs... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 86

A random sample of \(n=500\) observations from a binomial population produced \(x=240\) successes. a. Find a point estimate for \(p\), and find the margin of error for your estimator. b. Find a \(90 \%\) confidence interval for \(p\). Interpret this interval.

Short Answer

Expert verified
Answer: The 90% confidence interval for the population proportion (p) of successes is (0.4438, 0.5162).

Step by step solution

01

Find the point estimate for p

To find the point estimate for p (the proportion of successes in the population), we need to divide the number of successes (x) by the total number of observations (n). So, the point estimate for p is: $$ \hat{p} = \frac{x}{n} = \frac{240}{500} = 0.48 $$
02

Calculate the standard error for the point estimate

The standard error for the point estimate of a proportion is given by the formula: $$ SE(\hat{p}) = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} $$ Plugging in the values, we get: $$ SE(\hat{p}) = \sqrt{\frac{0.48(1-0.48)}{500}} = \sqrt{\frac{0.48(0.52)}{500}} = 0.022 $$
03

Find the critical value for a 90% confidence level

To find the critical value for a 90% confidence level (z-score), we can look it up in a standard normal distribution table or use statistical software. The critical value (z) for a 90% confidence level is 1.645.
04

Calculate the margin of error

The margin of error is calculated by multiplying the standard error by the critical value (z-score): $$ ME = SE(\hat{p}) \times z = 0.022 \times 1.645 = 0.0362 $$
05

Find the 90% confidence interval for p

To find the 90% confidence interval for p, we need to subtract and add the margin of error to the point estimate: $$ CI = (\hat{p} - ME, \hat{p} + ME) = (0.48 - 0.0362, 0.48 + 0.0362) = (0.4438, 0.5162) $$
06

Interpret the confidence interval

The 90% confidence interval for the population proportion (p) is (0.4438, 0.5162). This means that we are 90% confident that the true proportion of successes in the population is between 44.38% and 51.62%.

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 91影视!

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 researcher classified his subjects as innately right-handed or lefthanded by comparing thumbnail widths. He took a sample of 400 men and found that 80 men could be classified as left-handed according to his criterion. Estimate the proportion of all males in the population who would test to be left-handed using a \(95 \%\) confidence interval.

Does Mars, Incorporated use the same proportion of red candies in its plain and peanut varieties? A random sample of 56 plain M\&M'S contained 12 red candies, and another random sample of 32 peanut M\&M'S contained 8 red candies. a. Construct a \(95 \%\) confidence interval for the difference in the proportions of red candies for the plain and peanut varieties. b. Based on the confidence interval in part a, can you conclude that there is a difference in the proportions of red candies for the plain and peanut varieties? Explain.

State the Central Limit Theorem. Of what value is the Central Limit Theorem in large-sample statistical estimation?

The first day of baseball comes in late March, ending in October with the World Series. Does fan support grow as the season goes on? Two CNN/USA Today/Gallup polls, one conducted in March and one in November, both involved random samples of 1001 adults aged 18 and older. In the March sample, \(45 \%\) of the adults claimed to be fans of professional baseball, while \(51 \%\) of the adults in the November sample claimed to be fans. \({ }^{13}\) a. Construct a \(99 \%\) confidence interval for the difference in the proportion of adults who claim to be fans in March versus November. b. Does the data indicate that the proportion of adults who claim to be fans increases in November, around the time of the World Series? Explain.

Explain what is meant by "margin of error" in point estimation.
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Pure Maths

Read Explanation

Geometry

Read Explanation

Probability and Statistics

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation
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.