/*! 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} Q.2 It鈥檚 critical Find the appropr... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Q.2 (page 522)

It鈥檚 critical Find the appropriate critical value for constructing a confidence interval in each of the following settings.

(a) Estimating a population proportion p at a 94%confidence level based on an SRS of size 125.

(b) Estimating a population mean M at a 99%confidence level based on an SRS of size 58.

Short Answer

Expert verified

a). The critical value is 1.96.

b). The critical value is 2.576.

Step by step solution

01

Part (a) Step 1: Given Information

Population proportion (p)=94%=0.94.

Sample size (n)=125.

02

Part (a) Step 2: Explanation

The level of significance is:

Level of significance =1-Confidence level

=1-0.95

=0.05

The critical value using standard normal table is calculated as:

z伪/2=z0.05/2

=z0.025

=1.96

The critical value is 1.96.

03

Part (b) Step 1: Given Information

Confidence level =99%.

Sample size (n)=58.

04

Part (b) Step 2: Explanation

The level of significance is:

Level of significance =1-Confidence level

=1-0.99

=0.01

The critical value using the standard normal table is calculated as:

za/2=z0.01/2

=z0.005

=2.576

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

To assess the accuracy of a laboratory scale, a standard weight known to weigh 10grams is weighed repeatedly. The scale readings are Normally distributed with unknown mean (this mean is 10grams if the scale has no bias). In previous studies, the standard deviation of the scale readings has been about 0.0002gram. How many measurements must be averaged to get a margin of error of 0.0001with 98% confidence? Show your work.

60. Travel time to work A study of commuting times reports the travel times to work of a random sample of 20employed adults in New York State. The mean is x炉=31.25 minutes, and the standard deviation is sx=21.88minutes. What is the standard error of the mean? Interpret this value in context.

Interpret your interval in context.

Good wood? A lab supply company sells pieces of Douglas fir 4inches long and 1.5inches square for force experiments in science classes. From experience, the strength of these pieces of wood follows a Normal distribution with standard deviation 3000pounds. You want to estimate the mean load needed to pull apart these pieces of wood to within 1000pounds with 95%confidence. How large a sample is needed? Show your work.

- Understand how the margin of error of a confidence interval changes with the sample size and the level of confidence C.

In a poll,

I. Some people refused to answer questions.

II. People without telephones could not be in the sample.

III. Some people never answered the phone in several calls.

Which of these sources is included in the 卤2%margin of error announced for the poll?

(a) I only

(b) II only

(c) III only

(d) I, II, and III

(e) None of these

See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Statistics

Read Explanation

Pure Maths

Read Explanation

Decision Maths

Read Explanation

Geometry

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.