/*! 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 1. I want red!聽A candy maker offer... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Q 1. (page 639)

I want red!A candy maker offers Child and Adult bags of jelly beans with

different color mixes. The company claims that the Child mix has 30%red jelly beans, while the Adult mix contains 15%red jelly beans. Assume that the candy maker鈥檚 claim is true. Suppose we take a random sample of 50jelly beans from the Child mix and a separate random sample of 100jelly beans from the Adult mix. Let p^Cand p^Abe the sample proportions of red jelly beans from the Child and

Adult mixes, respectively.

a. What is the shape of the sampling distribution of p^C-p^A? Why?

b. Find the mean of the sampling distribution.

c. Calculate and interpret the standard deviation of the sampling distribution.

Short Answer

Expert verified

a. The shape in normal.

b. 渭p^C-p^A=0.15

c.蟽p^C-p^A=0.07399

Step by step solution

01

Given Information

It is given that nC=50

nA=100

pC=0.30

pA=0.15

02

Shape of p^C-p^A

Assuming p^C-p^Ais normal.

Condition is:

nCpC鈮�10

nC1-pC鈮�10

nApA鈮�10

nCpC=(50)(0.30)=15

nApA=(100)(0.15)=15

nA1-pA=(100)(1-0.15)=(100)(0.85)=85

The shape is approximately normal as it satisfies all the four conditions.

03

Mean of Sampling Distribution

Using 渭p^C-p^A=pC-pA

=0.30-0.15=0.15

Mean is0.17

04

Standard Deviation

Child jelly(50)<10%of all child jelly beansand

Adult jelly(100)<10%of all adult jelly beans

Standard deviation is 蟽p^C-p^A=pC1-pCnC+pA1-pAnA

=0.30(1-0.30)50+0.15(1-0.15)100

=0.30(0.70)50+0.15(0.85)100鈮�0.07399

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

Sports Illustrated planned to ask a random sample of Division I college athletes, 鈥淒o you believe performance-enhancing drugs are a problem in college sports?鈥� Which of the following is the smallest number of athletes that must be interviewed to estimate the true proportion who believe performance-enhancing drugs are a problem within 卤2% with 90% confidence?

a.17b.21c.1680d.1702e.2401

Thirty-five people from a random sample of 125 workers from Company A admitted

to using sick leave when they weren鈥檛 really ill. Seventeen employees from a random

sample of 68 workers from Company B admitted that they had used sick leave when

they weren鈥檛 ill. Which of the following is a 95% confidence interval for the difference

in the proportions of workers at the two companies who would admit to using sick

leave when they weren鈥檛 ill?

(a) 0.03卤(0.28)(0.72)125+(0.25)(0.75)68

(b) 0.03卤1.96(0.28)(0.72)125+(0.25)(0.75)68

(c) 0.03卤1.645(0.28)(0.72)125+(0.25)(0.75)68

(d) 0.03卤1.96(0.269)(0.731)125+(0.269)(0.731)68

(e)0.03卤1.645(0.269)(0.731)125+(0.269)(0.731)68

Which of the following describes a Type II error in the context of this study?

a. Finding convincing evidence that the true means are different for males and females when in reality the true means are the same

b. Finding convincing evidence that the true means are different for males and females when in reality the true means are different

c. Not finding convincing evidence that the true means are different for males and females when in reality the true means are the same

d. Not finding convincing evidence that the true means are different for males and females when in reality the true means are different

e. Not finding convincing evidence that the true means are different for males and females when in reality there is convincing evidence that the true means are different.

A beef rancher randomly sampled 42 cattle from her large herd to obtain a 95%confidence interval for the mean weight (in pounds) of the cattle in the herd. The interval obtained was (1010,1321). If the rancher had used a 98%confidence interval instead, the interval would have been

a. wider with less precision than the original estimate.

b. wider with more precision than the original estimate.

c. wider with the same precision as the original estimate.

d. narrower with less precision than the original estimate.

e. narrower with more precision than the original estimate.

A random sample of 200New York State voters included 88Republicans, while a random sample of 300California voters produced141Republicans. Which of the following represents the 95%confidence interval for the true difference in the proportion of Republicans in New York State and California?

a. (0.44鈭�0.47)卤1.96((0.44)(0.56)+(0.47)(0.53)200+300)

b. (0.44鈭�0.47)卤1.96((0.44)(0.56)200+(0.47)(0.53)300)

c. (0.44鈭�0.47)卤1.96(0.44)(0.56)200+(0.47)(0.53)300

d. (0.44鈭�0.47)卤1.96(0.44)(0.56)+(0.47)(0.53)200+300

e. (0.44鈭�0.47)卤1.96(0.45)(0.55)200+(0.45)(0.55)300
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Statistics

Read Explanation

Calculus

Read Explanation

Discrete Mathematics

Read Explanation

Logic and Functions

Read Explanation

Probability and Statistics

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.