/*! 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.8.81 The Ice Chalet offers dozens of ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The Ice Chalet offers dozens of different beginning iceskating classes. All of the class names are put into a bucket. The 5 P.M., Monday night, ages 8 to 12, beginning ice-skating class was picked. In that class were 64 girls and 16 boys. Suppose that we are interested in the true proportion of girls, ages 8 to 12, in all beginning ice-skating classes at the Ice Chalet. Assume that the children in the selected class are a random sample of the population.

Calculate the following:

a. x = _______

b. n = _______

c. p′ = _______

Short Answer

Expert verified
  1. The value of x=64.
  2. The value of the n=80.
  3. The value ofP′=0.8.

Step by step solution

01

Given information (Part a)

There are 64 girls and 16 boys in a ice skating class.

We have to calculate the value of x

02

Solution (Part a)

There are total 64 girls in the ice skating class.

The all girls are with the age of 8to 12

Therefore the valuex=64.

03

Given information (Part b)

There are 64 girls and 16 boys in a ice skating class.

We have to find the value ofn.

04

Solution (Part b)

The number of students for the ice-skating class is :

64+16=80

Therefore the n=80

05

Given information(Part c)

There are 64 girls and 16 boys in a ice skating class.

We have to findp'

06

Solution (Part c)

Let's consider the number of girls with ages 8to 12in the ice skating classes:

It is 64.

The total number of the students=80

Now, we can compute the population proportion as follow:

P′=6480

P′=0.8

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!

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

In complete sentences, explain why the confidence interval in Exercise 8.17 is larger than in Exercise 8.18.

Define the random variable X¯{"x":[[4,22,36],[6,35],[5.647456684472792,6.517307794504744,6.517307794504744,8.257010014568648,9.1268611246006,9.996712234632552,12.606265564728409,13.476116674760362,14.345967784792315,16.08567000485622,16.95552111488817,17.825372224920123,18.695223334952075,20.43492555501598,21.30477666504793,22.174627775079884,23.044478885111836,23.914329995143788,24.78418110517574,25.654032215207693,26.523883325239645,27.393734435271597,27.393734435271597,28.26358554530355,29.1334366553355,29.1334366553355,30.003287765367453,30.873138875399405,31.742989985431358,32.61284109546331]],"y":[[9,59,116],[115,9],[-7.088088802556169,-7.088088802556169,-7.957939912588121,-7.957939912588121,-7.957939912588121,-7.957939912588121,-7.957939912588121,-7.957939912588121,-7.957939912588121,-7.957939912588121,-7.957939912588121,-7.957939912588121,-7.957939912588121,-7.957939912588121,-7.957939912588121,-7.957939912588121,-7.957939912588121,-7.957939912588121,-7.957939912588121,-7.957939912588121,-7.957939912588121,-7.957939912588121,-8.827791022620074,-8.827791022620074,-8.827791022620074,-9.697642132652026,-9.697642132652026,-9.697642132652026,-9.697642132652026,-9.697642132652026]],"t":[[0,0,0],[0,0],[1648475418436,1648475418678,1648475418692,1648475418709,1648475418727,1648475418743,1648475418760,1648475418776,1648475418791,1648475418810,1648475418826,1648475418842,1648475418859,1648475418875,1648475418892,1648475418910,1648475418925,1648475418942,1648475418965,1648475419009,1648475419026,1648475419076,1648475419105,1648475419118,1648475419134,1648475419146,1648475419161,1648475419180,1648475419242,1648475419292]],"version":"2.0.0"}in words.

If you decreased the allowable error bound, why would the minimum sample size increase (keeping the same level of

confidence)?

Among various ethnic groups, the standard deviation of heights is known to be approximately three inches. We wish

to construct a 95% confidence interval for the mean height of male Swedes. Forty-eight male Swedes are surveyed. The

sample mean is 71 inches. The sample standard deviation is 2.8 inches.

a.

I. X=________

ii. σ =________

iii. n =________

b. In words, define the random variables X and X

c. Which distribution should you use for this problem? Explain your choice.

d. Construct a 95% confidence interval for the population mean height of male Swedes.

I. State the confidence interval.

ii. Sketch the graph.

iii. Calculate the error bound.

e. What will happen to the level of confidence obtained if 1,000 male Swedes are surveyed instead of 48? Why?

Suppose we have data from a sample. The sample mean is 15, and the error bound for the mean is 3.2. What is the confidence interval estimate for the population mean?

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.