Extra Practice Confidence Intervals

Extra Practice: Confidence
Intervals

Raja Almukkahal
Larry Ottman

Danielle DeLancey
Addie Evans
Ellen Lawsky
Brenda Meery

Say Thanks to the Authors
Click http://www.ck12.org/saythanks

(No sign in required)

To access a customizable version of this book, as well as other AUTHORS
interactive content, visit www.ck12.org Raja Almukkahal
Larry Ottman
CK-12 Foundation is a non-profit organization with a mission to Danielle DeLancey
reduce the cost of textbook materials for the K-12 market both in Addie Evans
the U.S. and worldwide. Using an open-source, collaborative, and Ellen Lawsky
web-based compilation model, CK-12 pioneers and promotes the Brenda Meery
creation and distribution of high-quality, adaptive online textbooks
that can be mixed, modified and printed (i.e., the FlexBook®
textbooks).

Copyright © 2017 CK-12 Foundation, www.ck12.org

The names “CK-12” and “CK12” and associated logos and the
terms “FlexBook®” and “FlexBook Platform®” (collectively
“CK-12 Marks”) are trademarks and service marks of CK-12
Foundation and are protected by federal, state, and international
laws.

Any form of reproduction of this book in any format or medium,
in whole or in sections must include the referral attribution link
http://www.ck12.org/saythanks (placed in a visible location) in
addition to the following terms.

Except as otherwise noted, all CK-12 Content (including CK-12
Curriculum Material) is made available to Users in accordance
with the Creative Commons Attribution-Non-Commercial 3.0
Unported (CC BY-NC 3.0) License (http://creativecommons.org/
licenses/by-nc/3.0/), as amended and updated by Creative Com-
mons from time to time (the “CC License”), which is incorporated
herein by this reference.

Complete terms can be found at http://www.ck12.org/about/
terms-of-use.

Printed: July 25, 2017

www.ck12.org Chapter 1. Extra Practice: Confidence Intervals

1CHAPTER Extra Practice: Confidence
Intervals

Learning Objectives

• Calculate the mean of a sample as a point estimate of the population mean.
• Construct a confidence interval for a population mean based on a sample mean.
• Calculate a sample proportion as a point estimate of the population proportion.
• Construct a confidence interval for a population proportion based on a sample proportion.
• Calculate the margin of error for a point estimate as a function of sample mean or proportion and size.
• Understand the logic of confidence intervals, as well as the meaning of confidence level and confidence

intervals.

Extra Practice: Confidence Intervals

1. In a local teaching district, a technology grant is available to teachers in order to install a cluster of four
computers in their classrooms. From the 6,250 teachers in the district, 250 were randomly selected and asked
if they felt that computers were an essential teaching tool for their classroom. Of those selected, 142 teachers
felt that computers were an essential teaching tool.

a. Calculate a 99% confidence interval for the proportion of teachers who felt that computers are an
essential teaching tool.

b. How could the survey be changed to narrow the confidence interval but to maintain the 99% confidence
interval?

2. Josie followed the guidelines presented to her and conducted a binomial experiment. She did 300 trials and
reported a sample proportion of 0.61.

a. Calculate the 90%, 95%, and 99% confidence intervals for this sample.
b. What did you notice about the confidence intervals as the confidence level increased? Offer an explana-

tion for your findings?
c. If the population proportion were 0.58, would all three confidence intervals enclose it? Explain.

3. Does replacing σ with s change your chance of capturing the unknown population mean? Is there a way to
increase the chance of capturing the unknown population mean?

4. A study was conducted to determine the mean birth weight of a certain breed of kittens. Consider the birth
weights of kittens to be normally distributed. A sample of 45 kittens was randomly selected from all kittens
of this breed at a large veterinary hospital. The birth weight of each kitten in the sample was recorded. The
sample mean was 3.56 ounces, and the sample standard deviation was 0.2 ounces. Set a 90% confidence
interval on the mean birth weight of all kittens of this breed.

5. In a study of seventh grade students, the mean number of hours per week that they watched television was
18.7 with a standard deviation of 4.5 hours. Assume the population has a normal distribution. Construct a
95% confidence interval for the mean number of hours of tv watched by seventh grade students.

6. A random sample of 40 college students has mean annual earnings of $3,245 and a standard deviation of
$567. Construct a 99% confidence interval for the population. Does the population have to follow a normal
distribution? Explain.

7. A random sample of 16 light bulbs has a mean life of 650 hours and a standard deviation of 32 hours. Assume
the population has a normal distribution. Construct a 90% confidence interval for the population mean.

1

www.ck12.org

8. A sample of 100 cans of peas showed an average weight of 14 ounces with a standard deviation of 0.7 ounces.
Construct a 95% confidence interval for the mean of the population.

9. What three factors affect the width of a confidence interval for a population mean? For each factor, indicate
how an increase in the numerical value of the factor affects the interval width.

10. For each of the following use the information given to calculate the standard error of the mean and find an
approximate 90% confidence interval for the population mean:

a. n = 81, x¯ = 64.2, s = 2.7
b. n = 100, x¯ = 123.5, s = 9
c. n = 324, x¯ = 123.5, s = 9

11. Suppose a random sample of 64 men has a mean foot length of 27.5 cm with a standard deviation of 2 cm.

a. Calculate the standard error of the sample mean.
b. Calculate an approximate 99% confidence interval for the mean foot length of men. Write a sentence

that interprets this interval.

12. For each combination of sample size and sample proportion find the approximate margin of error for the 90%
confidence interval:

a. n = 100, pˆ = 0.56
b. n = 400, pˆ = 0.56
c. n = 400, pˆ = 0.25
d. n = 400, pˆ = 0.75

13. Suppose a new cancer treatment is given to a sample of 300 patients. The treatment was successful for 210
of the patients. Assume that these patients are representative of the population of individuals who have this
cancer.

a. Calculate the sample proportion that was successfully treated.
b. Determine a 90% confidence interval for the proportion successful treated. Write a sentence that inter-

prets this interval.

14. Suppose a polling organization reports that the margin of error is 3% for a sample survey. Explain what this
indicates about the possible difference between a percent determined from the survey data and the population
value of the percent.

15. A poll conducted in the United States November 8 - 15, 2010 asked “The Secretary of Transportation recently
said that he may push Congress for a national ban on using a cell phone while driving. The ban would include
hands-free cell phones. Do you think that a national ban on using a cell phone while driving is a good idea or
a bad idea?" In the nationwide poll of n = 2, 424 registered voters 63% said they thought it was a good idea.
The margin of error was reported as ±2%. (source: wwwlpollingreport.com).

a. Find a 95% confidence interval estimate of the percent of American voters who believe banning cell
phones when driving is a good idea at the time of the poll.

b. Write a sentence that interprets the interval computed in part (a).

16. A Gallup Organization telephone poll of 511 adults, aged 18 and older, living in the continental United States
found that 70% of Americans feel confident in the accuracy of their doctor’s advice, and don’t feel the need
to check for a second opinion or do additional research. The margin of error for this survey was given as
±5percentage points.

a. Find a 95% confidence interval estimate of the percent of American adults who feel confident in the
accuracy of their doctor’s advice and don’t feel the need to check for a second opinion.

b. Based on the interval you found, is it reasonable to say that more than 65% of American voting adults
have confidence in their doctor’s advice?

17. Suppose 100 researchers each plan to independently gather data and construct 95% confidence interval for a
population mean. If X = the number of those intervals that actually cover the population mean, then Xis a
binomial random variable.

2

www.ck12.org Chapter 1. Extra Practice: Confidence Intervals

a. What is a “success” for this random variable?
b. What is the numerical value of the probability of success?
c. What is the expected number of intervals that will cover their population means?

18. In computing the confidence interval for a population mean, µ, explain whether the interval would be wider,
more narrow, or neither as a result of each of the following changes:

a. The level of confidence is changed from 85% to 90%.
b. The sample size is tripled.
c. A new random sample of the same size is taken and is increased by 10.

19. Calculate a 98% confidence interval for the proportion successfully treated in problem 13. Is this interval
wider or narrower than the interval computed in problem 12?

20. In a Gallup Youth Survey done in 2000, 501 randomly selected American teenagers were asked about how
well they get along with their parents.

a. According to the Gallup Organization, the margin of error for the poll was 5%. A survey result was that
54% of the sample said they get along “very well” with their parents. Using the reported margin of error,
calculate a 90% confidence interval for the population proportion that gets along “very well” with their
parents.

b. Using the more exact margin of error, calculate a 90% confidence interval. Compare your answer to part
(a).

3

www.ck12.org
4