Chapter 6 and 8 review
By Henry Mesa
Portland Community College
Here are a few problems to try out. Use the
arrow keys on your keyboard to move
forward or backwards on the slide.
Enjoy the animation.
p̂
What is the purpose of creating a confidence
interval?
1.
General answer: To estimate a parameter.
With respect to section 6.1: estimate the
population mean, .
With respect to section 8.1: to estimate the
population proportion parameter, p-hat.
If I create a 99% confidence interval to estimate μ, what
is the probability that the interval contains the parameter
μ?
In the long run, 99% of the time the interval created will contain the
parameter.
3. I want to estimate μ with a 90% confidence interval
so that my margin of error is at most 4 units. I know
that σ = 20 units. How big should be my sample size?
Z 
n= 

 m 
*
n = 68
2
4. I create a 95% confidence interval to estimate μ. If I
then construct a 99% confidence interval using the same
data, will my interval be larger or smaller than the 95%
confidence interval?
The 99% confidence interval will be wider compared
to the 95% confidence interval.
The news editor at KGW news wants to estimate the
number of Portland residents who would support same
sex civil unions. During the 5:00 pm broadcast of the
news, a story is done about the legalities of same sex
civil unions, and the end of the story segment, viewers
are asked to call in and vote on the question “Would you
support same sex civil unions?” Two phone numbers are
provided, 1-888- 789- 4000 for “Yes”, 1-888-789-4001
for “No”. At the10:00pm broadcast the news reported
that 55% of Portland residents are in support of same
sex civil unions, with a margin of error of plus/minus 4%
points. Is this result trustworthy? Explain.
No, the confidence interval created can not be used
to make an inference about the sentiment of the
entire population. The reason is that the method
used to gather the data is not a simple random
sample, but rather a volunteer response sample
(chapter 3 page 249)
The student union wants to estimate the mean age of
the student body at their school. A preliminary simple
random sample of 40 students is gathered. The
students will use the standard deviation from the
sample for σ, σ = 4.6 years. The sample mean is 27.4
years. Create a 95% confidence interval.
x  Z
*

n
4.6
27.4  1.96
40
(26.0, 28.8)
A person wants to estimate the probability that a die
tossed completely at random comes up six. The
person tosses a die 64 times, and the number of
times a six appears is 14. Create a 95% confidence
interval.
p̂  Z
*
ˆ  p)
ˆ
p(1
n
(0.2188) 1  0.2188 
14
 1.96
64
64
(0.1175, 0.3201)
1. A test for breast cancer is 90% effective in detecting the cancer in
women who do have breast cancer.
a. Suppose that 30 women who have breast cancer go in for testing.
What is the expected number test that would detect breast
cancer?
b. What is the probability having all the 30 test having a positive
result (all the cancers are detected)?
c.
What is the probability of having 25 or less of the tests detect the
cancer out of the 30?
1. A test for breast cancer is 90% effective in detecting the cancer in
women who do have breast cancer.
a. Suppose that 30 women who have breast cancer go in for testing.
What is the expected number test that would detect breast cancer?
= np
 = 30(0.9) = 27
We would expect 27 tests to come back with
positive results. Remember this is a long
run average.
1. A test for breast cancer is 90% effective in detecting the cancer in women
who do have breast cancer.
b. What is the probability having all the 30 test having a positive result (all the
cancers are detected)?
P(All 30 tests are positive) =
0.930
= 0.4239
1. A test for breast cancer is 90% effective in detecting the cancer in
women who do have breast cancer.
c.
What is the probability of having 25 or less of the tests detect the
cancer out of the 30?
Let X count the number of test that are positive out of the 30.
P(X ≤ 25) = Binomdist(25, 30, 0.9, true) = 0.1755