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Confidence Intervals
Mrs. Medina
Why would we want to estimate a
parameter and/or be confident about
your estimate?
• Estimate?
– We want to estimate because we cannot test the
entire population. It might be too large or
inaccessible.
• Confident?
– Because anyone can spout a predication, but is
the real value within a reasonable distance to
their prediction.
Population vs. Sample
• Identify the population and the sample:
– a) A survey of 1353 American households found
that 18% of the households own a computer.
– b) A recent survey of 2625 elementary school
children found that 28% of the children could be
classified obese.
– c) The average weight of every sixth person
entering the mall within 3 hour period was 146 lb.
Parameter vs. Statistic
• A recent survey by the alumni of a major
university indicated that the average salary of
10,000 of its 300,000 graduates was 125,000.
• The average salary of all assembly-line
employees at a certain car manufacturer is
$33,000.
• The average late fee for 360 credit card
holders was found to be $56.75.
identify the population, sample,
population parameters, and
sample statistics:
• a) In a USA Today Internet poll, readers
responded voluntarily to the question “Do you
consume at least one caffeinated beverage every
day?”
• b) Astronomers typically determine the distance
to galaxy (a galaxy is a huge collection of billions
of stars) by measuring the distances to just a few
stars within it and taking the mean (average) of
these distance measurements.
Determine whether the data are
qualitative or quantitative
• a) the colors of automobiles on a used car lot
• b) the numbers on the shirts of a girl’s soccer
team
• c) the number of seats in a movie theater
• d) a list of house numbers on your street
• e) the ages of a sample of 350 employees of a
large hospital
Population Distribution Probability vs.
Sampling Distribution Probability
• Assume that cholesterol levels for women
ages 20-34 are approximately normally
distributed with µ = 185 and σ = 39.
– What is the probability that a woman age 20-34
has a cholesterol level below 240 mg/dL?
– What is the probability that the average
cholesterol level of 20 women between the age of
20-34 is below 240 mg/dL?
Population Distribution Probability vs.
Sampling Distribution Probability
• Assume that the weights of 5-year old boys for
a normal distribution with µ = 43 lbs and σ = 5
lbs. What is more likely?
– That a randomly selected 5-year old will weigh
more than 49 lbs?
– Or, that the average of a 8 5-year old boys will
weigh more than 46 lbs?
Objective
• The objective of estimation is…
– to estimate the unknown value of a population
parameter, like the mean , or a proportion p, on
the basis of a sample statistic (sampling
distribution) calculated from sample data by giving
a good solid range.
We make point estimates
• Proportions, p
– What percent of the students at LHS vote for Jane Doe
for class president?
– Ex: 150 out 350 voted for Jane Doe.
How confident are
p = 150/350 = .429 or 42.9%
you about these
statistics?
• Means, 
– What is the average GPA of the students at LHS?
– If 100 students are sampled (n=100), the mean of the
GPAs is 2.98.
Confidence Interval
• A confidence interval is a range (or an interval) of
values used to estimate the unknown value of a
population parameter .
.95
p
ˆ will be in this interval
95% of the time p
Use Goldgar’s animation on
confidence intervals
When constructing confidence
intervals…
• Check conditions
– Is it a proportion? What is the sample size, n? Which
equations do you use?
– Is it a mean? What is the sample size, n? Which
equations do you use?
• Calculate the CI = confidence interval
– Follow through the calculations using the equations
• Interpret the results
– Write a statement that shows your understanding of
the Confidence interval.
Calculate confidence intervals for
proportions
Margin of error
CI %  pˆ  z%   pˆ
proportion
Standard Error
count
pˆ 
sample size
p  (1  p)
 pˆ 
n
CL/2 + 0.5
Make sure CL (confidence level)
is in decimal form
Medication side effects (confidence interval for p)
Arthritis is a painful, chronic inflammation of the joints.
An experiment on the side effects of pain relievers
examined arthritis patients to find the proportion of
patients who suffer side effects.
What are some side effects of
ibuprofen?
Serious side effects (seek medical
attention immediately):
Allergic reaction (difficulty breathing,
swelling, or hives), Muscle cramps,
numbness, or tingling, Ulcers (open sores)
in the mouth, Rapid weight gain (fluid
retention), Seizures, Black, bloody, or tarry
stools, Blood in your urine or vomit,
Decreased hearing or ringing in the ears,
Jaundice (yellowing of the skin or eyes), or
Abdominal cramping, indigestion, or
heartburn,
Less serious side effects (discuss with
your doctor):
Dizziness or headache, Nausea,
gaseousness, diarrhea, or constipation,
Depression, Fatigue or weakness, Dry
mouth, or Irregular menstrual periods
440 subjects with chronic arthritis were given ibuprofen for pain relief;
23 subjects suffered from adverse side effects.
Calculate a 90% confidence interval for the population proportion p
of arthritis patients who suffer some “adverse symptoms.”
CI %  pˆ  z%   pˆ
count
23
pˆ 

 .052  pˆ  p  (1  p)  .052  (1  .052)  .011
sample size 440
n
440
0.9 / 2  0.5  0.95
@ p  0.95; z90%  1.65
CI  .052  (1.65)(.011)
CI  .052  .018
INTERPRETATION OF THE CI:  We are 90% confident that the interval
(.034, .070) contains the true proportion of arthritis patients that experience
some adverse symptoms when taking ibuprofen.
Confidence Intervals for
sample means
Margin of error
CI %   x  z%   x
Sample mean
Standard Error

x 
n
CL/2 + 0.5
Make sure CL (confidence level)
is in decimal form
You are interested in the mean annual salary for registered nurses in
the U.S. Assume that  = $1,700. You test random samples of
35 nurses from hospitals and find the mean to be $38,000.
– Construct a 90% confidence interval.
– Construct a 95% confidence interval.
– Construct a 99% confidence interval.

x 
n
CI %   x  z%   x
CI 90%  38,000  (1.65)(
1700
)
35
CL =90%: 0.9/2 + 0.5 = .95
@p=0.95, Z90%=1.65
CI 90%  38,000  474
CI 95%
CI 95%
1700
 38,000  (1.96)(
)
35
 38,000  563
CI 99%  38,000  (2.58)(
CI 99%  38,000  741
1700
)
35
Write 1 interpretation
CL= 95%: 0.95/2+ 0.5 = .975
@p=.975; z95% = 1.96
CL=99%: 0.99/2 + 0.5 = .995
@p=.995, z99% = 2.58
The higher the confidence level, the wider the
interval
What sample size of nurses should you
use to have a confidence interval of
99% with only $500 from the mean.
Margin of error
• This is a backwards approach.
500  2.58  S
 x  500 / 2.58  194
CI %   x  z%   x

x 
n
CL=99%: 0.99/2
+ 0.5 = .995
@p=.995, z =
2.58
1700
194 
n
194  n  1700
n  1700 / 194  8.76 (square both sides)
n  76.7
The sample size tested has to be of
77 nurses in order to be 99%
confident that the mean salary of
nurses is 38,000±500