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6.1 Confidence Interval for the Mean
(n  30 or σ known w/normal population)
Often we do not know the Mean of an entire population, so we estimate
Point Estimate: A single value estimate – most unbiased estimate of the
Population mean is the sample mean x
Interval Estimate: Range of values to estimate a population parameter.
Forms an interval using a margin of error (E)
(x – E , x + E)
Margin of Error: A sampling error is the difference between the actual
mean and the point estimate of the mean. x – μ
Since we do not usually know the actual mean and the
sample mean varies from sample to sample, we construct a
maximum value of error (called margin of error)
within a level of confidence
Larson/Farber
1
Estimating the Population Mean μ
Market researchers use the number of sentences per advertisement as a
measure of readability for magazine advertisements. The following represents
a random sample of the number of sentences found in 50 advertisements. Find
a point estimate of the population mean, . (Journal of Advertising Research)
9 20 18 16 9 9 11 13 22 16 5 18 6 6 5 12 25
17 23 7 10 9 10 10 5 11 18 18 9 9 17 13 11 7
14 6 11 12 11 6 12 14 11 9 18 12 12 17 11 20
x
x 620

 12.4
n
50
Your point estimate for the mean length of all
magazine advertisements is 12.4 sentences.
Interval estimate : An interval, or range of values, used to estimate a population parameter.
Point estimate
How confident do we want
10.3
14.5
12.4
to be that the interval
(
)
estimate contains the
population mean μ? (90%,
95%, 80%??? - This
Interval estimate (with margin of error = 2.1)
determines what our margin
2
of error will be.
•
Larson/Farber
Confidence Intervals
Level of confidence c
• The probability that the interval estimate contains the population parameter.
c
½(1 – c)
c is the area under the standard normal
curve between the critical values.
½(1 – c)
-zc
z=0
Critical values
zc
z
Use the Standard Normal Table to
find the corresponding z-scores.
• If the level of confidence is 90% (c = .90) this means that we are 90%
confident that the interval contains the population mean μ.
• There is 10% left for the ‘tails’ of the distribution. (5% for each tail = .05)
• -zc = invnorm (.05) = -1.645 and zc = invnorm (.95) = 1.645
Larson/Farber 4th ed
3
Margin of Error (E)
• Greatest possible distance between the point estimate and the value of the
parameter it is estimating for a given level of confidence, c.
• Sometimes called the maximum error of estimate or error tolerance.
σ
E  zcσ x  zc
n
When n  30, the sample standard
deviation, s, can be used for .
Example: Use the magazine advertisement data and a 95% confidence level
to find the margin of error for the mean number of sentences in all magazine
advertisements. Assume the sample standard deviation (s) is about 5.0.
95% of the area under the curve
falls within 1.96 standard
deviations of the mean.
0.95
0.025
0.025
-zc = -1.96
E = 1.96 (5/√50) = 1.4 sentences.
Larson/Farber
z=0
zzcc= 1.96
Ti 83/84
Stat-Tests
7:Zinterval ()
z
11.0 < μ < 13.8
A 95% confidence interval for the population mean
(Probability confidence interval contains μ is 95%)
(x – E , x + E)
4
(12.4 – 1.4 , 12.4 + 1.4) = (11 , 13.8)
Constructing Confidence Intervals for μ
Finding a Confidence Interval for a Population Mean
(n  30 or σ known with a normally distributed population)
In Words
In Symbols
x
x
x
n
1.
Find the sample statistics n and
2.
Specify , if known. Otherwise, if n  30,
find the sample standard deviation s and use
it as an estimate for .
(x  x )2
s
n 1
3.
Find the critical value zc that corresponds to
the given level of confidence.
Std. Norm. Table -> InvNorm()
4.
Find the margin of error E.
5.
Find the left and right endpoints and form
the confidence interval.
E  zc

n
Interval: x  E    x  E
Interpretation: “If a large number of samples is collected and a confidence interval
is created for each sample, approximately C% of these intervals will contain μ.”
Larson/Farber
5
Practice
A publisher wants to estimate the mean length of time (in minutes)
All adults spend reading newspapers. To determine this estimate,
the publisher takes a random sample of 15 people and obtains the following results.
11, 9, 8, 10, 10, 9, 7, 11, 11, 7, 6, 9, 10, 8, 10
From past studies, the publisher assumes a population std deviation of 1.5 minutes
And that the population is Normally Distributed
Construct a 90% and 99% confidence interval for the population mean.
Larson/Farber
6
Sample Size
Given a c-confidence level & a margin of error E, the minimum
2
 zc 
n
sample size n needed to estimate the population mean  is
E 


(If  is unknown you can estimate it with ‘s’ if sample >= 30)
Example: You want to estimate the mean number of sentences in a magazine
advertisement. How many magazine advertisements must be included in the
sample if you want to be 95% confident that the sample mean is within one
sentence of the population mean? Assume the sample std. deviation is about 5.0.
0.95
0.025
0.025
-zc = -1.96 z = 0
Larson/Farber
zczc= 1.96
z
 z    1.96  5.0 
n c  
  96.04
E
1


 
You should include at least 97 magazine
7
advertisements in your sample
2
2
6.2 Confidence Interval for Mean
(σ unknown )
x -
t
s
n
• When the population standard deviation is unknown, the sample size is less
than 30, and the random variable x is approximately normally distributed, it
follows a t-distribution (critical values are denoted tc) w/properties below:
1.
2.
3.
4.
5.
The t-distribution is bell shaped and symmetric about the mean.
The t-distribution is a family of curves, each determined by a parameter
called the degrees of freedom. When estimating a population mean, the
degrees of freedom = n – 1
The total area under a t-curve is 1 or 100%.
The mean, median, and mode of the t-distribution are equal to zero.
As the degrees of freedom increase, t-distribution approaches the normal
distribution. After 30 d.f., t-distribution is very close to the standard
normal z-distribution.
The tails in the t-distribution
are “thicker” than those in the
standard normal distribution.
Larson/Farber
d.f. = 2
d.f. = 5
Standard normal curve
0
t
8
Example: Critical Values of t
Find the critical value tc for a 95% confidence when the sample size is 15.
Solution: d.f. = n – 1 = 15 – 1 = 14
95% of the area under the tdistribution curve with 14 degrees of
freedom lies between t = +2.145.
c = 0.95
tc = 2.145
-tc = -2.145
tc =2.145
t
A c-confidence interval for the population mean μ
(The probability that the confidence interval contains μ is c.
x E   x E
Larson/Farber
where E  tc
s
n
9
Confidence Intervals and t-Distributions
In Words
In Symbols
x
x
n
(x  x )2
s
n 1
1.
Identify the sample statistics n, x , and s.
2.
Identify the degrees of freedom, level of
confidence c, and the critical value tc.
d.f. = n – 1
3.
Find the margin of error E.
E  tc
4.
Find the left and right endpoints & find
the confidence interval.
Larson/Farber
s
n
xE  xE
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Example: Constructing a Confidence
Interval
You randomly select 16 coffee shops and measure the temperature of the coffee
sold at each. The sample mean temperature is 162.0ºF with a sample standard
deviation of 10.0ºF. Find the 95% confidence interval for the mean
temperature. Assume the temperatures are approximately normally distributed.
Use t-distribution (n < 30, σ unknown, approximate normal distribution)
•
•
•
n =16, x = 162.0 s = 10.0 c = 0.95
df = n – 1 = 16 – 1 = 15
tc = 2.131
Margin of Error
s  2.131 10  5.3
E  tc
n
16
Confidence Interval
(162 – 5.3, 162 +5.3) = (156.7, 167.3)
156.7 < μ < 167.3
With 95% confidence, you can say that the mean temperature of coffee sold is
between 156.7ºF and 167.3ºF.
Larson/Farber
Ti 83/84
Stat-Tests
11()
8:Tinterval
z-Normal Distribution OR t-Distribution?
z
t
Note: You must have reason to believe you are working with an approximately normal
distribution to use the t-distribution. If n  30, then according to the Central Limit Theorem the
sampling distribution of the sample means approximates a normal distribution, so you have met
this part of the criteria. Additionally, if n  30, the t-distribution is very close to the standard
normal z-distribution so actually you could use either the z or t distribution, though your book
guides you to use a “t” in this situation since the standard deviation of the population is unknown.
If n < 30 and the population is not normally distributed (or you do not know) then you cannot
use. the standard normal (z) or the t-distribution.
Example: Normal(z) or t-Distribution?
You randomly select 25 newly constructed houses. The
sample mean construction cost is $181,000 and the
population standard deviation is $28,000. Assuming
construction costs are normally distributed, should you
use the normal distribution, the t-distribution, or neither
to construct a 95% confidence interval for the
population mean construction cost?
Solution:
Use the normal (z) distribution (the population is
normally distributed and the population standard
deviation is known)
Larson/Farber
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Practice
Assume the variable is normally distributed and use (as appropriate)
a normal distribution or t-distribution to construct a 90% confidence interval
for the population mean.
A) In a random sample of 10 adults from the U.S., the mean waste generated
Per person per day was 4.54 pounds and the standard deviation was 1.21 pounds.
B) Repeat part (a), assuming the same statistics came from a sample size of 500.
Note that the sample size is quite large (much greater than 30).
Construct a 90% confidence interval for the population mean first using
a z-standard normal distribution, then again using a t-distribution. Compare
The two answers and consider the reason for your results.
Larson/Farber
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6.3 Confidence Interval for Population
Proportions
Recall that ‘p’ is the probability of success in a binomial experiment.
Just as we estimated the mean, we can estimate population proportions (p).
Point Estimate for p (“p-hat”)
Point Estimate for q (“q-hat”)
pˆ 
x number of successes in sample

n
number in sample
qˆ  1  pˆ
A binomial distribution can be approximated by the normal distribution.
If np >= 5 and nq >=5 p̂. is approximately ‘normal’.
Example: A survey of 1219 U.S.
adults: 354 said that their favorite
sport to watch is football. Find a
point estimate for the population
proportion of U.S. adults who say
their favorite sport to watch is
football. (The Harris Poll)
x 354
pˆ  
 0.290402  29.0%
n 1219
A c-confidence interval for the population
proportion p (The probability that the
confidence interval contains p is c.)
pˆ  E  p  pˆ  E
where E  zc
pq
ˆˆ
n
Constructing Confidence Intervals for p
In Words
1.
In Symbols
Identify the sample statistics n and x.
2. Find the point estimate p̂.
3.
Verify that the sampling distribution of p̂
can be approximated by the normal
distribution.
4.
Find the critical value zc that corresponds
to the given level of confidence c.
5.
Find the margin of error, E.
6.
Find left & right endpoints and find
confidence interval.
Larson/Farber
pˆ 
x
n
npˆ  5, nqˆ  5
Use Standard Normal Table
E  zc
pq
ˆˆ
n
pˆ  E  p  pˆ  E
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Example: Confidence Interval for p
In a survey of 1219 U.S. adults, 354 said that their favorite sport to watch is
football. Construct a 95% confidence interval for the proportion of adults in
the United States who say that their favorite sport to watch is football.
Solution: Recall
pˆ  0.290402
qˆ  1  pˆ  1  0.290402  0.709598
• Verify the sampling distribution of p̂ can be approximated by the normal
npˆ  1219  0.290402  354  5
distribution
nqˆ  1219  0.709598  865  5
•
Margin of error:
invNorm (.975) = 1.96
E  zc
•
Confidence Interval:
Larson/Farber
p̂
Ti 83/84
Stat-Tests
A:1-PropZInt
pq
(0.290402)  (0.709598)
ˆˆ
 1.96
 0.025
n
1219
-E
to
p̂
+E
0.265 < p < 0.315
With 95% confidence, you can say that the proportion
of adults who say football is their favorite sport is
17
between 26.5% and 31.5%.
Sample Size
• Given a c-confidence level and a margin of error E, the minimum sample
2
size n needed to estimate p is
z
 c
ˆ ˆ 
n  pq
E
• This formula assumes you have an estimate for p̂ and qˆ .
• If not, use pˆ  0.5 and qˆ  0.5.
Example: You are running a political campaign and wish to estimate, with
95% confidence, the proportion of registered voters who will vote for your
candidate. Your estimate must be accurate within 3% of the true population.
Find the minimum sample size needed if 1) no preliminary estimate is
available and 2) a preliminary estimate gives p̂ = .31
qˆ  0.5.
#1: Since you do not have a preliminary estimate, use: pˆ  0.5
2
2
 zc 
 1.96 
ˆ ˆ    (0.5)(0.5) 
n  pq
Answer: 1068 voters
  1067.11
 0.03 
E
#2: Use preliminary estimate:
2
Larson/Farber
pˆ  0.31
qˆ  1  pˆ  1  0.31  0.69
2
z 
 1.96 
ˆ ˆ  c   (0.31)(0.69) 
n  pq
  913.02
E
0.
03


 
Answer: 914 voters
18
Practice
You are a travel agent and wish to estimate with 95% confidence, the
Proportion of vacationers who plan to travel outside the U.S. in the next
12 months. Your estimate must be accurate within 3% of the true proportion.
(A) No preliminary estimate is available. Find the minimum sample size.
(A) Find the minimum sample size needed, using a prior study that found that
26% of the respondents said they planned to travel outside the
U.S. in the next 12 months.
Larson/Farber
19