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Understandable Statistics
Seventh Edition
By Brase and Brase
Prepared by: Lynn Smith
Gloucester County College
Chapter Eight Part 2
(Sections 8.4 & 8.5)
Estimation
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1
When estimating the mean,
how large a sample must be
used in order to assure a given
level of confidence?
Use the formula:
 zc 
n

 E 
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2
2
How do we determine the value
of the population standard
deviation, ?
Use the standard deviation, s, of a
preliminary sample of size 30 or
larger to estimate .
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3
Determine the sample size
necessary to determine (with 99%
confidence) the mean time it takes
to drive from Philadelphia to
Boston. We wish to be within 15
minutes of the true time. Assume
that a preliminary sample of 45
trips had a standard deviation of
0.8 hours.
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4
... determine with 99%
confidence...
z0.99 = 2.58
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5
... We wish to be within 15
minutes of the true time. ...
E = 15 minutes = 0.25 hours
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6
...a preliminary sample of 45
trips had a standard deviation
of 0.8 hours.
Since the preliminary sample is large
enough, we can assume that the
population standard deviation is
approximately equal to 0.8 hours.
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7
Minimum Sample Size =
 zc 
n
 
 E 
2
2
 2.58(0.8) 

  68.16
 .25 
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8
Rounding Sample Size
Any fractional value of n is always
rounded to the next higher whole
number.
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9
Minimum Sample Size
• n  68.16
• Round to the next higher whole number.
• To be 99% confident in our results, the
minimum sample size = 69.
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10
Formula for Minimum Sample
Size for Estimating p for the
Binomial Distribution
If p is an estimate of the true population
proportion,
z 
n  p(1  p)  
E
2
c
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11
Formula for Minimum Sample
Size for Estimating p for the
Binomial Distribution
If we have no preliminary estimate for p, the
probability is at least c that the point estimate
r/n for p will be in error by less than the
quantity E if n is at least:
1  zc 
n 


4 E 
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2
12
Formula for Minimum Sample
Size for Estimating p for the
Binomial Distribution
If we have no preliminary estimate for p, use
the following formula to determine minimum
sample size:
1  zc 
n 


4 E 
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2
13
The manager of a furniture store
wishes to estimate the proportion
of orders delivered by the
manufacturer in less than three
weeks. She wishes to be 95% sure
that her point estimate is in error
either way by less than 0.05.
Assume no preliminary study is
done to estimate p.
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14
She wishes to be 95% sure ...
z0.95 = 1.96
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15
... that her point estimate is in
error either way by less than
0.05.
E = 0.05
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16
... no preliminary study is done
to estimate p.
1  zc 
n  
4 E 
2
2
1  1.96 
n 
  384.16
4  0.05 
The minimum required sample size (if no
preliminary study is done to estimate p) is
385.
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17
If a preliminary estimate
indicated that p was
approximately equal to 0.75:
z 
n  p(1  p )  
E
2
c
2
 1.96 
n  .75(.25) 
  288.12
 0.05 
The minimum required sample size (if this
preliminary study is done to estimate p) is
289.
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18
How can we tell if two
populations are the same?
• Compare the difference in population
means.
• Compare the difference in population
proportions.
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19
Paired Data: Dependent
Samples
Members of each pair have a natural
matching of characteristics.
Example: weight before and weight after a
diet.
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20
Independent Samples
Take a sample from one population
and an unrelated random sample
from the other population.
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21
Testing the Differences of
Means for Large Independent
Samples
• Let x1 and x2 have normal distributions
with means 1 and 2 and standard
deviations 1 and 2 respectively.
• Take independent random samples of size
n1 and n2 from each distribution.
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22
Then the variable x1  x 2
as the following characteristics :
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23
Then the variable x1  x 2
as the following characteristics :
1. A normal distributi on.
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24
Then the variable x1  x 2
as the following characteristics :
1. A normal distributi on.
2. Mean    
1
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2
25
Then the variable x1  x 2
as the following characteristics :
1. A normal distribution.
2. Mean   1   2
3. Standard deviation 
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 12
n1

 22
n2
26
If both n1 and n2 are 30 or
larger
The Central Limit Theorem can be
applied even if the original
distributions are not normal.
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27
If both n1 and n2 are 30 or
larger
The sample standard deviations
(s1 and s2) are good
approximations of the population
standard deviations (1 and 2.)
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28
Confidence Intervals for the
Differences in Means for Large
Independent Samples
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29
A c Confidence Interval
for 1 – 2 for Large Samples
(n1 and n2 30)
( x1  x2 )  E  1   2  ( x1  x2 )  E
where E  zc
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2
1
2
2
s
s

n1 n2
30
Symbols Used
 1 , x1 , s1 , n1 are the population mean,
sample mean, sample standard deviation
and sample size for population 1.
 2 , x 2 , s2 , n2 are the population mean,
sample mean, sample standard deviation
and sample size for population 2.
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31
Symbols Used
zc  critical value for confidence level
c  confidence level
0c1
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32
Some Levels of Confidence and
Their Critical Values
Level of Confidence, c Critical Value, zc
0.75
1.15
0.80
1.28
0.85
1.44
0.90
1.645
0.95
1.96
0.99
2.58
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33
Determine a 95% Confidence
Interval for the Difference in
Population Mean Exam results:
Traditional
x
88
Distance
Learning
85
s
3.3
4.1
n
31
30
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34
A 95% Confidence Interval for
1 – 2 for Large Samples (n1
and n2 30)
( x1  x 2 )  E   1   2  ( x1  x 2 )  E
x1  x 2  88  85  3
where E  zc
2
1
2
2
2
2
3.3 4.1
s
s
  1.96

 1.87
n1 n2
31
30
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35
A c% Confidence Interval for
1 – 2 for Large Samples (n1
and n2 30)
x1  x 2  3, E  1.87
( x1  x 2 )  E   1   2  ( x1  x 2 )  E
3  1.87   1   2  3  1.87
1.13   1   2  4.87
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36
We conclude (with 95%
confidence) that the difference
in the mean exam results from
the two populations falls
between 1.13 and 4.87 points.
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37
Confidence Intervals for the
Differences of Two Means of
Small Independent Samples
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38
Assumptions
• Independent random samples are drawn
from two populations with means 1 and
2.
• The parent populations have normal (or
approximately normal) distributions.
• The standard deviations for the
populations (1 and 2) are
approximately equal.
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39
A c% Confidence Interval for
1 – 2 for Small Samples
where the Standard Deviations
are Approximately Equal
( x1  x 2 )  E   1   2  ( x1  x 2 )  E
1 1
where E  t c s

n1 n2
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40
Best Estimate of the Common
or Pooled Standard Deviation
for Two Populations
( n1  1) s  ( n2  1) s
s
n1  n2  2
2
1
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2
2
41
Other Symbols Used
 1 , x1 , s1 , n1 are the population mean,
sample mean, sample standard deviation
and sample size for population 1.
 2 , x 2 , s2 , n2 are the population mean,
sample mean, sample standard deviation
and sample size for population 2.
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42
Symbols Used
t c  critical value for confidence level
d.f.  n 1  n 2  2
c  confidence level
0c1
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43
Determine a 99% Confidence
Interval for the Difference in
Population Means:
x
M.p.h.
Group 1
54
M.p.h.
Group 2
52
s
2.3
2.1
n
14
17
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44
Find s = Pooled Standard
Deviation
( n1  1) s  ( n2  1) s
s

n1  n2  2
2
1
2
2
(14  1)2.3  (17  1)2.1
 2.192
14  17  2
2
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2
45
To Find a 99% Confidence
Interval
• d.f. = 14 + 17 – 2 = 29.
• Use the column headed by c = 0.990 in
Table 6 Appendix II to find t0.99 for d.f. =
29.
• t0.99 = 2.756.
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46
Find E, when
tc = 2.756 and s = 2.192
1 1
E  tc s

n1 n2
1 1
 2.756( 2.192)

 2.180
14 17
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47
The 99% Confidence Interval
for 1 – 2 for Small Samples
( x1  x 2 )  E   1   2  ( x1  x 2 )  E
x1  x 2  54  52  2
E  2.180
2  2.180   1   2  2  2.180
 0.18   1   2  4.180
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48
Conclusion
We are 99% confident that the
differences in speeds (m.p.h. for
Group 1 minus m.p.h. for Group 2)
between the two groups ranges
from negative 0.18 to positive 4.18
m.p.h.
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49
Confidence Intervals
for the difference of two
proportions from binomial
distributions
p1 – p2
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50
A c% Confidence Interval for
p1 – p2 for Large Samples
( pˆ 1  pˆ 2 )  E  p1  p2  ( pˆ 1  pˆ 2 )  E
where E  zcˆ  zc
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pˆ 1qˆ1 pˆ 2 qˆ 2

n1
n2
51
Symbols Used
zc  critical value for confidence level
c  confidence level
0c1
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52
Assume
• n1 and r1 = number of trials and number
of successes in binomial experiment 1.
• n2 and r2 = number of trials and number
of successes in binomial experiment 2.
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53
Assume
r1
r1
pˆ 1  and qˆ1  1 
n1
n1
r2
r2
pˆ 2  and qˆ 2  1 
n2
n2
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54
Assume that the following are
all larger than five:
n1 pˆ 1 , n1qˆ1 , n2 pˆ 2 , n2 qˆ 2
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55
Some Levels of Confidence and
Their Critical Values
Level of Confidence, c Critical Value, zc
0.75
1.15
0.80
1.28
0.85
1.44
0.90
1.645
0.95
1.96
0.99
2.58
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56
Find a 95% confidence interval
for the difference in the
proportions of students who
participate in early
registration. The two groups
are those who received a
reminder telephone call and
those who did not.
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57
The data:
n
Reminder No Reminder
(Group 1)
(Group 2)
600
600
Registered early
475
452
p̂
0.79
0.75
q̂
0.21
0.25
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58
We must assure that the
following are all larger than
five:
n1 pˆ 1 , n1qˆ1 , n2 pˆ 2 , n2 qˆ 2
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59
Checking
n1 pˆ 1  475
n1qˆ1  125
n2 pˆ 2  452
n2 qˆ 2  148
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60
Compute the sample statistic
p1  p2  0.79  0.75  0.04
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61
For a 95% confidence interval
zc = 1.96
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62
Calculate E
E  zc
pˆ 1qˆ1 pˆ 2 qˆ 2
.79(.21) .75(.25)

 1.96

n1
n2
600
600
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63
Calculate E
E  zc
pˆ 1qˆ1 pˆ 2 qˆ 2
.79(.21) .75(.25)

 1.96

n1
n2
600
600
 1.96 .000589  1.96(0.0243)  0.048
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64
Determine the Confidence
Interval
p1  p2  0.79  0.75  0.04
E  0.048
( pˆ 1  pˆ 2 )  E  p1  p2  ( pˆ 1  pˆ 2 )  E
(0.04)  0.048  p1  p2  (0.04)  0.048
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65
The 95% Confidence Interval
(0.04)  0.048  p1  p2  (0.04)  0.048
 0.008  p1  p2  0.088
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66
Interpreting a c% Confidence
Interval for p1 – p2
There are three cases.
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67
If the confidence interval
contains only negative values
Conclude that p1 – p2 < 0.
We are, therefore c% confident
that p1 < p2.
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68
If the confidence interval
contains only positive values
Conclude that p1 – p2 > 0.
We are, therefore c% confident
that p1 > p2.
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69
If the confidence interval
contains both positive and
negative values
We cannot conclude at c% confidence
that either p1 or p2 is larger.
A smaller c which would yield a
shorter confidence interval might
permit a conclusion.
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70
In our example, the 95%
Confidence Interval was:
 0.008  p1  p2  0.088
We cannot conclude that the reminder
telephone call made a difference in the
proportion of students who participate in
early registration.
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71