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Understandable Statistics
Seventh Edition
By Brase and Brase
Prepared by: Lynn Smith
Gloucester County College
Chapter Eight Part 1
(Sections 8.1 to 8.3)
Estimation
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1
Point Estimate
an estimate of a population
parameter given by a single
number
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2
Examples of Point Estimates
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3
Examples of Point Estimates
• x is used as a point estimate for  .
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4
Examples of Point Estimates
• x is used as a point estimate for  .
• s is used as a point estimate for .
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5
Error of Estimate
the magnitude of the difference
between the point estimate and
the true parameter value
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6
The error of estimate using
x as a point estimate for 
is
x
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7
Confidence Level
• A confidence level, c, is a measure of the
degree of assurance we have in our
results.
• The value of c may be any number
between zero and one.
• Typical values for c include 0.90, 0.95,
and 0.99.
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8
Critical Value for a Confidence
Level, c
the value zc such that the area
under the standard normal curve
falling between – zc and zc is
equal to c.
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9
Critical Value for a Confidence
Level, c
P(– zc < z < zc ) = c
This area = c.
– zc
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0
zc
10
Find z0.90 such that 90% of the
area under the normal curve
lies between z-0.90 and z0.90.
P(-z0.90 < z < z0.90 ) = 0.90
.90
– z.90
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0
z.90
11
Find z0.90 such that 90% of the
area under the normal curve
lies between z-0.90 and z0.90.
P(0< z < z0.90 ) = 0.90/2 = 0.4500
.4500
– z.90
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0
z.90
12
Find z0.90 such that 90% of the
area under the normal curve
lies between z-0.90 and z0.90.
P( z < z0.90 ) = .5 + 0.4500 = .9500
.9500
– z.90
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0
z.90
13
Find z0.90 such that 90% of the
area under the normal curve
lies between z-0.90 and z0.90.
• According to Table 5a in Appendix II,
0.9500 lies exactly halfway between two area
values in the table (.9495 and .9505).
• Averaging the z values associated with these
areas gives z0.90 = 1.645.
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14
Common Levels of Confidence
and Their Corresponding
Critical Values
Level of Confidence, c
Critical Value zc
0.90, or 90%
1.645
0.95, or 95%
1.96
0.98, or 98%
2.33
0.99, or 99%
2.58
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15
Confidence Interval for the
Mean of Large Samples (n  30)
xE
where
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   xE
x  Sample Mean
16
Confidence Interval for the
Mean of Large Samples (n  30)
xE
where
   xE
x  Sample Mean
s
Ez
n
c
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17
Confidence Interval for the
Mean of Large Samples (n  30)
   xE
x  Sample Mean
s
Ez
n
s  sample standard deviation
xE
where
c
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18
Confidence Interval for the
Mean of Large Samples (n  30)
   xE
x  Sample Mean
s
Ez
n
s  sample standard deviation
c  confidence level (0  c  1)
xE
where
c
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19
Confidence Interval for the
Mean of Large Samples (n  30)
xE
where



xE
x  Sample Mean
s
Ez
n
s  sample standard deviation
c
c  confidence level (0  c  1)
z  critical value for confidence level c
c
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20
Confidence Interval for the
Mean of Large Samples (n  30)
   xE
x  Sample Mean
s
Ez
n
s  sample standard deviation
c  confidence level (0  c  1)
z  critical value for confidence level c
xE
where
c
c
n  sample size
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21
Create a 95% confidence
interval for the mean driving
time between Philadelphia and
Boston.
Assume that the mean driving
time of 64 trips was 6.4 hours
with a standard deviation of 0.9
hours.
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22
x
= 6.4 hours
s = 0.9 hours
c = 95%,
so zc = 1.96
n = 64
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23
x = 6.4 hours
s = 0.9 hours
Approximate  as s = 0.9 hours.
95% Confidence interval will be from
x  E to
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xE
24
= 6.4 hours
x
s = 0.9 hours
c = 95%, so zc = 1.96
n = 64
E  zc
s
0.9
 1.96
 .2205
n
64
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25
95% Confidence Interval:
6.4 – .2205 <  < 6.4 + .2205
6.1795 <  < 6.6205
We are 95% sure that the true time is
between 6.18 and 6.62 hours.
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26
What if it is impossible or
impractical to use a large
sample?
Apply the Student’s t distribution.
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27
Student’s t Variable
x
t
s
n
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28
The shape of the t distribution
depends only only the sample
size, n, if the basic variable x
has a normal distribution.
When using the t distribution, we will
assume that the x distribution is normal.
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29
Table 6 in Appendix II gives
values of the variable t
corresponding to the number of
degrees of freedom (d.f.)
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30
Degrees of Freedom
d.f. = n – 1
where n = sample size
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31
The t Distribution has a Shape
Similar to that of the the
Normal Distribution
A Normal
distribution
A “t”
distribution
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32
Find the critical value tc for a
95% confidence interval if n = 8.
d.f.=7
c
'
''
d.f.
...
6
7
8
...0.90
...
0.95
...
0.98
...
0.99
...
...1.943
...1.895
...1.860
2.447
2.365
2.306
3.143
2.998
2.896
3.707
3.499
3.355
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33
Confidence Interval for the
Mean of Small Samples (n < 30)
from Normal Populations
xE xE
where x  Sample Mean
s
E  tc
n
c = confidence level (0 < c < 1)
tc = critical value for confidence level c, and
degrees of freedom = n - 1
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34
The mean weight of eight fish
caught in a local lake is 15.7
ounces with a standard deviation
of 2.3 ounces.
Construct a 90% confidence
interval for the mean weight of
the population of fish in the lake.
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35
Mean = 15.7 ounces
Standard deviation = 2.3 ounces.
• n = 8, so d.f. = n – 1 = 7
• For c = 0.90, Table 6 in Appendix II gives
t0.90 = 1.895.
s
2.3
E  tc
 1.895
 1.54
n
8
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36
Mean = 15.7 ounces
Standard deviation = 2.3 ounces.
E = 1.54
The 90% confidence interval is:
xE xE
15.7 - 1.54 <  < 15.7 + 1.54
14.16 <  < 17.24
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37
The 90% Confidence Interval:
14.16 <  < 17.24
We are 90% sure that the true
mean weight of the fish in the
lake is between 14.16 and 17.24
ounces.
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38
Review of the Binomial
Distribution
• Completely determined by the number of
trials (n) and the probability of success
(p) in a single trial.
• q=1–p
• If np and nq are both > 5, the binomial
distribution can be approximated by the
normal distribution.
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39
A Point Estimate for p, the
Population Proportion of
Successes
r
pˆ ( read as " p hat " ) 
n
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40
Point Estimate for q
(Population Proportion of
Failures)
qˆ ( read as " q hat " )  1  pˆ
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41
For a sample of 500 airplane
departures, 370 departed on time.
Use this information to estimate
the probability that an airplane
from the entire population departs
on time.
r 370
pˆ  
 0.74
n 500
We estimate that there is a 74% chance that any
given flight will depart on time.
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42
Error of Estimate for “p hat”
as a Point Estimate for p
pˆ  p
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43
A c Confidence Interval for p
for Large Samples
(np > 5 and nq > 5)
pˆ  E  p
 pˆ  E
r
where pˆ 
and E  z
n
c
pˆ (1  pˆ )
n
zc = critical value for confidence level c
taken from a normal distribution
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44
For a sample of 500 airplane
departures, 370 departed on
time. Find a 99% confidence
interval for the proportion of
airplanes that depart on time.
Is the use of the normal distribution
justified?
n  500
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pˆ  0.74
45
For a sample of 500 airplane
departures, 370 departed on
time. Find a 99% confidence
interval for the proportion of
airplanes that depart on time.
Can we use the normal distribution?
npˆ  370
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nqˆ  130
46
For a sample of 500 airplane
departures, 370 departed on
time. Find a 99% confidence
interval for the proportion of
airplanes that depart on time.
npˆ and
nqˆ are
both  5
so the use of the normal distribution is justified.
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47
Out of 500 departures, 370
departed on time. Find a 99%
confidence interval.
r 370
pˆ  
 0.74
n 500
.74(.26)
E  2.58
 0.0506
500
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48
99% confidence interval for the
proportion of airplanes that
depart on time:
E = 0.0506
Confidence interval is:
ˆ E  p p
ˆE
p
.74  0.0506  p  .74  0.0506
0.6894  p  0.7906
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49
99% confidence interval for the
proportion of airplanes that
depart on time
Confidence interval is
0.6894 < p < 0.7906
We are 99% confident that between 69%
and 79% of the planes depart on time.
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50
The point estimate and the
confidence interval do not
depend on the size of the
population.
The sample size, however, does
affect the accuracy of the statistical
estimate.
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51
Margin of Error
The margin of error is the maximal
error of estimate E for a
confidence interval.
Usually, a 95% confidence interval
is assumed.
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52
Interpretation of Poll Results
The proportion responding in a certain way is
p̂
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53
A 95% confidence interval for
population proportion p is:
pˆ  m arg in of error  p  pˆ  m arg in of error
pˆ  poll report
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54
Interpret the following poll
results:
“ A recent survey of 400 households
indicated that 84% of the households
surveyed preferred a new breakfast
cereal to their previous brand. Chances
are 19 out of 20 that if all households had
been surveyed, the results would differ by
no more than 3.5 percentage points in
either direction.”
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55
“Chances are 19 out of 20 …”
19/20 = 0.95
A 95% confidence interval is being
used.
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56
“... 84% of the households
surveyed preferred …”
84% represents the percentage of
households who preferred the
new cereal.
84% represents pˆ .
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57
“... the results would differ by
no more than 3.5 percentage
points in either direction.”
3.5% represents the margin of
error, E.
The confidence interval is:
84% - 3.5% < p < 84% + 3.5%
80.5% < p < 87.5%
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58
The poll indicates ( with 95%
confidence):
between 80.5% and 87.5% of the
population prefer the new cereal.
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59