Download Confidence Interval

Learning Objectives for Chapter 8
After careful study of this chapter, you should be able to do
the following:
1.
2.
3.
4.
5.
6.
Construct confidence intervals on the mean of a normal distribution,
using normal distribution or t distribution method.
Construct confidence intervals on variance and standard deviation of
normal distribution.
Construct confidence intervals on a population proportion.
Constructing an approximate confidence interval on a parameter.
Prediction intervals for a future observation.
Tolerance interval for a normal population.
Chapter 8 Learning Objectives
2
Introduction
Statistical Intervals For A Single Sample
Confidence Interval.
 We have learned in the previous chapter how a parameter can be estimated from sample
data
 It is important to understand how good is the estimate obtained.
 An interval estimate for population parameter is called a Confidence Interval.
•The confidence interval is a random interval
• The appropriate interpretation of a confidence interval (for example on ) is: The
observed interval [l, u] brackets the true value of , with confidence 100(1-).
 Tolerance interval is another important type of interval estimate.
Confidence interval
• A range of values, derived from sample statistics, that is likely to contain the value
of an unknown population parameter.
• Because of their random nature, it is unlikely that two samples from a given
population will yield identical confidence intervals.
• But if you repeated your sample many times, a certain percentage of the resulting
confidence intervals would contain the unknown population parameter.
• The percentage of these confidence intervals that contain the parameter is the
confidence level of the interval.
•
For example, suppose you want to know the average amount of time it takes for an
automobile assembly line to complete a vehicle. You take a sample of completed
cars, record the time they spent on the assembly line, and use the 1-sample t
procedure to obtain a 95% confidence interval for the mean amount of time all
cars spend on the assembly line. Because 95% of the confidence intervals
constructed from all possible samples will contain the population parameter, you
conclude that the mean amount of time all cars spend on the assembly line falls
between your interval's endpoints, which are called confidence limits.
8-1.1 Confidence Interval and its Properties
A confidence interval estimate for  is an interval of the form
l ≤  ≤ u,
where the end-points l and u are computed from the sample
data.
There is a probability of 1  α of selecting a sample for which
the CI will contain the true value of .
The endpoints or bounds l and u are called lower- and upperconfidence limits ,and 1  α is called the confidence
coefficient.
Sec 8-1 Confidence Interval on the Mean of a Normal, σ2 Known
5
Confidence interval
•
Creating confidence intervals is analogous to throwing nets over a target with an unknown, yet fixed,
location. Consider the graphic below, which depicts confidence intervals generated from 20 samples from
the same population. The black line represents the fixed value of the unknown population parameter;
the blue confidence intervals contain the value of the population parameter; the red confidence interval
does not.
•
A 95% confidence interval indicates that 19 out of 20 samples (95%) from the same population will
produce confidence intervals that contain the population parameter.
•
Confidence Interval on the Mean of a Normal
Distribution, Variance Known
Figure 8-1 Repeated construction of a confidence interval for .
Guidelines for Constructing Confidence Intervals
 Prediction Interval: Provides prediction on future observations.
Hw 2 groups
• Probability Tables
• Description, Objective and uses …5 ++ slides
Confidence Interval on the Mean of a Normal
Distribution, Variance Known
8-2.1 Development of the Confidence Interval and its Basic
Properties
•
The endpoints or bounds l and u are called lower- and upper-confidence limits,
respectively.
• Since Z follows a standard normal distribution, we can write:
Confidence Interval on the Mean of a Normal Distribution,
Variance Known
One sided confidence Interval
Confidence Interval on the Mean of a Normal
Distribution, Variance Known
Example 8-1
Confidence interval
• A range of values, derived from sample statistics, that is likely to contain the value of an
unknown population parameter.
•
Because of their random nature, it is unlikely that two samples from a given population will
yield identical confidence intervals.
• But if you repeated your sample many times, a certain percentage of the resulting confidence
intervals would contain the unknown population parameter.
• The percentage of these confidence intervals that contain the parameter is the confidence level
of the interval.
•
For example, suppose you want to know the average amount of time it takes for an
automobile assembly line to complete a vehicle. You take a sample of completed cars, record
the time they spent on the assembly line, and use the 1-sample t procedure to obtain a 95%
confidence interval for the mean amount of time all cars spend on the assembly line. Because
95% of the confidence intervals constructed from all possible samples will contain the
population parameter, you conclude that the mean amount of time all cars spend on the
assembly line falls between your interval's endpoints, which are called confidence limits.
Confidence Interval on the Mean of a Normal
Distribution, Variance Known
Figure 8-1 Repeated construction of a confidence interval for .
Confidence Interval on the Mean of a Normal Distribution,
Variance Known
Confidence Level and Precision of Error
The length of a confidence interval is a measure of the precision of
estimation.
Figure 8-2 Error in estimating  with
.
x
Sample Confidence Interval on the Mean of a Normal Distribution, Variance Known
8-2.2 Choice of Sample Size
Confidence Interval on the Mean of a Normal Distribution, Variance Known
EXAMPLE 8-2 Metallic Material Transition
Example 8-2
8-1.3 One-Sided Confidence Bounds
A 100(1 − α)% upper-confidence bound for  is
  x  z / n
(8-3)
and a 100(1 − α)% lower-confidence bound for 
is
x  z / n  l  
Sec 8-1 Confidence Interval on the Mean of a Normal, σ2 Known
(8-4)
26
Example 8-3 One-Sided Confidence Bound
The same data for impact testing from Example 8-1 are used
to construct a lower, one-sided 95% confidence interval for
the mean impact energy.
Recall that zα = 1.64, n = 10,  = l, and x  64.46 .
A 100(1 − α)% lower-confidence bound for  is

x  z

n

 

10
  
Sec 8-1 Confidence Interval on the Mean of a Normal, σ2 Known
27
Confidence Interval on the Mean of a Normal Distribution, Variance Known
8-2.5 A Large-Sample Confidence Interval for 
See example 8-3
Confidence Interval on the Mean of a Normal
Distribution, Variance Known
8-2.5 A Large-Sample Confidence Interval for 
Definition
Confidence Interval on the Mean of a Normal Distribution, Variance Known,
Example 8-5 Mercury Contamination
Example 8-3
Example 8-5 Mercury Contamination (continued)
The summary statistics for the data are as follows:
Variable
Concentration
N
53
Mean Median StDev Minimum Maximum
Q1
Q3
0.5250 0.4900 0.3486
0.0400
1.3300
0.2300 0.7900
Because n > 40, the assumption of normality is not necessary to use in Equation
8-5. The required values are n  53, x  0.5250, s  0.3486 , and z0.025 = 1.96.
The approximate 95 CI on  is
s
s
 μ  x  z0.025
n
n
0.3486
0.3486
0.5250  1.96
 μ  0.5250  1.96
53
53
0.4311  μ  0.6189
x  z0.025
Interpretation: This interval is fairly wide because there is variability in the
mercury concentration measurements. A larger sample size would have produced
a shorter interval.
Sec 8-1 Confidence Interval on the Mean of a Normal, σ2 Known
31
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•
•
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Text and Material
Exam 1
Probability Tables
Home Work Policy
Attendance policy
t-distribution
•
•
•
In probability and statistics, Student's t-distribution (or simply the t-distribution) is any member of a
family of continuousprobability distributions that arises when estimating the mean of a normally
distributed population in situations where thesample size is small and population standard deviation is
unknown.
he t-distribution plays a role in a number of widely used statistical analyses, including Student's t-test for
assessing thestatistical significance of the difference between two sample means, the construction
of confidence intervals for the difference between two population means, and in linear regression
analysis. The Student's t-distribution also arises in theBayesian analysis of data from a normal family.
The t-distribution is symmetric and bell-shaped, like the normal distribution, but has heavier tails,
meaning that it is more prone to producing values that fall far from its mean. This makes it useful for
understanding the statistical behavior of certain types of ratios of random quantities, in which variation in
the denominator is amplified and may produce outlying values when the denominator of the ratio falls
close to zero. The Student's t-distribution is a special case of the generalised hyperbolic distribution
Cumulative distribution function
Probability density function
Confidence Interval on the mean of a normal distribution,
Variance Unknown
8-2.2 The Confidence Interval on Mean, Variance Unknown
Table III
If and s are the mean and standard deviation of a random sample
from
x a normal distribution with unknown variance 2, a 100(1  )
confidence interval on  is given by
x  t /2, n 1s/ n    x  t /2, n 1s/ n
(8-7)
where t2,n1 the upper 1002 percentage point of the t distribution with
n  1 degrees of freedom.
One-sided confidence bounds on the mean are found by replacing
t/2,n-1 in Equation 8-7 with t ,n-1.
Sec 8-2 Confidence Interval on the Mean of a Normal, σ2 Unknown
36
Example
Example 8-6 Alloy Adhesion
Construct a 95% CI on  to the following data.
19.8
15.4
11.4
19.5
The sample mean is
10.1
18.5
14.1
8.8
14.9
7.9
17.6
13.6
7.5
12.7
16.7
11.9
15.4
11.9
15.8
11.4
15.4
11.4
x  13.71 and sample standard deviation is s = 3.55.
Since n = 22, we have n  1 =21 degrees of freedom for t, so t0.025,21 = 2.080.
The resulting CI is
x  t /2, n 1s / n    x  t /2, n 1s/ n
13.71  2.080(3.55) / 22    13.71  2.0803.55 / 22
13.71  1.57    13.71  1.57
12.14    15.28
Interpretation: The CI is fairly wide because there is a lot of variability in the
measurements. A larger sample size would have led to a shorter interval.
Sec 8-2 Confidence Interval on the Mean of a Normal, σ2 Unknown
39
Sec 8-7 to write
40
Sec 8-7 to write
41
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•
•
•
Exam : First week on March
No cell phones
Signature Cheat Sheet
First three Chapters + Sat 1 (Introduction)
(Chi-square or χ²-distribution)
•
•
The Chi Square test is the most important and most used method in statistical tests.
The purpose of Chi Square test
A-Test how well a sample fits a theoretical distribution
B- The difference between an observed frequency and expected frequency. (differences
between the two or more observed data).
C- Its value can be calculated by using the given observed frequency and expected frequency.
D- Test the independence between categorical variables. For example, a manufacturer wants to
know if the occurrence of four types of defects (missing pin, broken clamp, loose fastener, and
leaky seal) is related to shift (day, evening, overnight).
•
•
•
•
The chi-squared distribution (also It is a special case of the gamma distribution and is
one of the most widely used probability distributions in inferential statistics,
For example, you can use a goodness-of-fit test of an observed distribution to a
theoretical one and classification of qualitative data, and in confidence
interval estimation for a population standard deviation of a normal distribution from a
sample standard deviation. also to determine whether your sample data fit a Poisson
distribution.
The shape of the chi-square distribution depends on the number of degrees of freedom.
The distribution is positively skewed, but skewness decreases with more degrees of
freedom. When the degrees of freedom are 30 or more, the distribution can be
approximated by a normal distribution.
Goodness of Fit Test (observed and expected test)
Failure times,Bulbs
Poisson Distribution
(no of defects in finite space A/C Mech sys..lamda represent mean and variance
binomial and standard no upper bound of calls
Chi Square Formula
• The Chi Square is denoted by X2and the
formula is given as:
Here,
• O = Observed frequency
E = Expected frequency
∑ = Summation
X2 = Chi Square value
Confidence Interval on the
(Variance and Standard Deviation of a Normal Distribution)
Figure 8-8 Probability density
functions of several 2
distributions.
 Distribution
2
 Distribution
2
Let X1, X2, , Xn be a random sample from a normal distribution with
mean  and variance 2, and let S2 be the sample variance. Then the
random variable
X
2

n  1 S 2


2
(8-8)
has a chi-square (2) distribution with n  1 degrees of freedom.
Sec 8-3 Confidence Interval on σ2 & σ of a Normal Distribution
54
Confidence Interval on the Variance and Standard Deviation
If s2 is the sample variance from a random sample of n observations from a
normal distribution with unknown variance 2, then a 100(1 – )%
confidence interval on 2 is
(n  1) s 2

 
n 1

 
n 1



(n  1) s 2
 n 1
(8-9)

where
and  n 1 are the upper and lower 100/2
percentage points of the chi-square distribution with
n – 1 degrees of freedom, respectively.
A confidence interval for  has lower and upper limits that are the square
roots of the corresponding limits in
Equation 8–9.
Sec 8-3 Confidence Interval on σ2 & σ of a Normal Distribution
55
Sec 8-7 to write
56
Confidence Interval on the
Variance and Standard Deviation of a
Normal Distribution
One-Sided Confidence Bounds
Example 8-7 Detergent Filling
An automatic filling machine is used to fill bottles with liquid detergent. A random sample
of 20 bottles results in a sample variance of fill volume of s2 = 0.01532. Assume that the fill
volume is approximately normal. Compute a 95% upper confidence bound.
A 95% upper confidence bound is found from Equation 8-10 as follows:

n  1 s 2



20  1 0.0153


2
2
 
10.117
 2  0.0287
A confidence interval on the standard deviation  can be obtained by taking the square
root on both sides, resulting in
  0.17
Sec 8-3 Confidence Interval on σ2 & σ of a Normal Distribution
58
Sec 8-7 to write
59
8-6 Tolerance and Prediction Intervals
8-6.1 Prediction Interval for Future Observation
A 100 (1  )% prediction interval (PI) on a single future observation
from a normal distribution is given by
x  t n 1 s 1 
1
1
 X n  1  x  t n 1 s 1 
n
n
(8-15)
The prediction interval for Xn+1 will always be longer than the confidence
interval for .
Sec 8-6 Tolerance & Prediction Intervals
60
Repeated for comparison
Example 8-11 Alloy Adhesion
The load at failure for n = 22 specimens was observed, and found that x  13.71
and s  3.55. The 95% confidence interval on  was 12.14    15.28. Plan to test
a 23rd specimen.
A 95% prediction interval on the load at failure for this specimen is
x  t n1s 1 
1
1
 X n1  x  t n1s 1 
n
n
13.71  2.0803.55 1 
1
1
 X 23  13.71  2.0803.55 1 
22
22
6.16  X 23  21.26
Interpretation: The prediction interval is considerably longer than the CI. This is
because the CI is an estimate of a parameter, but the PI is an interval estimate of a
single future observation. See next slide
Sec 8-6 Tolerance & Prediction Intervals
61
Example 8-6 Alloy Adhesion
Construct a 95% CI on  to the following data.
19.8
15.4
11.4
19.5
The sample mean is
10.1
18.5
14.1
8.8
14.9
7.9
17.6
13.6
7.5
12.7
16.7
11.9
15.4
11.9
15.8
11.4
15.4
11.4
x  13.71 and sample standard deviation is s = 3.55.
Since n = 22, we have n  1 =21 degrees of freedom for t, so t0.025,21 = 2.080.
The resulting CI is
x  t /2, n 1s / n    x  t /2, n 1s/ n
13.71  2.080(3.55) / 22    13.71  2.0803.55 / 22
13.71  1.57    13.71  1.57
12.14    15.28
Interpretation: The CI is fairly wide because there is a lot of variability in the
measurements. A larger sample size would have led to a shorter interval.
Sec 8-2 Confidence Interval on the Mean of a Normal, σ2 Unknown
62
8-6.2 Tolerance Interval for a Normal Distribution
A tolerance interval for capturing at least % of the
values in a normal distribution with confidence level
100(1 – )% is
x  ks,
x  ks
where k is a tolerance interval factor found in
Appendix Table XII. Values are given for  = 90%,
95%, and 99% and for 90%, 95%, and 99%
confidence.
Sec 8-6 Tolerance & Prediction Intervals
63
Example 8-12 Alloy Adhesion
The load at failure for n = 22 specimens was observed, and found that x  13.71
and s = 3.55. Find a tolerance interval for the load at failure that includes 90% of
the values in the population with 95% confidence.
From Appendix Table XII, the tolerance factor k for n = 22,  = 0.90, and 95%
confidence is k = 2.264.
The desired tolerance interval is
 x  ks
[13.71   2.264  3.55
, x  ks 
, 13.71   2.264  3.55]
(5.67 , 21.74)
Interpretation: We can be 95% confident that at least 90% of the values of load at
failure for this particular alloy lie between 5.67 and 21.74.
Sec 8-6 Tolerance & Prediction Intervals
65
Important Terms & Concepts of Chapter 8
Chi-squared distribution
Confidence coefficient
Confidence interval
Confidence interval for a:
– Population proportion
– Mean of a normal
distribution
– Variance of a normal
distribution
Confidence level
Error in estimation
Large sample confidence interval
1-sided confidence bounds
Precision of parameter estimation
Prediction interval
Tolerance interval
2-sided confidence interval
t distribution
Chapter 8 Summary
67
Guidelines for Constructing Confidence Intervals
8-5 Guidelines for Constructing Confidence Intervals
Table 8-1 provides a simple road map for appropriate calculation of a
confidence interval.
Sec 8-5 Guidelines for Constructing Confidence Intervals
69
8-4 A Large-Sample Confidence Interval For a Population Proportion
Normal Approximation for Binomial Proportion
If n is large, the distribution of
Z 
X  np
np (1  p)

Pˆ  p
p (1  p)
n
is approximately standard normal.
The quantity
p(1  p) /isncalled the standard error of the point estimator
Sec 8-4 Large-Sample Confidence Interval for a Population Proportion
.
P̂
70
Approximate Confidence Interval on a Binomial Proportion
If p̂ is the proportion of observations in a random
sample of size n, an approximate 100(1  )%
confidence interval on the proportion p of the
population is
pˆ  z 
pˆ (1  pˆ )
pˆ (1  pˆ )
 p  pˆ  z 
n
n
(8-11)
where z/2 is the upper /2 percentage point of the
standard normal distribution.
Sec 8-4 Large-Sample Confidence Interval for a Population Proportion
71
Example 8-8 Crankshaft Bearings
In a random sample of 85 automobile engine crankshaft bearings, 10 have a
surface finish that is rougher than the specifications allow. Construct a 95%
two-sided confidence interval for p.
A point estimate of the proportion of bearings in the population that exceeds the
roughness specification is pˆ  x / n  10 / 85  0..12
A 95% two-sided confidence interval for p is computed from Equation 8-11 as
pˆ  z0.025
0.12  1.96
pˆ 1  pˆ 
 p  pˆ  z0.025
n
pˆ 1  pˆ 
n
0.12  0.88 
0.12  0.88 
 p  0.12  1.96
85
85
0.0509  p  0.2243
Interpretation: This is a wide CI. Although the sample size does not appear to
be small (n = 85), the value of is fairly small, which leads to a large standard
error for contributing to the wide CI.
Sec 8-4 Large-Sample Confidence Interval for a Population Proportion
72
Choice of Sample Size
Sample size for a specified error on a binomial proportion :
If we set
E  z p 1  p  / n
and solve for n, the appropriate sample size is
2
 z  
n
 p1  p 
 E 
(8-12)
The sample size from Equation 8-12 will always be a maximum for
p = 0.5 [that is, p(1 − p) ≤ 0.25 with equality for p = 0.5], and can be
used to obtain an upper bound on n.
 z  
n

 E 
2
0.25
Sec 8-4 Large-Sample Confidence Interval for a Population Proportion
(8-13)
73
Example 8-9 Crankshaft Bearings
Consider the situation in Example 8-8. How large a sample is required if we want
to be 95% confident that the error in using p̂ to estimate p is less than 0.05?
ˆ  0.12 as an initial estimate of p, we find from Equation 8-12 that the
Using p
required sample size is
2
2
z

 1.96 
n   0.025  pˆ 1  pˆ   
 0.120.88  163
E
0
.
05




If we wanted to be at least 95% confident that our estimate p̂ of the true proportion
p was within 0.05 regardless of the value of p, we would use Equation 8-13 to find
the sample size
2
2
 z 0.025 
 1.96 
n
 0.25  
 0.25  385
 0.05 
 E 
Interpretation: If we have information concerning the value of p, either from a
preliminary sample or from past experience, we could use a smaller sample while
maintaining both the desired precision of estimation and the level of confidence.
Sec 8-4 Large-Sample Confidence Interval for a Population Proportion
74
Approximate One-Sided Confidence Bounds on a Binomial Proportion
The approximate 100(1  )% lower and upper confidence
bounds are
pˆ  z
pˆ 1  pˆ 
p
n
and
p  pˆ  z
pˆ 1  pˆ 
n
(8-14)
respectively.
Sec 8-4 Large-Sample Confidence Interval for a Population Proportion
75
Example 8-10 The Agresti-Coull CI on a Proportion
Reconsider the crankshaft bearing data introduced in Example 8-8. In that
example we reported that pˆ  0.12 and n  85. The 95% CI was 0.0509  p  0.2243
.
Construct the new Agresti-Coull CI.
pˆ 1  pˆ  z2 2
pˆ 
 z 2
 2
2
n
n
4n
UCL 
2
1  z 2 n
z2 2
0.12  0.88  1.962
1.962
0.12 
 1.96

2(85)
85
4(85) 2

1  (1.962 / 85)
 0.2024
pˆ 1  pˆ  z2 2
pˆ 
 z 2
 2
2
n
n
4n
LCL 
2
1  z 2 n
z2 2
0.12  0.88  1.962
1.962
0.12 
 1.96

2(85)
85
4(85) 2

1  (1.962 / 85)
 0.0654
Interpretation: The two CIs would agree more closely if the sample size were
larger.
Sec 8-4 Large-Sample Confidence Interval for a Population Proportion
76
Probabilities Normal Distributions
•
•
•
•
Normal Distributions
Determining Normal Probabilities
Finding Values That Correspond to Normal Probabilities
Assessing Departures from Normality
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§7.1: Normal Distributions
• This pdf is the most popular distribution for
continuous random variables
• First described de Moivre in 1733
• Elaborated in 1812 by Laplace
• Describes some natural phenomena
• More importantly, describes sampling
characteristics of totals and means
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Normal Probability Density Function
• Recall: continuous
random variables are
described with
probability density
function (pdfs) curves
• Normal pdfs are
recognized by their
typical bell-shape
Figure: Age distribution
of a pediatric population
with overlying Normal
pdf
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Area Under the Curve
• pdfs should be viewed
almost like a histogram
• Top Figure: The darker bars
of the histogram
correspond to ages ≤ 9
(~40% of distribution)
• Bottom Figure: shaded area
under the curve (AUC)
corresponds to ages ≤ 9
(~40% of area)
7: Normal Probability Distributions
 x 

 
 12 
1
f ( x) 
e 
2 
80
2
Parameters μ and σ
• Normal pdfs have two parameters
μ - expected value (mean “mu”)
σ - standard deviation (sigma)
μ controls location
σ controls spread
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Mean and Standard Deviation of
Normal Density
σ
μ
7: Normal Probability Distributions
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Standard Deviation σ
• Points of inflections one
σ below and above μ
• Practice sketching Normal
curves
• Feel inflection points
(where slopes change)
• Label horizontal axis with
σ landmarks
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Two types of means and standard deviations
• The mean and standard deviation from the pdf (denoted μ
and σ) are parameters
• The mean and standard deviation from a sample (“xbar”
and s) are statistics
• Statistics and parameters are related, but are not the same
thing!
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68-95-99.7 Rule for
Normal Distributions
• 68% of the AUC within ±1σ of μ
• 95% of the AUC within ±2σ of μ
• 99.7% of the AUC within ±3σ of μ
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Example: 68-95-99.7 Rule
Wechsler adult
intelligence scores:
Normally distributed with
μ = 100 and σ = 15; X ~
N(100, 15)
• 68% of scores within
μ±σ
= 100 ± 15
= 85 to 115
• 95% of scores within
μ ± 2σ
= 100 ± (2)(15)
= 70 to 130
• 99.7% of scores in
μ ± 3σ =
100 ± (3)(15)
= 55 to 145
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Symmetry in the Tails
Because the Normal
curve is symmetrical and
the total AUC is exactly
1…
95%
… we can easily
determine the AUC in
tails
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Example: Male Height
• Male height: Normal with μ = 70.0˝ and σ = 2.8˝
• 68% within μ ± σ = 70.0  2.8 = 67.2 to 72.8
• 32% in tails (below 67.2˝ and above 72.8˝)
• 16% below 67.2˝ and 16% above 72.8˝ (symmetry)
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Reexpression of Non-Normal Random
Variables
• Many variables are not Normal but can be
reexpressed with a mathematical
transformation to be Normal
• Example of mathematical transforms used for
this purpose:
– logarithmic
– exponential
– square roots
• Review logarithmic transformations…
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§7.2: Determining Normal
Probabilities
When value do not fall directly on σ landmarks:
1. State the problem
2. Standardize the value(s) (z score)
3. Sketch, label, and shade the curve
4. Use Table B
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§7.2: Determining Normal
Probabilities
When value do not fall directly on σ landmarks:
1. State the problem
2. Standardize the value(s) (z score)
3. Sketch, label, and shade the curve
4. Use Table B
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Step 1: State the Problem
• What percentage of gestations are less
than 40 weeks?
• Let X ≡ gestational length
• We know from prior research:
X ~ N(39, 2) weeks
• Pr(X ≤ 40) = ?
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Step 2: Standardize
• Standard Normal
variable ≡ “Z” ≡ a
Normal random
variable with μ = 0 and
σ = 1,
• Z ~ N(0,1)
• Use Table B to look up
cumulative
probabilities for Z
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Example: A Z variable
of 1.96 has cumulative
probability 0.9750.
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Step 2 (cont.)
Turn value into z score:
z
x

z-score = no. of σ-units above (positive z) or below
(negative z) distribution mean μ
For example, the value 40 from X ~ N (39,2) has
40  39
z
 0.5
2
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Steps 3 & 4: Sketch & Table B
3. Sketch
4. Use Table B to lookup Pr(Z ≤ 0.5) = 0.6915
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Probabilities Between Points
a represents a lower boundary
b represents an upper boundary
Pr(a ≤ Z ≤ b)
=
Pr(Z ≤ b)
7: Normal Probability Distributions
−
Pr(Z ≤ a)
97
Between Two Points
Pr(-2 ≤ Z ≤ 0.5) =
.6687
=
.6687
-2
0.5
Pr(Z ≤ 0.5) −
.6915
−
.6915
0.5
Pr(Z ≤ -2)
.0228
.0228
-2
See p. 144 in text
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§7.3 Values Corresponding to Normal
Probabilities
1. State the problem
2. Find Z-score corresponding to percentile
(Table B)
3. Sketch
4. Unstandardize:
x    z p
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z percentiles




zp ≡ the Normal z variable with cumulative
probability p
Use Table B to look up the value of zp
Look inside the table for the closest
cumulative probability entry
Trace the z score to row and column
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e.g., What is the 97.5th
percentile on the Standard
Normal curve?
z.975 = 1.96
Notation: Let zp
represents the z score
with cumulative
probability p,
e.g., z.975 = 1.96
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Step 1: State Problem
Question: What gestational length is smaller
than 97.5% of gestations?
• Let X represent gestations length
• We know from prior research that
X ~ N(39, 2)
• A value that is smaller than .975 of gestations
has a cumulative probability of.025
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Step 2 (z percentile)
Less than 97.5%
(right tail) = greater
than 2.5% (left tail)
z lookup:
z.025 = −1.96
z
.00
–1.9 .0287
.01
.02
.03
.04
.05
.06
.07
.08
.09
.0281
.0274
.0268
.0262
.0256
.0250
.0244
.0239
.0233
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Unstandardize and sketch
x    z p  39  (1.96)( 2)  35
The 2.5th percentile is 35 weeks
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Example: A Z variable of 1.96 has cumulative
probability of ……
z lookup: z.025 = ----What is the 97.5th percentile on the Standard
Normal curve? z.975 =
Between Two Points
Q1: z lookup: z.025 = ----Q 2:A Z variable of 1.96 has
cumulative probability of ……
Q3: Find P(Z less than 1.96)
.6915
.6687
-2
q2 What is the 97.5th percentile on the
Standard Normal curve? z.975 =
Find P(Z greater than 1.96)
0.5
0.5
7: Normal Probability Distributions
.0228
-2
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