NG BB 33 Hypothesis Testing Basics

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National Guard
Black Belt Training
Module 33

Hypothesis Testing Basics

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CPI Roadmap – Analyze
8-STEP PROCESS
6. See
1.Validate 2. Identify 3. Set 4. Determine 5. Develop 7. Confirm 8. Standardize
Counter-
the Performance Improvement Root Counter- Results Successful
Measures
Problem Gaps Targets Cause Measures & Process Processes
Through

Define Measure Analyze Improve Control

ACTIVITIES TOOLS
• Value Stream Analysis
• Identify Potential Root Causes • Process Constraint ID
• Reduce List of Potential Root • Takt Time Analysis
Causes • Cause and Effect Analysis
• Brainstorming
• Confirm Root Cause to Output
• 5 Whys
Relationship
• Affinity Diagram
• Estimate Impact of Root Causes • Pareto
on Key Outputs • Cause and Effect Matrix
• FMEA
• Prioritize Root Causes
• Hypothesis Tests
• Complete Analyze Tollgate • ANOVA
• Chi Square
• Simple and Multiple
Regression

Note: Activities and tools vary by project. Lists provided here are not necessarily all-inclusive. UNCLASSIFIED / FOUO

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Learning Objectives
 Review the terms “Parameters” and “Statistics” as
they relate to Populations and Samples.
 Introduce Confidence Intervals for expressing the
uncertainty when predicting a population parameter
using a sample statistic, and how to calculate CI’s
for some common situations for different sample
sizes.
 Show how the Central Limit Theorem and the
Standard Error of the Mean applies to the use of
Confidence Intervals and Tests

Hypothesis Testing - Basic UNCLASSIFIED / FOUO 3

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Learning Objectives (Cont.)
 Introduce statistical tests for some common tests and
introduce the t-distribution with testing
 Learn about Hypothesis Testing to prove a statistical
difference in process performance in applications of
Minitab
 Understand the tradeoffs and influences of sample
sizes on statistical tests.
 Apply knowledge of different classes of statistical
errors to the decisions used in process improvement
to minimize risk.


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Application Examples
 Transactional – A Black Belt has just finished a pilot of a new
process for handling blanket Purchase Orders and wants to
know if it has a statistically significant: a) shorter cycle time
and b) increased accuracy over the old process.
 Administrative – The manager of an AAFES1 order entry
department wants to compare two order entry procedures to
see if one is faster than the other.
 Service – Medical diagnostic imaging services are provided
from two different medical treatment facilities to a central
hospital which wants to know if there are differences in the
quality of service, particularly: a) the number of lost records
and re-takes, and b) average waiting time for MRIs and X-rays.
1AAFES, Army and Air Force Exchange System

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Population vs. Sample

Population Sample
All U.S. registered voters 10,000 people are asked
who they will vote for
President
All sufferers of a certain 3,000 people are given a
disease that might be new treatment in a clinical
given the new treatment study
All appraisals completed 25 appraisals chosen at
that month random from a given
month

Since it is not always practical or possible to measure/query every
item/person in the population, you take a random sample.


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Terms and Labels: Population vs. Sample

Population = Sample =
Term
Parameter Statistic

Count of items N n
m 
x
Mean
~ ~
Median m x
Standard Dev. s S
 
m x
Estimators = 
s s


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Population Parameters vs. Sample Statistics
Random Samples
of Size, n = 4
Population
x1 , s1

x2 , s2

x3 , s3

m, s x4 , s4
Population Parameters; Sample Statistics; Mean, x-bar,
Mean, m (mu), and Standard Deviation, s (sigma) and Standard Deviation, s


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Central Limit Theorem
 If:
x1, x2, …, xn are independent measurements (i.e., a random sample of size n)
from a population, where the mean of x is m, when
the standard deviation of x is given as s,

 Then:

The distribution of x X  X 1  X 2 x3  X n

n
has mean and standard
deviation given by:  s Standard Error
mX  m and sX  of the Mean
n

In addition, when n is sufficiently large, then the distribution of x- bar is
approximately normal (“bell-shaped curve”). More on sample sizes later...


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Variability of Means
 Sample statistics estimate population parameters by inference:


For a given sample ( x, s, n ), we can estimate population
 
parameters of m  s by inference.
 As the sample size increases we are more confident that our
sample statistic is a more valid estimator of the population
parameter.

n=5

sx
n=3
sx  sx
n

n=1


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National Guard
Black Belt Training
Confidence Intervals

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What Is a Confidence Interval?
 We know that when we take the average of a sample,
it is probably not exactly the same as the average of
the population.
 Confidence intervals help us determine the likely
range of the population parameter.
 For example, if my 95% confidence interval is 5 +/-
2, then I have 95% confidence that the mean of the
population is between 3 and 7.


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What Is a Confidence Interval? (Cont.)
 Usually, confidence intervals have an additive
uncertainty:

Estimate ± Margin of Error

Sample Statistic ± [ ___ X ___ ]

Example: Confidence Measure of
x, s Factor Variability

Note: Detailed formulas may be found in the appendix.

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Why Do We Need Confidence Intervals?
 Sample statistics, such as Mean and Standard
Deviation, are only estimates of the population’s
parameters.

 Because there is variability in these estimates from
sample to sample, we can quantify our uncertainty
using statistically-based confidence intervals.
Confidence intervals provide a range of plausible
values for the population parameters (m and s).

 Any sample statistic will vary from one sample to
another and, therefore, from the true population or
process parameter value.

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Exercise
 Let’s look at a population that has a normal distribution with:
 known mean value = 65
 standard deviation = 4
(This has been generated in dataset Confidence.mtw)
 Each member in the class will randomly sample 25 data points from
this population. (In Minitab, use Calc>Random Data>Sample from
Columns.)
 Sample 25 rows of data from C1 and store the results in C2.
 Use graphical descriptive statistics to calculate the 95% confidence
interval for the mean and sigma based on your sample of 25 data
points. Do they include the mean, 65, and the sigma, 4?
 Based on a class size of 25, we would expect 1 confidence interval to
not contain 65 for the mean, and 1 that does not include 4 for sigma.


Confidence Interval for the Mean (m) with
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Population Standard Deviation (s) Known
Example
A random sample of size, n = 36, is taken and the distribution of
x
is normal. We are given that the population standard deviation
(s) is 18.0. The value of x-bar is an estimator of the population
mean (m),
and the standard error of x-bar is:
s x bar s / n  18.0 / 36  3.0
 From the properties of the standardized normal distribution,
there is a 95% chance that m is within the range of ( x-bar +
and - 1.96 times the Standard Error of x-bar).

This is known as the Standard Error of the Mean


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What Values of x-bar Can I Expect?
Distribution of x-bar
.95
95% of all x-bars will
fall into the shaded region,
defined by m ± 1.96(3.0)

.025 .025

Standard Error
m1 - 1.96(3.0) m1 m1+ 1.96(3.0) of the Mean


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But I Don’t Know m, I Only Know x-bar!
 We can turn it around.
 x-bar lying in the interval m ±
1.96(3.0) is the same thing as
m lying in the interval x-bar ±
(----------- x-barsample A -----------)
1.96(3.0).
(---------- x-barsample B-----------)
 Because there is a 95% chance (---------- x-barsample C-----------)
that x-bar lies in the interval
m ± 1.96(3.0), there is a 95%
chance that the interval x-bar m1 - 1.96(3.0) m1 m1+ 1.96(3.0)
± 1.96(3.0) encloses m.

 The interval we construct using Observed sample mean, x-barsample C
the observed sample mean is
called a 95% confidence
interval for m.

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Population Standard Deviation (s) Known
Another Example
An airline needs an estimate of the average number of
passengers on a newly scheduled flight. Its experience is that
data for the first month of flights is unreliable, but thereafter the
passenger loading settles down.
Therefore, the mean passenger load is calculated for the first 20
weekdays of the second month after initiation of this particular
new flight. If the sample mean (x-bar) is 112.0 and the
population standard deviation (s) is assumed to be 25, find a
95% confidence interval for the true, long-run average number
of passengers on this flight.


Confidence Interval for the Mean (m)
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with Standard Deviation (s) Known
Solution
We assume that the hypothetical population of daily passenger loads for
weekdays is not badly skewed. Therefore, the sampling distribution of
x-bar is approximately normal and the confidence interval results are
approximately correct, even for a sample size of only 20 weekdays.
x -bar  112.0
s  25
s
s x -bar   5.59
20

For a 95% confidence interval, we use z.025= 1.96 in the formula to
obtain 112  1.965.59 or 101.04 to 122.96

We are 95% confident that the long-run mean, m , lies in this interval.

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Population Standard Deviation (s) Unknown

A very important point to remember is that for this
example we assumed that we knew the population
standard deviation, and many times that is not the case.
Often, we have to estimate both the mean and the
standard deviation from the sample.
 When s is not known, we use the t-distribution rather
than the normal (z) distribution. The t-distribution will
be explained next.
 In many cases, the true population s is not known, so
we must use our sample standard deviation (s) as an
estimate for the population standard deviation (s


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Standard Deviation (s) Unknown (Cont.)
 Since there is less certainty (not knowing m or s ),
the t-distribution essentially “relaxes” or
“expands” our confidence intervals to allow for this
additional uncertainty.
 In other words, for a 95% confidence interval, you
would multiply the standard error by a number
greater than 1.96, depending on the sample size.
 1.96 comes from the normal distribution, but the
number we will use in this case will come from the
t-distribution.


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What Is This t-Distribution?
 The t-distribution is actually a family of distributions.

 They are similar in shape to the normal distribution
(symmetric and bell-shaped), although wider, and flatter in
the tails.

 How wide and flat the specific t-distribution is depends on
the sample size. The smaller the sample size, the wider
and flatter the distribution tails.

 As sample size increases, the t-distribution approaches the
exact shape of the normal distribution.


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An Example of a t-Distribution

0.4

t-distribution
0.3 (n = 5)
frequency

0.2

Area =
0.025
0.1

0.0
-3 -2 -1 0 1 2 2.78 3
t


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Some Selected t-Values
 Here are values from the t-distribution for various
sample sizes (for 95% confidence intervals):
Sample Size t-value (.025)*
2 12.71
3 4.30
5 2.78
10 2.26
20 2.09
30 2.05
100 1.98
1000 1.96

* For a 95% CI,  = .05. Therefore, for a two tail distribution: /2= .05/2= .025


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Population Standard Deviation (s) Unknown
Example
The customer expectation when phoning an order-out pizza shop is that
the average amount of time from completion of dialing until they hear
the message indicating the time in queue is equal to 55.0 seconds (less
than a minute was the response from customers surveyed, so the
standard was established at 10% less than a minute). You decide to
randomly sample at 20 times from 11:30am until 9:30pm on 2 days to
determine what the actual average is. In your sample of 20 calls, you
find that the sample mean, x-bar, is equal to 54.86 seconds and the
sample standard deviation, s, is equal to 1.008 seconds.
The actual data was as follows:
54.1, 53.3, 56.1, 55.7, 54.0, 54.1, 54.5, 57.1, 55.2, 53.8,
54.1, 54.1, 56.1, 55.0, 55.9, 56.0 ,54.9, 54.3, 53.9, 55.0
What is a 95% confidence interval for the true mean call completion
time?

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95% Confidence Interval
for Mean Call Completion Time
x = 54.860
We’re 95% confident that the actual mean
s = 1.008 call completion time is somewhere between
54.389 seconds and 55.331 seconds,
n = 20 based on our sample of 20 calls.

 t.025,19 = 2.09 our sample of 20 calls

s
Luckily, we don’t
x  t α/2, n1
have to worry about
n
the details of how to
calculate the t-value. 1.008
54.860  2.09 
Minitab takes care of 20
that for us.
54.389, 55.331


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Now Let Minitab Calculate
the Confidence Interval
1. Open the Minitab file
PizzaCall.mtw


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the Confidence Interval (Cont.)
2. Select Stat>
Basic Statistics>
Graphical Summary


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3. Double click on C-1 to place it
in the Variables box

4. Click on OK


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Summary for C1
A nderson-Darling Normality Test
We’re 95% confident that
A -Squared 0.60 the actual mean is
P-V alue 0.105
between
Mean
StDev
54.860
1.008
54.388 and 55.332
V ariance 1.016
Skew ness 0.560026 We’re also taking a 5%
Kurtosis -0.509797
N 20 chance that we’re wrong.
Minimum 53.300
1st Q uartile 54.100
Median 54.700

54 55 56 57
3rd Q uartile
Maximum
55.850
57.100
95% Confidence Interval
95% C onfidence Interv al for Mean for Mean (m:
54.388 55.332
54.388 55.332
95% C onfidence Interv al for Median
54.100 55.582

95% Confidence Intervals 95% C onfidence Interv al for StDev 95% Confidence Interval
Mean
0.767 1.472
for Standard Deviation (s:
0.767 1.472
Median

54.00 54.25 54.50 54.75 55.00 55.25 55.50


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Other Types of Confidence Intervals
 There are other types of confidence intervals that are
based on the same principles we have learned:
 Standard Deviation
 Proportions
 Median
 Others
 We will discuss some of these later.


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National Guard
Black Belt Training
Hypothesis Testing

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Extending the Concept
of Confidence Intervals
 Extending the concept of confidence intervals allows
us to set-up and interpret statistical tests.
 We refer to these tests as Hypothesis Tests.
 One way to describe a hypothesis test:
 Determining whether or not a particular value of
interest is contained within a confidence interval.
 Hypothesis testing also gives us the ability to
calculate the probability that our conclusion is wrong.


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The New Car
 You buy a one-year old car from the Lemon Lot in
order to save money on gas. The previous owner still
had the original features sticker and you were pleased
to note that the EPA mileage estimate indicated that
the car should get 31 miles per gallon overall.


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The New Car (Cont.)
 As soon as you buy the car, you fill up the tank so
that you’ll be ready to take the family for a drive and
to go to work the next day. A few days later, you fill
up again and calculate your gas mileage for that tank.
After you push the “=“ key on your calculator, the
number 27.1 appears.


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The New Car (Cont.)
 Should you send the car to a mechanic to check for
problems?
 Do you conclude that the EPA estimate is simply wrong?
 Do you leave cruel messages on the seller’s answering
machine?
 What ARE your conclusions?


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Continuing the Car Situation
 At what value of gas consumption should you become
alarmed that you are experiencing anything more
than just random variation?


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The Car Situation (Cont.)
 What if we knew this?

Distribution of gas
consumption for this
car

12.8 %

s = 3.46


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Hypothesis Testing
Hypothesis Testing:
 Allows us to determine statistically whether or not a
value is cause for alarm (or is simply due to random
variation)
 Tells us whether or not two sets of data are different
 Tells us whether or not a statistical parameter
(mean, standard deviation, etc.) is statistically
different from a test value of interest
 Allows us to assess the “strength” of our conclusion
(our probability of being correct or wrong)

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Hypothesis Testing (Cont.)
Hypothesis Testing Enables Us to:
 Handle uncertainty using a commonly accepted
approach
 Be more objective (2 persons will use the same
techniques and come to similar conclusions almost all
of the time)
 Disprove or “fail to disprove” assumptions
 Control our risk of making wrong decisions or coming
to wrong conclusions


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Hypothesis Testing (Cont.)
Some Possible Samples
Sample A
True
Sample B
Population
Distribution
Sample C

Sample D

m
Population
Mean

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Sample Size Concerns
 If we sample only one item, how close do we expect
to get to the true population mean?
 How well do you think this one item represents the
true mean?
 How much ability do we have to draw conclusions
about the mean?
 What if we sample 900 items? Now, how close
would we expect to get to the true population
mean?


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Sample Size (Cont.)
The larger our sample, the
closer x-bar is likely to be to
Population the true population mean.

Likely value of x-bar
with a small sample
size

m
Likely value of x-bar
x with a large sample
size
x

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Standard Deviation
 What effect would a lot of variation in the population
have on our estimate of the population mean from a
sample?
 How would this affect our ability to draw conclusions
about the mean?
 What if there is very little variation in the population?


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Standard Deviation (Cont.)

Population with a lot
of variation

m Likely value of x-bar
x with sample size, n

Population with less
variation

m Likely value of x-bar
with sample size, n
x

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Statistical Inferences and Confidence
 How much confidence do we have in our estimates?
 How close do you think the true mean, m, is to our
estimate of the mean, x-bar?
 How certain do we want/need to be about conclusions
we make from our estimates?
 If we want to be more confident about our sample
estimate (i.e., we want a lower risk of being wrong),
then we must relax our statement of how close we
are to the true value.


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Statistical Inferences and Confidence
(Cont.)

Population

If we want to have
high confidence in our
conclusions, we must
relax the range in
which we say the true
m mean lies
As we tighten our
estimate of the mean,
our risk of being wrong
x
increases. Thus, our
confidence decreases. x

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Three Factors Drive Sample Sizes
 Three concepts affect the conclusions drawn from a single
sample data set of (n) items:
 Variation in the underlying population (sigma)
 Risk of drawing the wrong conclusions (alpha, beta)
 How small a Difference is significant (delta)
Risk

(n)
Variation Difference

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Three Factors: Variation, Risk, Difference
These 3 factors work together. Each affects the others.
 Variation: When there’s greater variation, a larger sample
is needed to have the same level of confidence that the
test will be valid. More variation reduces our confidence
interval.
 Risk: If we want to be more confident that we are not
going to make a decision error or miss a significant event,
we must increase the sample size.
 Difference: If we want to be confident that we can identify
a smaller difference between two test samples, the sample
size must increase.


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Three Factors (Cont.)
 Larger samples improve our confidence interval.
 Lower confidence levels allow smaller samples.
 All of these translate into a specific confidence
interval for a given parameter, set of data, confidence
level and sample size.
 They also translate into what types of conclusions
result from hypothesis tests.
 Testing for larger differences between the samples,
reduces the size of the sample. This is known as delta
(D).

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An Example
A unit has several quick response forces, QRF. Some forces have over
700 members, with at least 300 on the site at any time.
 By regulation, all forces must have a quick response plan, the critical
first phase of which is required to be completed in 10 minutes (600
seconds) or less.
 There are two teams that are vying for “most responsive.” They have
taken somewhat different approaches to implementing their quick
response plans and management wants to know which approach is
better: Team 1 or Team 2
 Each one has 100 data points for actual responses and drills (Minitab
file Response.mtw)


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The Data from Team 1
598.0 598.8 600.2 599.4 599.6
599.8 598.8 599.6 599.0 601.2
600.0 599.8 599.6 598.4 599.6
599.8 599.2 599.6 599.0 600.2
600.0 599.4 600.2 599.6 600.0
600.0 600.0 599.2 598.8 600.0
598.8 600.2 599.0 599.2 599.4
598.2 600.2 599.6 599.6 599.8
599.4 599.6 600.4 598.6 599.2
599.6 599.0 600.0 599.8 599.6
599.4 599.0 599.0 599.6 599.4
599.4 599.8 599.6 599.2 600.0
600.0 600.8 599.4 599.6 600.0
598.8 598.8 599.2 600.2 599.2
599.2 598.2 597.8 599.8 599.4
599.4 600.0 600.4 599.6 599.6
599.6 599.2 599.6 600.0 599.8
599.0 599.8 600.0 599.6 599.0
599.2 601.2 600.8 599.2 599.6
600.6 600.4 600.4 598.6 599.4

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The Data from Team 2
601.6 600.8 599.4 599.8 601.6
600.4 598.6 598.0 602.8 603.4
598.4 600.0 597.6 600.0 597.0
600.0 600.4 598.0 599.6 599.8
596.8 600.8 597.6 602.2 597.8
602.8 600.8 601.2 603.8 602.4
600.8 597.2 599.0 603.6 602.2
603.6 600.4 600.4 601.8 600.6
604.2 599.8 600.6 602.0 596.2
602.4 596.4 599.0 603.6 602.4
598.4 600.4 602.2 600.8 601.4
599.6 598.2 599.8 600.2 599.2
603.4 598.6 599.8 600.4 601.6
600.6 599.6 601.0 600.2 600.4
598.4 599.0 601.6 602.2 598.0
598.2 598.2 601.6 598.0 601.2
602.0 599.4 600.2 598.4 604.2
599.4 599.4 601.8 600.8 600.2
599.4 600.2 601.2 602.8 600.0
600.8 599.0 597.6 597.6 596.8


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Descriptive Statistics – Team 1
Summary for Team 1
A nderson-D arling N ormality Test
A -S quared 0.84
P -V alue 0.029
M ean 599.55
S tD ev 0.62
V ariance 0.38
S kew ness -0.082566
Kurtosis 0.745102
N 100
M inimum 597.80
M edian 599.60
3rd Q uartile 600.00
597.75 598.50 599.25 600.00 600.75 M aximum 601.20
95% C onfidence Interv al for M ean
599.43 599.67
95% C onfidence Interv al for M edian
599.40 599.60
95% C onfidence Interv al for S tD ev
9 5 % C onfidence Inter vals
0.54 0.72
Mean

Median

599.40 599.45 599.50 599.55 599.60 599.65 599.70


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Descriptive Statistics – Team 2
Summary for Team 2
A nderson-D arling N ormality Test
A -S quared 0.29
P -V alue 0.615
M ean 600.23
S tD ev 1.87
V ariance 3.51
S kew ness 0.051853
Kurtosis -0.518286
N 100
M inimum 596.20
M edian 600.20
3rd Q uartile 601.60
597.0 598.5 600.0 601.5 603.0 M aximum 604.20
95% C onfidence Interv al for M ean
599.86 600.60
95% C onfidence Interv al for M edian
599.80 600.60
95% C onfidence Interv al for S tD ev
9 5 % C onfidence Inter vals
1.65 2.18
Mean

Median

599.8 600.0 600.2 600.4 600.6


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Example
 The average cycle time for Team 1 is 599.55
seconds.
 The average cycle time for Team 2 is 600.23
seconds.
 The target cycle time for Phase 1 response is 600
seconds.
 Is the difference between the two average cycle times
statistically significant?


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Example (Cont.)
 The unit wants to determine if the true averages of
the two teams are really different.
 The unit thinks that the 600.23 average of team 2 is
little too high, so there is a need to determine if the
data indicates that the true average is really not equal
to the target of 600 seconds.
 The unit will use hypothesis testing to answer these
questions.


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Example
 The first hypothesis test to be performed is to determine
whether there is a statistically significant difference between
the means of the two teams. This is called a 2-Sample t
Test.
 The real question is whether or not the means are different
enough to indicate that the approaches taken by the two
teams really are centered differently, or are they close
enough that the difference could simply be a result of random
variation?
 After that, hypothesis testing can tell us if there is evidence
indicating whether or not each team’s average is different
from the target of 600 seconds.
 First, we need to introduce some terminology.

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The Null Hypothesis for a 2-Sample t Test
 The 2-Sample t Test is used to test whether or not
the means of two populations are the same.
 The null hypothesis is a statement that the
population means for the two samples are equal.
Ho: μ1 = μ2
 We assume the null hypothesis is true unless we have
enough evidence to prove otherwise. We say – we
“fail to reject the null”.
 If we can prove otherwise, then we “reject the null”
hypothesis and accept the Alternative Hypothesis
HA: μ1 ≠ μ2

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Null Hypothesis for 2-Sample t Test (Cont.)
 This is analogous to our judicial system principle of
“innocent until proven guilty”
 The symbol used for the null hypothesis is Ho:
H 0 : m1  m2 OR H 0 : m1  m2  0


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The Alternative Hypothesis
for a 2-Sample t Test
 The alternative hypothesis is a statement that
represents reality if there is enough evidence to reject
Ho.
 If we reject the null hypothesis then we accept the
alternative hypothesis.
 This is analogous to being found “guilty” in a court of
law.
 The symbol used for the alternative hypothesis is Ha:
H a : m1  m2 OR H a : m1  m2  0


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Our Emergency Response Team Example
 In our example, the first hypothesis test will take this
form:
H o : m1  m 2
H a : m1  m 2 Reminder:
We are conducting a 2-
Sample t test to determine if
We can rewrite it in this form: the average cycle time of the
Phase 1 response from our two
H o : m1  m 2  0 teams are different.

H a : m1  m 2  0


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(Cont.)
 If we wanted to specifically test only whether or not
there was enough evidence to indicate that team 2’s
average was greater than team 1’s, it would take this
form:
H o : m1  m 2  0
H a : m1  m 2  0 This is still a 2-Sample t-Test


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(Cont.)
 The second hypothesis test will be a 1-Sample t. It
will take this form for each team:

H o : m1  600
H a : m1  600

When you are testing whether or not a
population mean is equal to a given or
Target value, you use a 1-Sample t


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Hypothesis Test in Minitab
 We will use Minitab to conduct our hypothesis tests.
 Open the Minitab file Response.mtw


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Hypothesis Test in Minitab:
2-Sample t-Test
Select Stat>
Basic
Statistics>
2-Sample t
to compare
Team 1 to
Team 2


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Hypothesis Test in Minitab (Cont.)
Team 1 and Team 2
are in different
columns, so select
Samples in
different columns

Double click on C1-Supp1
Then double click on
C2-Supp2 to place them
In First and Second boxes
Select Graphs to get the
Graphs dialog box


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In the Graphs dialog
box, check both
Boxplots of data
and Dotplots of
data

Click OK here, and
then click on OK in
the previous dialog
box


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Boxplot of Team 1, Team 2
605

604

603

602

601
Data

600

599

598

597

596
Team 1 Team 2


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Individual Value Plot of Team 1, Team 2
605

604

603

602

601
Data

600

599

598

597

596
Team 1 Team 2


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This descriptive output shows up
in your Session Window


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The null hypothesis states that the difference
between the two means is zero

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We will cover
p-values in more
detail a little later

The p-value here is less than 0.05, so
we can reject the null hypothesis

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Assumptions
 The Hypothesis Tests we have discussed make certain
assumptions:
 Independence between and within samples
 Random samples
 Normally distributed data
 Unknown Variance
 In our example, we did not assume equal variances.
This is the safe choice. However, if we had reason to
believe equal variances, then we could have checked
the “Assume equal variances” box in the dialogue box.


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The Risks of Being Wrong
Error Matrix

Conclusion Drawn
Accept Ho Reject Ho

Type I
Ho True Correct Error
The -Risk)
True
State Type II Error
Correct
Ho False  -Risk)


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Type I and Type II Errors
 Type I Error
I’ve discovered
 Alpha Risk something that really
 Producer Risk
isn’t here!

 The risk of rejecting the null, and taking action, when
no action was necessary
 Type II Error
I’ve missed a
 Beta Risk
significant effect!
 Consumer Risk
 The risk of failing to reject the null when you should
have rejected it.
 No action is taken when there should have been action.

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Type I and Type II Errors (Cont.)
 The Type I Error is determined up front.
 It is the alpha value you choose.
 The confidence level is one minus the alpha level.

 The Type II Error is determined from the circumstances of the
situation.
 If alpha is made very small, then beta increases (all else being
equal).
 Requiring overwhelming evidence to reject the null increases the
chances of a type II error.
 To minimize beta, while holding alpha constant, requires increased
sample sizes.
 One minus beta is the probability of rejecting the null hypothesis
when it is false. This is referred to as the Power of the test.


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Type I and Type II Errors (Cont.)
 What type of error occurs when an innocent man is
convicted?
 What about when a guilty man is set free?
 Does the American justice system place more
emphasis on the alpha or beta risk?


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Exercise
 Draw the Type I & II error matrix for airport security.
 Do you think the security system at most airports
places more emphasis on the alpha or beta risk?


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The p-Value
 If we reject the null hypothesis, the p-value is the
probability of being wrong.
 In other words, if we reject the null hypothesis, the p-
value is the probability of making a Type I error.
 It is the critical alpha value at which the null
hypothesis is rejected.
 If we don’t want alpha to be more than 0.05, then we
simply reject the null hypothesis when the p-value is
0.05 or less.
 As we will learn later, it isn’t always this simple.

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National Guard
Black Belt Training
Power, Delta and Sample
Size

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Beta, Power, and Sample Size
 If two populations truly have different means, but only by a
very small amount, then you are more likely to conclude they
are the same. This means that the beta risk is greater.
 Beta only comes into play if the null hypothesis truly is false.
The “more” false it is, the greater your chances of detecting it,
and the lower your beta risk.
 The power of a hypothesis test is its ability to detect an effect
of a given magnitude.

Power  1  
 Minitab will calculate beta for us for a given sample size, but
first let’s show it graphically….


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Beta and Alpha

95% Confidence Limit
(alpha = .05) for mean, m1
(critical value)

Alpha Risk

m1
Here is our first population and its corresponding alpha risk.

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Beta and Alpha (Cont.)

95% Confidence Limit (alpha
= .05) for mean, m1 (critical
value)

m1 m2
D
We want to compare these two populations. Do you think that
we will easily be able to determine if they are different?

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Beta and Alpha (Cont.)
95% Confidence Limit (alpha
= .05) for mean, m1 (critical
value)

Beta Risk

m1 m2
D
If our sample from population 2 is in this grey area, we will
not be able to see the difference. This is called Beta Risk.


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Beta and Delta
 If we are trying to see a larger change, we have less Beta
Risk.

(alpha = .025) for
Beta Risk mean, m1 (critical value)

m1 m2
D

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Beta and Sigma
 Now we’re back to our original graphic. What do you think
happens to Beta Risk if the standard deviations of the
populations decrease?
(alpha = .05) for mean,
m1 (critical value)

Beta Risk

m1 m2
D

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Beta and Sigma (Cont.)
 If the standard deviation decreases, Beta Risk decreases.
 Reducing variability has the same effect on Beta Risk as
increasing sample size.

Beta Risk
(alpha = .05) for mean, m1
(critical value)

m1 m2
D

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How Can Power Be Increased?
 Power is related to risk, variation, sample size, and
the size of change that we want to detect.
 If we want to detect a smaller delta (effect), we
typically must increase our sample size.


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Example:
Power
 Let’s use Minitab to determine the beta risk of the hypothesis
test we performed on the two teams.
 First, we’ll have to make some assumptions.
 We don’t know the TRUE difference in the means, so we’ll assume that
it’s 0.682, the differences in the sample averages.
 A variance hypothesis test shows that the variances are not equal.
 We will average the variances from Minitab to determine the combined
variance using the following formula:

2 2
s1  s 2
s 
2


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Example:
Power (Cont.)
Select; Stat>
Power and Sample Size>
2-Sample t...


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Example:
Power (Cont.)
To calculate Power, we need three things;
1. Sample Size
2. The Difference between the two Means
3. The Average Standard Deviation of the two samples

We can get all this information from our 2-Sample t-Test conducted earlier:

1. Sample Size = 100

2. Difference Between Means = 0.682
(600.230 – 599.548 = 0.682)

3. Average Standard Deviation ??
(See Next Slide)


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Example:
Power (Cont.)
To Calculate Average Standard Deviation

Remember that Standard Deviations are
the Square Roots of the Variance. Since
square roots are not additive (we cannot
add them and divide by two) we have to
convert them back to Variances which are
additive.

StDev Squared = Variance
Team 1 0.619 squared = 0.3832
Team 2 1.870 squared = 3.4969
Sum = 3.8801
Divide by 2 to get Average = 1.9401
And Square Root of Average = 1.3929

So the Average Standard Deviation for the two samples is 1.3929


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Example:
Power (Cont.)

1. Type in Sample Size of 100 here

2. Type in Difference Between
Means of 0.682 here

3. Type in Average
Standard Deviation
of 1.393 here

4. Click on OK


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Example:
Power (Cont.)

The Power = 0.9312
And since Beta = (1 –Power)
Beta = 0.0688.

If the TRUE difference between the two support orgs. was
0.682, we would have a 6.88% chance of not observing this
and therefore concluding that they are the same.

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Example:
Power (Cont.)
 In practice, we evaluate the power of a test to
determine its ability to detect a difference of a given
magnitude that we deem important, or practically
significant.
 For example, we could calculate the power of a
hypothesis test to see if we could measure a one
minute difference in responsiveness between the two
teams.


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Example:
Power (Cont.)
 Let’s say that if the two support organizations’ cycle
times differ by as little as 0.4 seconds, then we need
to analyze the reasons for the differences.
 What is the power of our test to detect this
difference?
 What is the probability of making a type II error
(concluding that there is no difference when one
exists)?
 Use Minitab to individually answer these questions.


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Exercise:
Sample Size
 Now that we understand the relationship between
Beta, Power, Delta, and Sample Size, we can use this
information to calculate the sample size necessary to
give us the information we want.
 We simply use the same function in Minitab to solve
for sample size rather than power.
 This is a very useful and common extension of
Hypothesis Testing.


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Exercise:
Sample Size (Cont.)
Here we enter the
Difference (delta) we wish
to detect, and the minimum
Power value that we are
willing to live with.
We leave Sample sizes
blank.


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Exercise:
Sample Size
 Let’s extend our response team cycle time example
 Determine what sample size we would need to detect
a difference of 0.4 seconds at a power of 0.90.
 What about at a power of 0.95?
 What about at a power of 0.95 and an alpha of
0.025?
 Hint: Click the Options button in the Power and
Sample Size dialogue box.


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Other Power and Sample Size Scenarios

We can perform these
calculations not only for
the difference between
two means, but for other
tests as well.


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1-Sample t-test in Minitab
 Now, we will return to Minitab to test the following hypothesis
about our two support organizations cycle times:

H o : m1  600
H a : m1  600


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Back to the Support Organization Example:
One Sample t-Test
Choose Stat>Basic
Statistics>1-Sample t
to test the mean of
each response team against
a standard or spec


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Double click on C1 Team 1
and C2 Team 2 to place
them in the dialog box
here.
Type in the Hypothesized
mean, or standard we are
comparing to. Here it is
600.
Click the Graphs
button to get to the
Graphs dialog box.


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1-Sample t-test in Minitab (Cont.)

Select Histogram
of data and
Boxplot of data

Click OK here and on
the previous Screen


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Histogram of Team 1
(with Ho and 95% t-confidence interval for the mean)
35

30 This shows the Target
we are testing, along with
25
the Average and the
20 Confidence Interval
Frequency

from the data.
15

10

5

0 _
X

-5 Ho
598.0 598.5 599.0 599.5 600.0 600.5 601.0
Team 1


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1-Sample t-test in Minitab (Cont.) - adj
Boxplot of Team 1

_
X
Ho

598.0 598.5 599.0 599.5 600.0 600.5 601.0 601.5
Team 1


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Histogram of Team 2
15.0

12.5

10.0
Frequency

7.5

5.0

2.5

0.0 _
X
Ho

597.0 598.5 600.0 601.5 603.0
Team 2


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Boxplot of Team 2

_
X
Ho

596 597 598 599 600 601 602 603 604 605
Team 2


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Here is the descriptive output for the 1-Sample t-Test
found in Session Window


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2-Sided and 1-Sided Hypothesis Tests
 We have concentrated on 2-sided hypothesis tests.
 2-Sided tests determine whether or not two items are equal
or whether a parameter is equal to some value.
 Whether an item is less than or greater than another item or
a value is not sought up front. A 2-sided test is a less specific
test.
 The alternative hypothesis is “Not Equal”.
 Everything we have learned also applies to 1-sided tests.
 1-Sided tests determine whether or not an item is less than
(<) or greater than (>) another item or value.
 The alternative hypothesis is either (<) or (>).
 This makes for a more powerful test (lower beta at a given
alpha and sample size).

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More Detailed Information

Remember to use the Stat Guide button to learn more
about the results and to help you interpret them.


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Hypothesis Test Summary Template
Hypothesis Test Factor (x)
(ANOVA, 1 or 2 sample t - test, Chi Squared, p Value Observations/Conclusion
Regression, Test of Equal Variance, etc) Tested
Significant factor - 1 hour driving time from DC
Example: ANOVA Location 0.030 to Baltimore office causes ticket cycle time to
generally be longer for the Baltimore site
Significant factor - on average, calls requiring
Example: ANOVA Part vs. No Part 0.004 parts have double the cycle time (22 vs 43
hours)
Significant factor - Department 4 has digitized
Example: Chi Squared Department 0.000 addition of customer info to ticket and less
human intervention, resulting in fewer errors
South region accounted for 59% of the defects
Example: Pareto Region n/a due to their manual process and distance from
the parts warehouse

- Example - Optional BB Deliverable
Describe any other observations about the root cause (x) data

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One-Way ANOVA Template Boxplots of Net Hour by Part/No
(means are indicated by solid circles)

 After further investigation, possible 150
Boxplot: Part/ No Part Impact on Ticket Cycle Time
reasons proposed by the team are
OEM backorders, lack of technician
- Example -
Net Hours Call Open

certifications and the distance from
the OEM to the client site. It is also 100

caused by the need for technicians to
make a second visit to the end user
to complete the part replacement. 50
Next step will be for the team to
confirm these suspected root causes.

0
Part/No Part

Part
No Part
Analysis of Variance for Net Hour  Because the p-value <=
Source DF SS MS F P
Part/No 1 7421 7421 8.65 0.004 0.05, we can be confident
Error 69 59194 858 that calls requiring parts
Total 70 66615 do have an impact on the
Individual 95% CI's For Mean
Level N Mean StDev --+---------+---------+---------+----
ticket cycle time.
No Part 27 21.99 19.95 (--------*---------)
Part 44 43.05 33.70 (------*------)
--+---------+---------+---------+----
Pooled StDev = 29.29 12 24 36 48

Optional BB Deliverable UNCLASSIFIED / FOUO

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Linear Regression Template
 95% confident that 94.1% of the variation in “Wait Time” is from the “Qty of Deliveries”

Fitted Line Plot
Wait Time = 32.05 + 0.5825 Deliveries
55
S 1.11885
R-Sq 94.1%
R-Sq(adj) 93.9%
50
Wait Time

45

40
- Example -
35
10 15 20 25 30 35
Deliveries

Optional BB Deliverable UNCLASSIFIED / FOUO

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Takeaways
 Since it is not always practical or possible to measure
every item in the population, you take a random
sample.
 A basic understanding of the terms: Population,
Sample, Population Parameter, Sample Statistic,
Sample Mean, and Sample Standard Deviation
 How to calculate a confidence interval with the
population standard deviation known
 How to calculate a confidence interval with the
population standard deviation unknown


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Takeaways (Cont.)
 How Hypothesis tests help us handle uncertainty
 The role of sample size, variation, and confidence
level
 The null and alternative hypotheses
 Type I and Type II errors
 Hypothesis tests in Minitab
 Stat Guide
 p-value


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Takeaways (Cont.)
 How to conduct a 1-way and 2-way t-test
 How to conduct a Variance test (see Appendix)
 How to conduct a Paired t-test (see Appendix)
 Understanding of 1-way and 2-way test of proportions
(see Appendix)
 Understanding the relationship between Power and
sample size and detectable difference (delta)


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What other comments or questions
do you have?

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References
 Hildebrand and Ott, Statistical Thinking for Managers,
4th Edition
 Kiemele, Schmidt, and Berdine, Basic Statistics,
4th Edition


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National Guard
Black Belt Training

APPENDIX

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Hypothesis Testing - Steps
Step 1: Define the problem objective
Step 2: Determine what data to collect (continuous or attribute)
Step 3: Based on data type, determine the appropriate hypothesis test to use
Step 4: Specify the null (H0) hypothesis and the alternative (H1) hypothesis
Step 5: Select a significance level (degree of risk acceptable), usually 0.05
Step 6: Execute Data Collection plan from step 2
Step 7: From the sample, conduct the hypothesis test using a statistical tool
Step 8: Identify the p-value
Step 9: Compare the p-value to the significance level - if the p-value is less than or
equal to your acceptable risk (your alpha), then the null hypothesis is rejected
Step 10: Translate the decision to the situation

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Decision Tree Matrix
Data Type
Hypothesis to be Tested (Step 3) Tree
(Step 2)
Variable Testing equality of population MEAN (average) to a specific value 1

Variable Testing equality of population MEANS (averages) from two populations 2

Variable Testing equality of population MEANS (averages) from more than two populations 3

Testing equality of population VARIANCES (standard deviation) from more than two
Variable
populations 4

Attribute - Binomial
"Go/No-Go" Testing equality of population PROPORTIONS (binomial data; e.g., pass/fail, go/no
"Pass/Fail" or go, is/is not, etc.) from one or more populations 5
"Defective" Data

Attribute - Poisson
Testing equality of population PROPORTIONS (Poisson data; i.e., frequency of
"Count" or
occurence in time or space) from two or more populations 6
"Defects" data

Testing for ASSOCIATION (not necessarily causal)
Attribute (Contingency
Table Data)
Note: For use with attribute data only. For variable data, use correlation 7
or regression. No decision tree required.

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Decision Tree # 1
Testing Equality of Population Mean
Application:
to a Specific Value
Type of Data: Variable (Continuous)
Has the average button diameter from the welder
Example:
changed from its historical value?

Start

Is
Yes
1-Sample Z-test
n > 30? Stat > Basic Statistics > 1-Sample Z

No

Is
population 1-Sample Wilcoxon test
normally random sample from a continuous,
No
distributed? symmetric population
(Anderson- Stat > Nonparametrics > 1-Sample Wilcoxon
Darling)

Yes

1-Sample t-test
(reasonably robust
against normality assumption)
Stat > Basic Statistics > 1-Sample t

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Decision Tree # 2
Testing Equality of Means
Application:
from Two Populations
Is the average button diameter from Welder A
Example:
different from that of Welder B?

Start

Are the Paired t-te st
two samples Yes (samples from normal distribution)
dependent? Stat > Basic Statistics > Paired t

No

Do
n1 and n2 Yes
2-Sample Z-te st
both exceed Stat > Basic Statistics > 2-Sample t
30?

No
2-Sample M ann Whitne y
(independent, random variables from two
Are both populations with same shape, same variance)
populations Stat > Nonparametrics > Mann-Whitney
normally No
distributed?
(Anderson- Note: If the two populations have different shapes
Darling) or different standard deviations, then use:
2-Sample t-test
without pooling variances
Yes

Equal
2-Sample t-te st
Variances? No without pooling variances
(F-test) Stat > Basic Statistics > 2-Sample t
(Do not assume equal variances)

Yes

2-Sample t-te st
w ith pooled variances
(reasonably robust against normality
assumption)
Stat > Basic Statistics > 2-Sample t
 Assume equal variances

Example2.vsd 6-1-00

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Decision Tree # 3
Testing Equality of Means from
Application:
More than Two Populations
Do the average button diameters from
Example:
Welders A, B and C differ from one another?

Start

Are the One-Way Analysis of Variance
populations
normally (ANOVA)
Yes
distrubted? (reasonably robust against assumptions
(Anderson- of normality and equal v ariances)
Darling) Stat > ANOVA > One-way

Did the test show
signif icance?
No Stop
No

Tukey's test
Yes to conduct pairwise comparisons
Stat > ANOVA > One-way
Comparisons:  Tukeys

Do samples
Mood's Median test
contain outliers? Yes (independant, random samples f rom continuous
(Box Plot) distributions hav ing same shape)
Stat > Nonparametrics > Mood's Median test

No

Note: Use Dunnett's Method if
Kruskal-Wallis comparing treatments to a control.
(independant, random samples
f rom continuous distributions
hav ing same shape)
Stat > Nonparametrics > Kruskal-Wallis

Example3.v sd 6-1-00

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Decision Tree # 4
Application: Testing Equality of Variances

Do the variances in button diameter
Example:
from the three welders differ from one another?

Start

Are the
populations
How many
populations are 2
normally
No
Levene's test
being compared? distributed? Stat > Basic Statistics > 2 Variances
(Anderson-
Darling)

Yes

More than 2
F-test
Stat > Basic Statistics >2 Variances

Are the
populations
normally
No
Levene's test
distributed? Stat > ANOVA > Test for Equal Variances
(Anderson-
Darling)

Yes

Note: The F-test and Bartlett's test are not robust
Bartlett's test against the normality assumption.
Stat > ANOVA > Test for Equal Variances

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Example4.v sd 6-1-00

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Decision Tree # 5
Testing Equality of Population
Application:
Proportions
Type of Data: Attribute (Discrete) - Binomial Distribution

Case 2:
Case 1:
Te sting Equality of
Te sting Population Proportion
Proportions from Two
Against a Spe cific Value
Populations
Example: Has the % defective rate
on Line 1 changed Example: Are Lines 1 and 2
from its historical value? running at the same
% defective rate?

Stat > Basi c Stati sti cs > 2-Proporti ons
Ho:P1=P2 no di fference i n popluati on
Stat > Basi c Stati sti cs > 1-Proporti on
proporti ons
M i niTab - Opti ons sel ect pool ed p

Case 3:
Te sting Equality of
Proportions from M ore than
Two Populations
Example: Are Lines 1, 2 and 3
running at the same
% defective rate?

Use Chi-Square test
MiniTab
Stat>Tables>Chi-square test

Example5.vsd 5-10-01 UNCLASSIFIED / FOUO

UNCLASSIFIED / FOUO

Decision Tree # 6
Testing Equality of Population
Application:
Defect Rates
Type of Data: Attribute (Discrete) - Poisson Distribution

1) Is the number of errors on invoices different
between Dept. A and Dept. B?
Examples:
2) Does the number of seat defects
differ among shifts 1, 2 and 3?

Comparing two Poission Comparing more than two Poisson
Distributions Distributions

Stat >Use 2 Sample t-test
Basic Stats > 2-sample Poisson Rate Use One-Way Analysis of Variance
Stat > Basic Stat > 2 Sample t Stat > ANOVA > One-way

Caution
No Extreme Outliers

UNCLASSIFIED / FOUO

UNCLASSIFIED / FOUO

Decision Tree # 7

Application: Testing for Association

Type of Data: Attribute (Contingency Table Data)

Does the type of defect that occurs
Example:
depend on which product is being produced?

Chi-square test
Minitab:
Stat > Tables > Chi-square test

UNCLASSIFIED / FOUO

UNCLASSIFIED / FOUO

Group 1:
Hypothesis Tests for Variation
 Use the Minitab electronic docs, stat guide, and help
to learn about performing hypothesis tests for equality
of variance among two populations. You may also use
your textbooks if you wish.
 Prepare a 10-15 minute teachback on hypothesis tests
for variation.
 Be sure to work an example in your teachback.
 Hint: Using Minitab to conduct a hypothesis test to
determine if there is a difference in the amount of
variation exhibited by each support organization would
be a good example to use.


UNCLASSIFIED / FOUO

Group 2:
Paired t-Test
 Use the Minitab electronic docs, stat guide, and help
to learn about performing paired t-tests. You may
also use your textbooks if you wish.
 Prepare a 10-15 minute teachback on paired t-tests.
 Be sure to illustrate the difference between a
standard 2-way t-test and a paired t-test.
 Go through a sample size calculation in your example.


UNCLASSIFIED / FOUO

Group 3:
Hypothesis Tests with Proportions
 Use the Minitab electronic docs, stat guide, and help to
learn about performing hypothesis tests with proportions.
You may also use your text books if you wish.
 Prepare a 10-15 minute teachback on hypothesis tests with
proportions.
 Be sure to illustrate the main difference between
hypothesis tests of proportions and the other hypothesis
tests we have talked about.
 Include both 1-way and 2-way proportion hypothesis tests.
 Go through a sample size calculation in your example.


UNCLASSIFIED / FOUO

Confidence Interval Formulas
Confidence Intervals for:
 Mean (s Known) Mean (s Unknown)

σ s
x  Z α/2 x  t α/2,n1
n n
 Standard Deviation
n1 n1
s σs
 α/2, n1
2
 1α/2, n1
2

 Proportions (Approximate)
p 1  p 
ˆ ˆ p 1  p 
ˆ ˆ
ˆ
p  Z  /2 ˆ
 p  p  Z  /2
n n

UNCLASSIFIED / FOUO

Table of Normal Curve Areas
z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
0.00 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239 0.0279 0.0319 0.0359
0.10 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.0753
0.20 0.0793 0.0832 0.0871 0.0910 0.0948 0.0987 0.1026 0.1064 0.1103 0.1141
0.30 0.1179 0.1217 0.1255 0.1293 0.1331 0.1368 0.1406 0.1443 0.1480 0.1517
0.40 0.1554 0.1591 0.1628 0.1664 0.1700 0.1736 0.1772 0.1808 0.1844 0.1879
0.50 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088 0.2123 0.2157 0.2190 0.2224

0.60 0.2257 0.2291 0.2324 0.2357 0.2389 0.2422 0.2454 0.2486 0.2517 0.2549
0.70 0.2580 0.2611 0.2642 0.2673 0.2704 0.2734 0.2764 0.2794 0.2823 0.2852
0.80 0.2881 0.2910 0.2939 0.2967 0.2995 0.3023 0.3051 0.3078 0.3106 0.3133
0.60 0.3159 0.3186 0.3212 0.3238 0.3264 0.3289 0.3315 0.3340 0.3365 0.3389
0 z 1.00 0.3413 0.3438 0.3461 0.3485 0.3508 0.3531 0.3554 0.3577 0.3599 0.3621

1.10 0.3643 0.3665 0.3686 0.3708 0.3729 0.3749 0.3770 0.3790 0.3810 0.3830
1.20 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3980 0.3997 0.4015
1.30 0.4032 0.4049 0.4066 0.4082 0.4099 0.4115 0.4131 0.4147 0.4162 0.4177
1.40 0.4192 0.4207 0.4222 0.4236 0.4251 0.4265 0.4279 0.4292 0.4306 0.4319
1.50 0.4332 0.4345 0.4357 0.4370 0.4382 0.4394 0.4406 0.4418 0.4429 0.4441

1.60 0.4452 0.4463 0.4474 0.4484 0.4495 0.4505 0.4515 0.4525 0.4535 0.4545
1.70 0.4554 0.4564 0.4573 0.4582 0.4591 0.4599 0.4608 0.4616 0.4625 0.4633
1.80 0.4641 0.4649 0.4656 0.4664 0.4671 0.4678 0.4686 0.4693 0.4699 0.4706
1.90 0.4713 0.4719 0.4726 0.4732 0.4738 0.4744 0.4750 0.4756 0.4761 0.4767
2.00 0.4772 0.4778 0.4783 0.4788 0.4793 0.4798 0.4803 0.4808 0.4812 0.4817

2.10 0.4821 0.4826 0.4830 0.4834 0.4838 0.4842 0.4846 0.4850 0.4854 0.4857
2.20 0.4861 0.4864 0.4868 0.4871 0.4875 0.4878 0.4881 0.4884 0.4887 0.4890
2.30 0.4893 0.4896 0.4898 0.4901 0.4904 0.4906 0.4909 0.4911 0.4913 0.4916
2.40 0.4918 0.4920 0.4922 0.4925 0.4927 0.4929 0.4931 0.4932 0.4934 0.4936
2.50 0.4938 0.4940 0.4941 0.4943 0.4945 0.4946 0.4948 0.4949 0.4951 0.4952

2.60 0.4953 0.4955 0.4956 0.4957 0.4959 0.4960 0.4961 0.4962 0.4963 0.4964 z area
2.70 0.4965 0.4966 0.4967 0.4968 0.4969 0.4970 0.4971 0.4972 0.4973 0.4974 3.5 0.49976737
2.80 0.4974 0.4975 0.4976 0.4977 0.4977 0.4978 0.4979 0.4979 0.4980 0.4981 4.0 0.49996833
2.90 0.4981 0.4982 0.4982 0.4983 0.4981 0.4984 0.4985 0.4985 0.4986 0.4986 4.5 0.49999660
3.00 0.4987 0.4987 0.4987 0.4988 0.4988 0.4989 0.4989 0.4989 0.4990 0.4990 5.0 0.49999971

Source: Computed by P.J. Hildebrand.

Source: Statistical Thinking for Managers, Hildebrand and Ott, 4th Edition, page 800.

UNCLASSIFIED / FOUO

Calculation of t Test Statistic
 The t test statistic is calculated as follows:
(X 1  X 2 )  D0
t
sX 1  X 2
where D0 is the hypothesized difference between the
two population means.
 For an assumption of unequal variances:
2 2
s1 s2
sX  
1 X 2
n1 n 2


UNCLASSIFIED / FOUO

Calculation of t Test Statistic
 For an assumption of equal variances:

1 1
sX  sp 
1 X 2
n1 n2

where

2 2
(n1  1)s 1  (n 2  1)s 2
sp 
n1  n 2  2


NG BB 33 Hypothesis Testing Basics

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