8. testing of hypothesis for variable & attribute data

QUALITY TOOLS &
TECHNIQUES
By: -
Hakeem–Ur–Rehman
IQTM–PU 1
TQ T
ANALYZE PHASE
STATISTICAL INFERENCE:
HYPOTHESIS OF TESTING FOR
VARIABLE & ATTRIBUTE DATA

STATISTICAL METHODS
Statistical
Methods
Descriptive
Statistics
Inferential
Statistics
Estimation
Hypothesis
Testing

NATURE OF INFERENCE
in·fer·ence (n.) “The act or process of deriving logical conclusions from
premises known or assumed to be true. The act of reasoning from factual
knowledge or evidence.” 1
1. Dictionary.com
Inferential Statistics – To draw inferences about the process or population
being studied by modeling patterns of data in a way that account for
randomness and uncertainty in the observations. 2
2. Wikipedia.com
Putting the pieces of
the puzzle
together….

HYPOTHESIS TESTING
Population
I believe the
population mean
age is 50
(hypothesis).
Mean
X = 20
Reject
hypothesis! Not
close.
Random
sample







 

WHAT’S A HYPOTHESIS?
A Belief about a Population
Parameter
 Parameter Is Population
Mean, Proportion,
Variance
 Must Be Stated
Before Analysis
I believe the mean GPA of this
class is 3.5!

HYPOTHESIS
The hypotheses to be tested consists of two
complementary statements:
1) The null hypothesis (denoted by H0) is a statement
about the value of a population parameter; it must
contain the condition of equality.
2) The alternative hypothesis (denoted by H1) is the
statement that must be true if the null hypothesis
is false.
e.g.:
H0: μ = some value vs H1: μ ≠ some value
H0: μ ≤ some value vs H1: μ > some value
H0: μ ≥ some value vs H1: μ < some value
6

NULL Vs. ALTERNATIVE
HYPOTHESIS
A contractor is interested in finding out whether a new
material to be used in foundation of towers will have
any effects on the strength of the tower foundation. Will
the foundation strength increase, decrease, or remain
uncharged? If the mean foundation strength of towers
is 5000 lbs/sq-in, the hypothesis for this situation are:
H0 : µ = 5000 and H1 : µ ≠ 5000
This is called a TWO-TAILED HYPOTHESIS since the
possible effects of the new material could be to raise or
lower the strength.
EXAMPLE:
7

HYPOTHESIS (cont…)
A design engineer develops an additive to
increase the life of an automobile battery. If
the mean lifetime of the automobiles battery
is 36 months, then his hypothesis are :
H0 : µ ≤ 36 and H1 : µ > 36
This is called a ONE-TAILED HYPOTHESIS
(RIGHT-TAILED) since the interest is in an
increase only.
EXAMPLE:
8

A contractor wishes to lower the heating bills
by using a special type of insulation in site
cabins. If the average of the monthly heating
bills is $78, his hypothesis about heating costs
with the use of insulation are:
H0 : µ ≥ 78 and H1 : µ < 78
This is called a ONE-TAILED HYPOTHESIS
(LEFT-TAILED) since the contractor is
interested only in lowering the heating costs.
EXAMPLE:
9

EXERCISES:
1. An engineer hypothesizes that the mean number of defects
can be decreased in a manufacturing process of compact
disc by using robots instead of humans for certain tasks.
The mean number of defective discs per 1000 is 18.
2. A Safety Engineer claims that by regularly conducting a
specific safety orientation program for riggers, the accident
rate during tower erection will reduce. The mean accident
rate is 9 accidents per year.
3. A contractor wants to investigate whether using safety gears
affects the quality performance of workers erecting the
towers. The contractor is not sure whether using safety
gears increases or decreases the quality performance. In the
past, the average defect rate was 7 per worker.
10

SAMPLING RISK
 α - Risk, also referred as Type I Error or Producer’s Risk:
 Is the risk of rejecting H0 when H0 is true.
 i.e. concluding that the process has drifted when it really has not.
 β - Risk, also referred to Type II Error or Consumer’s Risk:
 Is the risk of accepting H0 when H0 is false.
 i.e. failing to detect the drift that has occurred in a process.
HYPOTHESIS STATEMENT:
H0 : μ = some value
H1 : μ ≠ some value
Criteria for “Accepting” & “Rejecting” a Null Hypothesis: 1. For any fixed α, an increase in
the sample size will cause a
decrease in β.
2. For any fixed sample size, a
decrease in α will cause an
increase in β. Conversely, an
increase in α will cause a
decrease in β.
3. To decrease both α and β,
increase the sample size. 11

What is P-Value?
This is the probability that a value as extreme as X–
Bar (i.e. ≥ X–Bar) is observed, given that H0 is true.
We reject H0 if the obtained P-Value is less than α.
Interpreting P-Value:
H0 : μ = 5
H1 : μ ≠ 5
α = 0.05
A low p-value for the statistical test points to rejection
of Null hypothesis because it indicates how unlikely it
is that a test statistic as extreme as or more extreme
than a observed from this population if Null
Hypothesis is true.
If a p-value = 0.005, this means that if the
population means were equal (as hypothesized),
there is only 5 in 1000 chance that a more extreme
test statistic would be obtain using data from this
population and there is significant evidence to
support the Alternative Hypothesis (H1).
P-value > α, Accept Ho
P-value < α, Reject Ho
12

HYPOTHESIS TESTING
What Do You Do? If……
You have:
 Different types of Materials. (Stainless,
Carbon Steel & Aluminum)
 Different types of oils. (Shell & Mobil)
 Different type of Cleaning solutions.
(Hydrocarbon & Water base)
You want to find which method of cleaning
yield the best results for all these materials?
13

ANALYZE PHASE
HYPOTHESIS TESTING FOR
CONTINUOUS DATA

PARAMETRIC STATISTICAL
INFERENCE
Comparing Two
groups
Data Normally
Distributed
Equality of Variances
Equal Variances if P
≥ 0.05
Unequal Variances if
P<0.05
Indep. Samp. T Tests Indep. Samp. T Tests
(Weltch
Approximation)
Comparing one
group with a Target
One Sample
Measured once
One Sample
Measured Twice
Data Distribution
Normal Data (P≥0.05)
One sample T Test
Data Distribution
Normal (P≥0.05)
Paired Sample T-Test
Comparing More
than Two groups
Data Distribution
One Way Anova Test *Welch Test
Testing of Hypothesis
Decision Making
1.Data is Normal when p ≥ 0.05 ,Use Anderson test
2.The Variance of groups are equal when p ≥ 0.05 Use the Levenes Test
3.Accept the Null Hypothesis when P≥0.05 otherwise accept the alternative hypothesis
Levenes Test
Normal (P≥0.05)
Equality of Variances
Equal Variances if P
≥ 0.05
Unequal Variances if
P<0.05
Levenes Test
* Not Available in Minitab

TEST OF MEANS (t-tests): 1 Sample t
Measurements were made on nine widgets. You know that the distribution of
widget measurements has historically been close to normal, but suppose
that you do not know Population Standard deviation. To test if the
population mean is 5 and to obtain a 90% confidence interval for the mean,
you use a t-procedure.
1. Open the worksheet EXH_STAT.MTW.
2. Check the Normality of the data using Normality Test “VALUES”.
3. Choose Stat > Basic Statistics > 1-Sample t.
4. In Samples in columns, enter Values.
5. Check Perform hypothesis test. In Hypothesized mean, enter 5.
6. Click Options. In Confidence level, enter 95. Click OK in each dialog box.
Target
A 1-sample t-test is used to compare an
expected population Mean to a target.
μsample

1 Sample t: HISTOGRAM & BOX PLOT OF VALUES
Values
5.15.04.94.84.74.64.54.4
_
X
Ho
Individual Value Plot of Values
(with Ho and 95% t-confidence interval for the mean)
17
Values
Frequency
5.15.04.94.84.74.64.54.4
2.0
1.5
1.0
0.5
0.0 _
X
Ho
Histogram of Values
Note our target Mean (represented
by red Ho) is outside our population
confidence boundaries which tells
that there is significant difference
between population and target
Mean.
Values
5.15.04.94.84.74.64.54.4
_
X
Ho
Boxplot of Values
INDIVIDUAL VALUE
PLOT (DOT PLOT)

One-Sample T: Values
Test of mu = 5 vs not = 5
Variable N Mean StDev SE Mean 95% CI T P
Values 9 4.78889 0.24721 0.08240 (4.59887, 4.97891) -2.56 0.034
1 Sample t: SESSION WINDOW
Ho
Ha
n
S
MeanSE 
 


n
1i
i
1n
)X(X
s
2
Since the P-value of 0.034 is less than 0.05,
reject the null hypothesis.
Based on the samples given there is a
difference between the average of the sample
and the desired target. X Ho
CONCLUSIONS: The new supplier’s claim that they can meet the target of 5 for
the hardness is not correct.

TEST OF MEANS (t-tests): 2-Sample
(Independent) t Test
Practical Problem:
We have conducted a study in order to determine the effectiveness of a new heating
system. We have installed two different types of dampers in home ( Damper = 1 and
Damper = 2).
We want to compare the BTU.In data from the two types of dampers to determine if
there is any difference between the two products.
 Open the MINITABTM worksheet: “Furnace.MTW”
Statistical Problem:
Ho:μ1 = μ2
Ha:μ1 ≠ μ2
2-Sample t-test (population Standard Deviations unknown).
α = 0.05
No, not that kind of damper!

2-Sample (Independent) t Test:
Follow the Roadmap…
NORMALITY TEST

Follow the Roadmap…
TEST OF EQUAL VARIANCE
Stat ANOVA  Test for
Equal Variances…
Damper
95% Bonferroni Confidence Intervals for StDevs
2
1
4.03.53.02.52.0
Damper
BTU.In
2
1
2015105
F-Test
0.996
Test Statistic 1.19
P-Value 0.558
Levene's Test
Test Statistic 0.00
P-Value
Test for Equal Variances for BTU.In
Sample 1
Sample 2

Equal Variance

Box Plot
State Statistical Conclusions: Fail to reject the null hypothesis.
State Practical Conclusions: There is no difference between the dampers for BTU’s in.
Equal Variance
Damper
BTU.In
21
20
15
10
5
Boxplot of BTU.In by Damper

2-Sample (Independent) t
Test: EXERCISE
A bank with a branch located in a commercial district of a city has the business
objective of developing an improved process for serving customers during the
noon- to-1 P.M. lunch period. Management decides to first study the waiting
time in the current process. The waiting time is defined as the time that
elapses from when the customer enters the line until he or she reaches the
teller window. Data are collected from a random sample of 15 customers, and
the results (in minutes) are as follows (and stored in Bank-I):
4.21 5.55 3.02 5.13 4.77 2.34 3.54 3.20
4.50 6.10 0.38 5.12 6.46 6.19 3.79
Suppose that another branch, located in a residential area, is also concerned
with improving the process of serving customers in the noon-to-1 P.M. lunch
period. Data are collected from a random sample of 15 customers, and the
results are as follows (and stored in Bank-II):
9.66 5.90 8.02 5.79 8.73 3.82 8.01 8.35
10.49 6.68 5.64 4.08 6.17 9.91 5.47
Is there evidence of a difference in the mean waiting time between the two
branches? (Use level of significance = 0.05)

TEST OF MEANS (t-
tests): PAIRED T-TEST A Paired t-test is used to compare the Means of two measurements from the same
samples generally used as a before and after test.
 MINITABTM performs a paired t-test. This is appropriate for testing the difference between
two Means when the data are paired and the paired differences follow a Normal
Distribution.
 Use the Paired ‘t’ command to compute a confidence interval and perform a Hypothesis
Test of the difference between population Means when observations are paired. A paired t-
procedure matches responses that are dependent or related in a pair-wise manner. This
matching allows you to account for variability between the pairs usually resulting in
a smaller error term, thus increasing the sensitivity
of the Hypothesis Test or confidence interval.
– Ho: μδ = μo
– Ha: μδ ≠ μo
 Where μδ is the population Mean of the differences and μ0 is the hypothesized Mean of the
differences, typically zero.
Stat > Basic Statistics > Paired t
mbefore
delta
(d)
mafter

TEST OF MEANS (t-tests):
PAIRED T-TEST
Practical Problem:
• We are interested in changing the sole material for a
popular brand of shoes for children.
• In order to account for variation in activity of children
wearing the shoes, each child will wear one shoe of each
type of sole material. The sole material will be randomly
assigned to either the left or right shoe.
Statistical Problem:
Ho: μδ = 0
Ha: μδ ≠ 0
Paired t-test (comparing data that must remain paired).
α = 0.05
Just checking your souls,
er…soles!
EXH_STAT.MTW

PAIRED T-TEST
NORMALITY TEST: “Delta”
Calc Calculator
AB Delta
Percent
1.51.00.50.0-0.5
99
95
90
80
70
60
50
40
30
20
10
5
1
Mean
0.622
0.41
StDev 0.3872
N 10
A D 0.261
P-Value
Probability Plot of AB Delta
Normal

PAIRED T-TEST Using 1-Sample t
Stat > Basic Statistics > 1-Sample t-test… Since there is only one column,
AB Delta, we do not test for equal
variance per the Hypothesis
Testing roadmap.
 Check this data for statistical
significance in its departure from
our expected value of zero.

TEST OF MEANS (t-tests): PAIRED
T-TEST Using 1-Sample t…
State Statistical Conclusions: Reject the null hypothesis
State Practical Conclusions: We are 95% confident that there is a difference in
wear between the two materials.
Box Plot of AB Delta
One-Sample T: AB Delta
Test of mu = 0 vs not = 0
Variable N Mean StDev SE Mean
AB Delta 10 0.410000 0.387155 0.122429
95% CI T P
(0.133046, 0.686954) 3.35 0.009
MINITABTM Session Window

PAIRED T-TEST
Another way to analyze this data is to use the paired t-test
command.
Stat  Basic Statistics  Paired T-test
Click on Graphs and select
the graphs you would like
to generate.

PAIRED T-TEST
Differences
0.0-0.3-0.6-0.9-1.2
_
X
Ho
Boxplot of Differences
Paired T-Test and CI: Mat-A, Mat-B
Paired T for Mat-A - Mat-B
N Mean StDev SE Mean
Mat-A 10 10.6300 2.4513 0.7752
Mat-B 10 11.0400 2.5185 0.7964
Difference 10 -0.410000 0.387155 0.122429
95% CI for mean difference: (-0.686954, -0.133046)
T-Test of mean difference = 0 (vs not = 0): T-Value = -3.35 P-Value = 0.009
The P-value of from this
Paired T-Test tells us the
difference in materials is
statistically significant.

EXERCISE: PAIRED T-TEST
Nine experts rated two brands of Colombian coffee in a taste-testing experiment. A
rating on a 7- point scale (1 = extremely unpleasing, 7 = extremely pleasing) is given
for each of four characteristics: taste, aroma, richness, and acidity. The following
data (stored in coffee) display the ratings accumulated over all four characteristics.
Brand
Expert A B
C.C. 24 26
S.E. 27 27
E.G. 19 22
B.L. 24 27
C.M. 22 25
C.N. 26 27
G.N. 27 26
R.M. 25 27
P.V. 22 23
At the 0.05 level of significance, is there evidence of a difference in the mean
ratings between the two brands?

PURPOSE OF ANOVA
Analysis of Variance (ANOVA) is used to investigate and model
the relationship between a response variable and one or more
independent variables.
Analysis of variance extends the two sample t-test for testing
the equality of two population Means to a more general null
hypothesis of comparing the equality of more than two Means,
versus them not all being equal.
– The classification variable, or factor, usually has three or
more levels (If there are only two levels, a t-test can be
used).
– Allows you to examine differences among means using
multiple comparisons.
TEST FOR MORE THAN TWO MEANS
(F – Test): ANOVA

ANOVA: EXAMPLE
We have three potential suppliers that claim to have equal levels of quality.
Supplier B provides a considerably lower purchase price than either of the
other two vendors. We would like to choose the lowest cost supplier but we
must ensure that we do not effect the quality of our raw material.
Supplier A Supplier B Supplier C
3.16 4.24 4.58
4.35 3.87 4.00
3.46 3.87 4.24
3.74 4.12 3.87
3.61 3.74 3.46
We would like test the data to determine whether there is a difference between the
three suppliers.
(F – Test): ANOVA...

FOLLOW THE ROADMAP…TEST FOR NORMALITY
36
Supplier C
Percent
5.04.54.03.53.0
99
95
90
80
70
60
50
40
30
20
10
5
1
Mean
0.910
4.03
StDev 0.4177
N 5
AD 0.148
P-Value
Probability Plot of Supplier C
Normal
Supplier B
Percent
4.504.254.003.753.50
99
95
90
80
70
60
50
40
30
20
10
5
1
Mean
0.385
3.968
StDev 0.2051
N 5
AD 0.314
P-Value
Probability Plot of Supplier B
Normal
Supplier A
Percent
4.54.03.53.02.5
99
95
90
80
70
60
50
40
30
20
10
5
1
Mean
0.568
3.664
StDev 0.4401
N 5
AD 0.246
P-Value
Probability Plot of Supplier A
Normal All three suppliers samples are
Normally Distributed.
Supplier A P-value = 0.568
Supplier B P-value = 0.385
Supplier C P-value = 0.910

TEST FOR EQUAL VARIANCE
STACK DATA FIRST:
Data  stack  Columns…
37

TEST FOR EQUAL
VARIANCE…
38

ANOVA Using Minitab
39
Click on “Graphs…”,
Check “Boxplots of data”
Data
Supplier CSupplier BSupplier A
4.6
4.4
4.2
4.0
3.8
3.6
3.4
3.2
3.0
Boxplot of Supplier A, Supplier B, Supplier C

ANOVA: Session window
40
Test for Equal Variances: Suppliers vs ID
One-way ANOVA: Suppliers versus ID
Analysis of Variance for Supplier
Source DF SS MS F P
ID 2 0.384 0.192 1.40 0.284
Error 12 1.641 0.137
Total 14 2.025
Individual 95% CIs For Mean
Based on Pooled StDev
Level N Mean StDev ----------+---------+---------+------
Supplier 5 3.6640 0.4401 (-----------*-----------)
Supplier 5 3.9680 0.2051 (-----------*-----------)
Supplier 5 4.0300 0.4177 (-----------*-----------)
----------+---------+---------+------
Pooled StDev = 0.3698 3.60 3.90 4.20
Normal data P-value > .05
No Difference

ANOVA Assumptions
In one-way ANOVA, model adequacy can be checked by either of the
following:
1. Check the data for Normality at each level and for homogeneity of
variance across all levels.
2. Examine the residuals (a residual is the difference in what the model
predicts and the true observation).
i. Normal plot of the residuals
ii. Residuals versus fits
iii. Residuals versus order
41

42
Residual
Frequency
0.60.40.20.0-0.2-0.4-0.6
5
4
3
2
1
0
Histogram of the Residuals
(responses are Supplier A, Supplier B, Supplier C)
The Histogram of
residuals should show a
bell shaped curve.
ANOVA Assumptions
Residual
Percent
1.00.50.0-0.5-1.0
99
95
90
80
70
60
50
40
30
20
10
5
1
Normal Probability Plot of the Residuals
 Normality plot of the
residuals should follow a
straight line.
 Results of our example look
good.
 The Normality assumption is
satisfied.

43
Fitted Value
Residual
4.054.003.953.903.853.803.753.703.65
0.75
0.50
0.25
0.00
-0.25
-0.50
Residuals Versus the Fitted Values
 The plot of residuals versus fits examines constant variance.
 The plot should be structureless with no outliers present.
 Our example does not indicate a problem.
ANOVA Assumptions

ANOVA EXERCISE
EXERCISE OBJECTIVE: Utilize what you have learned to
conduct and analyze a one way ANOVA using MINITABTM.
You design an experiment to assess the durability of four experimental
carpet products. You place a sample of each of the carpet products in
four homes and you measure durability after 60 days. Because you
wish to test the equality of means and to assess the differences in
means, you use the one-way ANOVA procedure (data in stacked form)
with multiple comparisons. Generally, you would choose one multiple
comparison method as appropriate for your data.
1. Open the worksheet EXH_AOV.MTW.
2. Choose Stat > ANOVA > One-Way.
3. In Response, enter Durability. In Factor, enter Carpet.
4. Click OK in each dialog box.
44

NON–PARAMETRIC STATISTICAL
INFERENCE
Comparing Two
groups
Data Distribution
Non-normal
(P<0.05)
Mann Whitney / U
Test
Comparing one
group with a Target
One Sample
Measured once
One Sample
Measured Twice
Data Distribution
Non-normal
(P<0.05)
Sign Test
Data Distribution
Non-normal
(P<0.05)
Wilcoxon Signed
Rank Test
Comparing More
than Two groups
Data Distribution
Non-normal
(P<0.05)
Kruskal Wallis (H)
Test
Testing of Hypothesis
Decision Making
1. Data is Normal when p ≥ 0.05 ,Use Anderson test
2. Accept the Null Hypothesis when P≥0.05 otherwise accept the alternative
hypothesis
Mood’s Median Test
OR

SIGN TEST: EXAMPLE
Price index values for 29 homes in a suburban area in the Northeast were
determined. Real estate records indicate the population median for similar
homes the previous year was 115. This test will determine if there is
sufficient evidence for judging if the median price index for the homes was
greater than 115 using level of significance = 0.10.
1. Open the worksheet EXH_STAT.MTW
2. Check the normality of the variable “Priceindex”
3. Choose Stat > Nonparametrics > 1-Sample Sign.
4. In Variables, enter PriceIndex.
5. Choose Test median and enter 115 in the text box.
6. In Alternative, choose greater than. Click OK.

WILCOXON SIGNED RANK
TEST: EXAMPLE
This following dataset consists of cholesterol levels in patients two and four days after a heart
attack. Note: these data are paired because the two measurements have been made on the
same individuals.
2 days after 4 days after Difference
270 218 52
236 234 2
210 214 -4
142 116 26
280 200 80
272 276 -4
160 146 14
245 236 9
257 225 32
178 180 -2
Choose Stat > Nonparametrics > 1-Sample
Wilcoxon…

MANN-WHITNEY TEST: EXAMPLE
Samples were drawn from two populations and diastolic blood pressure was measured. You will
want to determine if there is evidence of a difference in the population locations without
assuming a parametric model for the distributions. Therefore, you choose to test the equality
of population medians using the Mann-Whitney test with level of significance = 0.05 rather
than using a two-sample t-test, which tests the equality of population means.
2. Choose Stat > Nonparametrics > Mann-Whitney.
3. In First Sample, enter DBP1. In Second Sample, enter DBP2. Click OK.
Interpreting the results
 The sample medians of the ordered data
as 69.5 and 78.
 The 95.1% confidence interval for the
difference in population medians (ETA1-
ETA2) is [-18 to 4].
 The test statistic W = 60 has a p-value of
0.2685 or 0.2679 when adjusted for ties.
Since the p-value is not less than the
chosen a level of 0.05, you conclude that
there is insufficient evidence to reject H0.
Therefore, the data does not support the
hypothesis that there is a difference
between the population medians.

KRUSKAL-WALLIS TEST: EXAMPLE
Measurements in growth were made on samples that were each given one of three
treatments. Rather than assuming a data distribution and testing the equality of
population means with one-way ANOVA, you decide to select the Kruskal-Wallis
procedure to test H0: h1 = h2 = h3, versus H1: not all h's are equal, where the h's
are the population medians.
2. Choose Stat > Nonparametrics > Kruskal-Wallis.
3. In Response, enter Growth.
4. In Factor, enter Treatment. Click OK.

MOOD'S MEDIAN TEST: EXAMPLE
One hundred seventy-nine participants were given a lecture with cartoons to
illustrate the subject matter. Subsequently, they were given the OTIS test, which
measures general intellectual ability. Participants were rated by educational level:
0 = preprofessional, 1 = professional, 2 = college student. The Mood's median test
was selected to test H0: ɳ1 = ɳ2 = ɳ3, versus H1: not all h's are equal, where the
h's are the median population OTIS scores for the three education levels.
1. Open the worksheet CARTOON.MTW.
2. Choose Stat > Nonparametrics > Mood's Median Test.
3. In Response, enter Otis. In Factor, enter ED. Click OK.

HYPOTHESIS TESTING ROADMAP
ATTRIBUTE DATA
Attribute Data
One Factor Two Factors
One Sample
Proportion
Two Sample
Proportion
MINITABTM:
Stat - Basic Stats - 2 Proportions
If P-value < 0.05 the proportions
are different
Chi Square Test
(Contingency Table)
MINITABTM:
Stat - Tables - Chi-Square Test
If P-value < 0.05 the factors are not
independent
Chi Square Test
(Contingency Table)
MINITABTM:
Stat - Tables - Chi-Square Test
If P-value < 0.05 at least one
proportion is different
Two or More
Samples
Two
SamplesOne Sample
54

Test for association (or dependency) between two
classifications
(Chi–Square Test) “TWO FACTORS”….
Contingency Tables
55
Exercise objective: To practice solving problem presented using
the appropriate Hypothesis Test.
 You are the quotations manager and your team thinks that the reason
you don’t get a contract depends on its complexity.
 You determine a way to measure complexity and classify lost contracts
as follows:
1. Write the null and alternative hypothesis.
2. Does complexity have an effect?
Low Med High
Price 8 10 12
Lead Time 10 11 9
Technology 5 9 16

Test for association (or dependency) between
two classifications
Contingency Tables
56
First we need to create a table in
MINITABTM
Secondly, in MINITABTM perform a
Chi-Square Test

Test for association (or dependency) between
two classifications
Contingency Tables
57
Are the factors independent of
each other?
Yes; Both factors are independent

8. testing of hypothesis for variable &amp; attribute data

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