Chapter Nine
Inferences Based
on Two Samples
Hypothesis Test 2-Sample Means (Known )
Both Normal pdf (known )
Null Hypothesis: H0: u1 – u2 = 0
Test Statistic: z = x – y – 0
 21/m + 22/n
Alternative Hypothesis:
Reject Region
Ha: u1 – u2 > 0 (Upper Tailed)
z  z
Ha: u1 – u2 < 0 (Lower Tailed)
z  -z
Ha: u1 – u2  0 (Two-Tailed) either z  z/2
or z  -z/2
P-Value computed the same as 1-Sample Mean.
Example HT 2-Sample Means (Known )
During a total solar eclipse the temperature drops
quickly as the moon passes between the earth and
the sun. During the June 2001 eclipse in Africa,
data was collected on the drop in temperature in
degrees F at two types of locations. The average
drop in temperature for 9 samples taken in
Mountainous terrain was 15.0. The average drop in
temperature for 12 samples taken in River-level
terrain was 17.5. Assume the variance in
temperature drop is known to be 9 for this type of
terrain-temperature drop experiment and that
experiments of this type follow a Normal pdf. Is
there evidence at the  = .10 level to conclude that
there is a difference in temperature drop between
the two types of terrain in this experiment?P-value?
Determining  ( known)
Alternative
Hypothesis
Ha: u1 - u2 > 0
Type II
Error ()
 z-  - 0

Ha: u1 - u2 < 0
1 -  -z-  - 0

Ha: u1 - u2  0  z/2 - - 0 -  -z/2- - 0


Where  = X-Y = (21/m) + (22/n)
Example HT 2-Sample Means (Known )
A study of report writing by student engineers was
conducted at Watson School. A scale that measures the
intelligibility of student engineers’ English is devised. This
scale, called an “index of confusion,” is devised, to the
delight of the students, so that low scores indicate high
readability. Data are obtained on articles randomly
selected from engineering journals and from unpublished
reports. A sample of 16 engineering journals yielded an
average score of 1.75 while a sample of 25 unpublished
reports yielded an average score of 2.5. Variance for this
type of scale is known to be 0.48 and the scores are
known to follow a Normal pdf. At a significance level of
.05, does there appear to be a difference between the
average scores of the two types of reports? What is the
Beta error when the true averages differ by as much as
0.5?
Hypothesis Test 2-Sample Means (Large n)
Null Hypothesis: H0: u1 – u2 = 0
Test Statistic: z = x – y – 0
 s21/m + s22/n
Alternative Hypothesis:
Reject Region
Ha: u1 – u2 > 0 (Upper Tailed)
z  z
Ha: u1 – u2 < 0 (Lower Tailed)
z  -z
Ha: u1 – u2  0 (Two-Tailed) either z  z/2
or z  -z/2
Both m > 40 & n > 40.
Example HT 2-Sample Means (Large n)
Aseptic packaging of juices is a method of
packaging that entails rapid heating followed by
quick cooling to room temperature in an air-free
container. Such packaging allows the juices to be
stored un-refrigerated. A new & old machine used
to fill aseptic packages is being compared. The
mean number of containers filled per minute on
the new machine was 114.1 for 50 observations
with a standard deviation of 5.0. The mean
number of containers filled per minute on the old
machine was 112.7 for 72 observations with a
standard deviation of 3.0. Is there evidence that
the new machine is faster than the old machine?
Use a test with  = .01. (Include the P-value).
Hypothesis Test 2-Sample Means (Small n)
Both Normal pdf (unknown )
Null Hypothesis: H0: u1 – u2 = 0
Test Statistic: t = x – y – 0
 s21/m + s22/n
Alternative Hypothesis:
Reject Region
Ha: u1 – u2 > 0 (Upper Tailed)
t  t, 
Ha: u1 – u2 < 0 (Lower Tailed)
t  -t, 
Ha: u1 – u2  0 (Two-Tailed) either t  t/2, 
or t  -t/2, 
Estimating the Degrees of Freedom

s21 + s22 2
=
m
n
(s21/m)2 + (s22/n)2
m –1
n -1
Example HT 2-Sample Means (Small n)
The slant shear test is used for evaluating the bond of
resinous repair materials to concrete. The test utilizes
cylinder specimens made of 2 identical halves bonded at
300C.
Twelve specimens were prepared using wire brushing. The
sample mean shear strength (N/mm2) and sample standard
deviation were 19.20 & 1.58, respectively. Twelve specimens
were prepared using hand-chiseled specimens; the
corresponding values were 23.13 & 4.01.
Does the true average strength appear to be different for the
two methods of surface preparation? Use a significance level
of .05 to test the relevant Hypothesis & assume the shear
strength distributions to be Normal. (Include the P-value).
Parameters of interest:
R. R.:
Null Hypothesis:
Calculation:
Alternative:
Decision:
Test Statistic:
P-value:
Example HT 2-Sample Means (Small n)
A manufacturer of power-steering components
buys hydraulic seals from two sources. Samples
are selected randomly from among the seals
obtained from these two suppliers, and each seal
is tested to determine the amount of pressure that
it can withstand. These data result:
Supplier I
Supplier II
x = 1342 lb/in2
y = 1338 lb/in2
s2 = 100
s2 = 33
m = 10
n = 11
Is there evidence at the  = .05 level to suggest
that the seals from supplier I can withstand higher
pressures than those from supplier II? (P-value)
Assume measurements of this type are Normal.
HT 2-Sample Means-Paired Data (Small n)
Both Normal X and Y (unknown )
Null Hypothesis: H0: uD = 0
Test Statistic: t =
d – 0
sD /  n
Alternative Hypothesis:
Reject Region
Ha: uD > 0 (Upper Tailed)
t  t, 
Ha: uD < 0 (Lower Tailed)
t  -t, 
Ha: uD  0 (Two-Tailed) either
t  t/2, 
or
t  -t/2, 
D = X – Y within each paired observation.
Example HT 2-Sample Means-Paired Data (Small n)
One important aspect of computing is the CPU time
required by an algorithm to solve a problem. A new
algorithm is developed to solve zero-one multiple objective
problems in linear programming. It is thought that a new
algorithm will solve problems faster than the algorithm
currently used. To obtain statistical evidence to support
this research hypothesis, a number of problems will be
selected at random. Each problem will be solved twice;
once using the current algorithm and once using the newly
developed one. These CPU times are not independent; they
are based on the same problems solved by two different
methods and so are paired by design. The mean difference
between the (16) paired data points was 2.7 seconds with a
standard deviation calculated at 6.0 seconds. Does the data
support this hypothesis at a  = .025 level of significance?
Assume measurements of this type are known to be
Normal. (Give the P-value) Let X = old & Y = new.
Example HT 2-Sample Means-Paired Data (Large n)
Highway engineers studying the effects of
wear on dual-lane highways suspect that
more cracking occurs in the travel lane of the
highway than in the passing lane. To verify
this contention, 64 one-hundred-feet-long test
strips are selected, paved, and studied over a
period of time. It is found that the mean
difference in the number of major cracks is
3.3 with a sample deviation of 8.8. Does this
data support the research hypothesis at a
significance level of .05? (Include P-value).
Let RV X = Travel lane & RV Y = Passing lane
The F Distribution
 = W/ =  +  /2 x(/2)-1
Y/
2 
(+)/2
   x+1
2 2 
0<x<
W & Y are independent Chi-Square RV’s
with  &  degrees of freedom.
Hypothesis Test on 2 Population Variances
Both Normal (unknown u1 & u2)
Null Hypothesis: H0:
Test Statistic:
2

1=
2

2
 = S21 / S22
Alternative Hypothesis:
Reject Region
Ha: 21 > 22 (Upper Tail)
  F, m-1, n-1
Ha: 21 < 22 (Lower Tail)
  F1- , m-1, n-1
Ha: 21  22 (Two-Tail) either   F/2, m-1, n-1
or
  F1- /2, m-1, n-1
F1- , m-1, n-1 = 1 / F, n-1, m-1 F-Tables pg. 730735
Example HT 2-Population Variances
The cost of repairing a fiberoptic component may
depend of the stage of production at which it fails.
The following data are obtained on the cost of
repairing parts that fail when installed in the system
and on the cost of repairing parts that fail after the
system is installed in the field:
System failure
Field failure
Sample Size = 21
Sample Size = 25
Mean = $65
Mean = $120
s2 = 25
s2 = 100
It is thought that the variance in cost of repairs
made in the field is larger than the variance in cost
of repairs made when the component is placed into
the system. Test at the  = .10 level to see if there
is statistical evidence to support this contention.
Example HT 2-Population Variances
Oxide layers on semiconductor wafers are etched
in a mixture of gases to achieve the proper
thickness. The variability in the thickness of these
oxide layers is a critical characteristic of the
wafer, and low variability is desirable for
subsequent processing steps. Two different
mixtures of gases are being studied to determine
whether one is superior in reducing the variability
of the oxide thickness. Twenty wafers are etched
in each gas. The sample standard deviation of
oxide thickness are s1= 1.96 angstroms and s2=
2.13 angstroms, respectively. Is there any
evidence to indicate that either gas is preferable?
Use  = 0.10 & assume measurements of this type
are Normal.
Example HT 2-Population Variances
Two companies supply raw materials to the
manufacturer of paper products. The concentration
of hardwood in these materials is important for the
tensile strength of the products. The mean
concentration of hardwood for both suppliers is
the same; however, the variability in concentration
may differ between the two companies. The
standard deviation of concentration in a random
sample of 26 batches produced by company X is
4.7 g/l, while for company Y a random sample of 21
batches yields 6.1 g/l. Is there sufficient evidence
to conclude that the two population variances
differ? Use  = .05 & list any assumptions that you
make.