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Introduction to the Statistical Analysis
Using SPSS
Lecture # 2
By: Dr. Nahed Mohammad Hilmy
Department of Statistical and Operations Research
T-Test
Hypothesis testing involves making a decision concerning
some hypothesis or statement about a population parameter
such as the population mean,  using the sample mean, X to
decide whether this statement about the value of  is valid or
not.
 The steps of the hypothesis testing :
1- The first step is to formulate a null hypothesis written H 0 . The
statement for H 0 is usually expressed as an equation or
inequality as follows: H 0:   given value
H 0:   given value
H 0:   given value

2
Also in this step it is stated an alternative hypothesis, written H a ,
a statement that indicates the opinion of the conductor of the test
as to the actual value of  . H a is expressed as follows:
H a:   given value
H a:   given value
H a:   given value
We conduct a hypothesis test on a given value  to find out if
actual observation would lead us to reject the stated value.
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T-Test
The alternative hypothesis suggests the direction of the actual
value of the parameter relative to the stated value. The statement
of H a in the form of an inequality that indicates that the
investigator has no opinion as to whether the actual value of  is
more than or less than the stated value but the feeling is that the
stated value is incorrect. In this case the test is two-tail test.
Statements in the form of strictly greater than or strictly less than
relationship indicate that the investigator has an opinion as to the
direction of the value of the parameter relative to the stated
value. In this case it is called one-tail test.
4
T-Test



2- State the level of significance of the test and the
corresponding Z values (for large sample tests), or the
corresponding T values ( for small sample tests). The
hypothesis test is frequently conducted at the 5%, 1% and 10%
levels of significance. Some can use the Z values. For a test
conducted at any other level of significance, we simply use the
normal distribution table to determine a corresponding Z value.
3- Calculate the test statistic for the sample that has taken.
There are three cases:
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T-Test

Case 1: The variable has a normal distribution and  2 is known.
In this case the test statistic is
Z 
x  0

n
which has a standard normal distribution if   0 in H 0.
2
 Case 2: The variable has a normal distribution and  is
unknown. The test statistic is
x  0
Z 
S
n
which has a t n 1distribution if H 0 is true.
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T-Test
Case 3: The variable is not normal but n is large (which n>30),
 2 may be known or unknown.
 The test statistic is

Z 
x  0

if  2 is known
n
x  0
or Z 
if  2 is unknown
s
n

By central limit theorem it has approximately standard normal
distribution (0,1) if H 0 is true.
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T-Test
4- Determine the boundary (or boundaries) for the area of
rejection regions using either X c or Z c values. A critical value is
the boundary or limit value that requires as to reject the
statement of the null hypothesis.
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T-Test
Rejection region
Lower X C
Rejection region

upper
XC
In directional test there are two critical values when:
H a :   o
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T-Test
Rejection region

upper
XC
In directional test there is one critical value (upper boundary ) when:
H a :   o
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T-Test
Rejection region
Lower
XC

In directional test there is one critical value (lower boundary ) when:
H a :   o
11

The critical value X is simply the maximum or minimum value that we are
willing to accept as being consistent with the stated parameter . The mean of the
distribution is given by:
x  

The standard deviation of the distribution is given by:
x 




n
5- Formulate a decision rule on the basis of the boundary values obtained in step
4. When we conduct an hypothesis test, we are required to make one of two
decisions:
a- Reject Ho or
B- Accept Ho
12
It is possible to make two errors in decision . One error is
called a type I error or   error .We make a type I
error whenever we reject the statement of H 0 ,when is in
fact true. The probability of making a type I error is the
level of significance of the test. The second error we can
make in an hypothesis test is called a type II error, or Berror. We commit a type II error if we fail to reject the
statement of ,when
is in fact false. The four combinations
H
0
of truth values of
and the resulting
decisions are
H
0
summarizing below:
13
H
0
H
0
True
False
Reject Type I
Correct
H
0
error
Decision
Accept
Correct Type II
H
Decision error
0
14
When we lower the level of significance of an
hypothesis test we always increase the possibility of
committing a B-error.
6- State a conclusion for the hypothesis test based on the
sample data obtained and the decision rule stated in steps.
15







P-value of a test:
The p- value is the probability of getting a value more
extreme than one observed value of the test statistic, it is
denoted by Z
When H is as follows: H a 
obs
a
P-value= 2p (Z >| Z obs |)
When H a is :>
p-value= p (Z > Z obs )
When H a is :<
P-value = p (Z < Z obs )

16




If we have a T statistic with a t n  1
distribution and
observe value t
, these p-values becomes:
obs
>| t obs |)
 alternative :p-value = 2p (t n  1
> alternative :p-value = p ( t n  1 > t obs )
< alternative :p-value = p( t
< t
)
n 1
obs
17

Thus H is rejected if p-value <  . When data is
o
collected from a normally distributed population and the
sample size is small, the t values of the student t
distribution must be used in the hypothesis test not the Z
values of the normal distribution. This is due to the fact
that her central limit theorem does not apply when n < 30.
18




Ex:
Suppose we measure the sulfur content (as a percent) of
15 samples of crude oil from a particular Middle Eastern
area obtaining:
1.9,2.3,2.9,2.5,2.1,2.7,2.8,2.6,2.6,2.5,2.7,2.2,2.8,2.7,3.
Assume that sulfur content are normally distributed . Can
we conclude that the average sulfur content in this area is
less than 2.6? Use a level of significance of .05.
19
n  15
X  2.533
S  .3091
  .05
H 0 :   2.6
H a :   2.6
20
Rejection region
.95
.05
-1.6
21
One-Sample Statistics
N
X
15
Mean
2.5533
Std. Error
Std. Deviation
Mean
.3091 7.980E-02
22
One-Sample Test
Test Value = 2.6
X
t
-.585
Mean
df
Sig. (2-tailed) Difference
14
.568 -4.667E-02
95% Confidence
Interval of the
Difference
Lower
Upper
-.2178
.1245
23


Testing for the Difference in Two Population means:
Often we have two populations for which we would
like to compare the means. Independent random
samples of sizes n1 and n 2 are selected from the two
populations with no relationship between the elements
we drawn from the two populations. The statistical
hypothesis are given by:
24
H 0 : 1  2 vs
H a : 1   2
or
H a : 1   2
or
H a : 1   2
25



There are three cases which depend on what is known
about the the population variances.
 12
and
 22
Case1:
Population variances are known for normal populations
(or non normal populations with both n1and n 2 large).
In this case the test statistic is to be :
Z

X1  X 2
 12
n1

2
2
n2
26

Case2:

Populations are unknown but are to be equal 12   22   2
in normal populations. In this case, we pool our estimates
to get the pooled two- sample variance


( n1  1) S12  ( n2  1) S 22
2
S 
p
n1  n2  2
27

And the test statistic is to be
T 

Which has a
X1  X 2
1
1
2
Sp(

)
n1 n2
t n n 2
1 2
distribution if
H
0
is true.
28


Case 3:
1
2


are unknown and unequal normal
2 and
2
populations . In this case the test statistic is given by:
T 
X1  X 2
S12
n1

2
S2
n2
which does not have a known distribution.
29
Ex:
The amount of solar ultraviolet light of wavelength from 290 to 320
nm which reached the earths surface in the Riyadh area was
measured for independent samples of days in cooler months
(October to March) and in warmer months (April to September):
 Cooler:5.31,4.36,3.71,3.74,4.51,4.58,4.64,3.83,3.16,3.67,4.34,2.95,
3.62,3.29,2.45.
 Warmer:4.07,3.83,4.75,4.84,5.03,5.48,4.11,4.15,3.9,4.39,4.55,4.91,
4.11,3.16,2.99,3.01,3.5,3.77.
30

Assuming normal distributions with equal variances ,
test whether there is a difference in the average ultraviolet
light reaching Riyadh in the cooler and warmer months .
Use a level of significance of .05.
31
n
1
X
 15
n
1  3.877
S1
2
X2
 .751
S2
H0
:
1

2
Ha
:
1

2
 18
 4.142
 .709
32

The pooled two sample variance is
2
( n1  1) S1

(
n

1
)
S
2
2
2
2
S

 .531
p
n1  n2  1

And the test statistic is to be
T 
X1  X 2
S 2p (
1
n1

 1.033
1
)
n2
33
.95
.025
.025
t
31.025
 2.0423
t
31, 025  2.0423
34
Group Statistics
VAR00001
VAR00002
1.00
2.00
N
15
18
Mean
3.8773
4.1417
Std. Deviation
.7507
.7088
Std. Error
Mean
.1938
.1671
35
Independent Samples Test
Levene's Test for
Equality of Variances
F
VAR00001 Equal variances
assumed
Equal variances
not assumed
.091
Sig.
.764
t-test for Equality of Means
t
Mean
Sig. (2-tailed) Difference
df
Std. Error
Difference
95% Confidence
Interval of the
Difference
Lower
Upper
-1.039
31
.307
-.2643
.2545
-.7834
.2548
-1.033
29.238
.310
-.2643
.2559
-.7875
.2588
36


Since the value of the test statistic is in the
acceptance region , then H 0 is accepted at   05
.
It means that there is no difference in the average
ultraviolet light reaching Riyadh in the cooler and
warmer months .
37


Dependent Samples:
The method of comparing parameters of populations
using paired dependent samples requires that we pair the
items of data as we sample them from the two
populations. .Further more , the size of the two
populations selected from both populations is the same,
that is
n
1
 n
2
 n
38

For each X (the elements of the sample before the
i
Y
experiment) and i (the elements of the sample after the
experiment) we obtain in the two samples, we compute a
value d of a random variable D which represents the
i
difference between the two populations and n is the
number of items of data obtained in each of the two
samples .
39
The samples drawn from the two populations are
therefore converted to single sample –a sample of d i ' s
The mean , d , and the standard deviation, S
, of the
d
distribution of d i ' s are obtained as follows:

d 
 di

 ( xi
n
Sd 
 yi )
n
 ( di
 d )2
n 1
40

We are interested in testing one of the tests of hypothesis:
H 0 : d  0

T

H a : d  0
or
H a : d  0
or
H a : d  0
Thus the quantity
d
vs

d
Sd
n

has a t n  1 distribution.
41
Ex:
In an experiment comparing two feeding methods for
calves, eight pairs of twins were used-one twin receiving
Method A and the other twin receiving Method B. At the
end of a given time, the calves were slaughtered and
cooked, and the meat was rated for its taste (with a higher
number indicating a better taste
42
Twin pair
1
2
3
4
5
6
7
8
Method A
27
37
31
38
29
35
41
37
Method B
23
28
30
32
27
29
36
31
43
Assuming approximate normality, test if the average taste
score for calves fed by Method B is less than the average
  .05A. Use
taste for calves fed by Method
.
44
d
i
4
9
1
6
2
6
5
6
39
d
2
i
16
81
1
36
4
36
25
36
235
45
H 0 : d  0
d 
di
Sd 
n
1
(
n
vs
H a : d  0
 4.875

d i2  n d 2 )  2.542
46
The test statistic is
T 
d  d
 5.447
Sd
n
47
.95
rejection region
.05
t
n  1, 
 1.8946
48
Paired Samples Statistics
Pair
1
VAR00001
VAR00002
Mean
34.3750
29.5000
N
8
8
Std. Deviation
4.8679
3.8173
Std. Error
Mean
1.7211
1.3496
49
Paired Samples Correlations
N
Pair 1 VAR00001 & VAR00002
Correlation
8
.857
Sig.
.007
50
Paired Samples Test
Paired Differences
Pair 1 VAR00001 - VAR00002
Std. Error
Mean Std. Deviation Mean
4.8750
2.5319
.8952
95% Confidence
Interval of the
Difference
Lower
Upper
2.7582 6.9918
t
5.446
df
Sig. (2-tailed)
7
.001
51
Quality Control

A “defect” is an instance of a failure to meet a requirement
imposed on a unit with respect to single quality characteristic . In
inspection or testing , each unit is checked to see if it does or dose
not contain any defects. For example
, if every dosage unit could

be tested , the expense would probably be prohibitive both to
manufacturer and consumer. Also it is may cause misclassification
of items and other errors . Quality can be accurately and precisely
estimated by testing only part of the total material (a sample) .It
requires small samples for inspection or analysis .
52

Data obtained from this sampling can then be treated
statistically to estimate population parameters. After
inspection (n) units we will have found say (d) of them to
be defectives and (n - d) of them to be good ones. On the
other hand we may count and record the number of
defects, c, we find on single unit. This count may be
0,1,2,…. Such an approach of counting of defects on a
unit becomes especially useful if most of the units contain
one or more defects.
53

Control charts can be applied during in - process
manufacturing operations, for finished product
characteristics and in research and development for
repetitive procedures.We may always convert a
measurable characteristics of a unit to an attribute by
setting limits, say L (lower bound) and U (upper bound)
for x. Then if x lies between, the unit is a good one, or if
outside, it is a defective one. As an example for the control
chart the tablet weight.
54

We are interested in ensuring that tablet weight remain
close to a target value under “statistical control”. To
achieve this object , we will periodically sample a group
of tablets, measuring the mean weight and variability.
Variability can be calculated on the basis of the standard
deviation or the range. The range is the difference between
the lowest and highest value.
55

If the sample size is not large (<10) the range is an
efficient estimator of the standard deviation. The mean
weight and variability of each sample (subgroup) are
plotted sequentially as a function of time. The control
chart is a graph that has time or order of submission of
sequential lots on the x axis and the average test result on
the Y axis. The subgroups should be as homogeneous as
possible relative to overall process. They are usually ( but
not always) taken as units manufactured close in time.
56

Four to five items per subgroup is usually as adequate
sample size. In our example (10) tablets are individually
weighted at approximately (1) hour intervals. The mean
and range are calculated for each of the subgroups
samples. As long as the mean and range of the 10 tablet
samples do not vary “ too much” from subgroup to
subgroup, the product is considered to be in control (it
means that the observed variation is due only to the
random, uncontrolled variation inherent in the process).
57

We will define upper and lower limits for the mean and
range of the subgroups. The construct of these limits is
based on normal distribution. In particular, a value more
than (3) standard deviations from the mean is highly
unlikely and can be considered to be probably due to some
systematic, assignable cause. The average line (the target
value) may be determined from the history of the product
regular updating or may be determined from the product
specifications .
58



The action lines (the limits) are constructed to represent
 3 limits) from the
 3 standard deviations (
target value. The upper and lower limits for the mean X
chart are given by:
X  AR ,
R


R
K
is the average range , K is the number of samples
(subgroups).A is a factor which is obtained from a table
according to the sample size .
59


The central line, the upper and lower limits for the range chart are
given by:
Central line =
R 

R
K

Lower limit = D

Upper limit =
L
R
D
R
U
60




Where D
and D
are factors which are
L
U
obtained from a table according to the sample size. It is
noticed that the sample size is constant.
Ex:
Tablet weights and ranges from a tablet Manufacturing
Process (Data are the average and range of 10 tablets):
61
Date
Time
Mean
X
3/1
11 a.m.
12 p.m.
302.4
298.4
1 p.m.
2 p.m.
11 a.m.
300.2
299
300.4
Range
R
16
13
10
9
13
12 p.m.
302.4
300.3
299
5
12
17
3/5
1 p.m.
2 p.m.
62
Date
Time
Mean
Range R
X
3/9
3/11
11 a.m.
12 p.m.
1 p.m.
300.8
301.5
301.6
18
6
7
2 p.m.
11 a.m.
12 p.m.
301.3
301.7
303
8
12
9
1 p.m.
2 p.m.
300.5
299.3
9
11
63
Date
Time
Mean X
Range R
3/16
11 a.m.
12 p.m.
1 p.m.
300
299.1
300.1
13
8
8
2 p.m.
11 a.m.
12 p.m.
303.5
297.2
296.2
10
14
9
1 p.m.
2 p.m.
297.4
296
11
12
3/22
64
X chart
The central line ( the t arg et value) is X  300
A  .31 at n  10, R  10.833
L , U  X  A R  300  (.31)(10.833)  300  3.358
Lower Limit  296.642
Upper Limit  303.358
65
R chart
The central line  R  10.833
D
L
 .22,
D
U
 1.78
Lower Limit  D
L
Upper Limit  D
U
at
n  10
R  2.383
R  19.283
66
X
304
302
U c L=303.358
C L=300
300
L c L=296.642
298
296
294
292
290
3\1
3\5
3\9
3\11
3\16
3\22
67
R
U c L=19.283
18
16
14
12
C L=10.833
10
8
6
4
L c L=2.383
3\1
3\5
3\9
3\11
3\16
3\22
68