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Dr. Nafez M. Barakat
Lecture no.1
Descriptive measures
 Measure of center
o Mean
Definition: the mean of the data set is the sum of the
observations divided by the number of observation symbolized
n
by
X 
x
i 1
i
n
Example1 : ( exam scores) let x1 = 88, x2 = 75, x3=95, and
x4=100
Compute the mean
n
Solution :
X 
x
i 1
i
n

358
 89.5
4
o Median
Definition: median of a data set
Arrange the data in increasing order.
-if the number of observation is odd then the median is the
observation exactly in the middle of the ordered lest
- if the number of observation is even then the median is the
mean of the two middle observations in the ordered list.
Example2 : weekly salaries
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Dr. Nafez M. Barakat
Find the median for the tow weekly salaries data
Table (1.1)
Data set I( n=13) odd
300 300 300 940 300
300 400 300 400 450
800 450 1050
Data set II( n= 10)even
300 300 940 450 400
400 300 300 1050 300
SOLUTION : Median for data set (I) = 400$, and for data set (II)
= 350$
o Mode
Definition: the mode is the value that occurs most frequently in
a data set
Example : find the mode in data set (I) in table 1.1
Solution: the frequency distribution of the data shown in table
1.2
Table 1.2
Salary
300
400
450
800
Frequency
6
2
2
1
2
940 1050
1
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Dr. Nafez M. Barakat
NOTE:
Median< mean
Median =mean
Right-skewed
symmetric
mean< Median
left-skewed
 Measure of variation (measure of spread)
o Range
Definition: the range of the data set is the difference between
its maximum and minimum observations : Range = Max - Min
o Standard deviation
Definition : Standard deviation equal to the square
root of the arithmetic mean of the squares of the
deviations from the arithmetic mean denoted by S.
S
(X  X )
n 1
2
or
S
x
2
 ( x) 2 / n
n 1
o Variance
Definition : the variance equal the square of Standard
deviation
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Dr. Nafez M. Barakat
Example : the height, in inches of five players on team II are 67,
72, 76, 76 and 84. Obtain the Standard deviation Of these
height
Solution :
X 
 x  75
n
X x x
67 -8
72 -3
76
1
76
1
84
9
S
(X  X )
n 1
2

156
 6.2
4
Try to use the formula
(x  x)2
64
9
1
1
81
156
inches
S
x
2
 ( x) 2 / n
n 1
to commute Standard
deviation
o Inter quartile range
Definition : inter quartile range or ( IQR), is the difference
between the first and third quartiles, that is Standard deviation
IQR = Q3 – Q1
Example : find the IQR fore these data
25
41
27
32
43
66
35
31
15
5
34
26
32
38
16
30
38
30
20
21
Solution : Q1 = 23
,
Q3 = 36.5
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Dr. Nafez M. Barakat
IQR = 36.5- - 23 = 13.5
Hypothesis Test for One Population Mean
Definition : Null hypotheses and Alternative hypothesis
Null hypotheses : a hypothesis to be tested, We use the symbol
H0 to represent the null hypothesis.
Alternative hypothesis: a hypothesis to be conceder as
alternative to null hypothesis, We use the symbol Ha to
represent the alternative hypothesis.
One sample t test for population Mean
We present two step by step procedure for performing a one
sample t-test. Procedure (I) covers the critical-value approach,
and Procedure (II) covers the p-value approach.
 One sample t test for population Mean
(critical-value approach)
Assumptions
1. Normal population or large sample
2.  unknown
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Dr. Nafez M. Barakat
Step 1: the null hypothesis is
H O :   O
and the alternative
hypothesis is
H a :    O (two  tailed ) H O :   O (left tailed )
H O :    O ( right tailed )
Step 2 : decide on the significance level, 
Step 3: compute the value of the test statistic
T
X  O
S/ n
Step 4: the critical value (s) are
 t / 2 (two  tailed ) or
 t (left  tailed )
or
t (right  tailed ) with degrees of freedom (df= n-1)
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Dr. Nafez M. Barakat
Step 5 : if the value of the t test statistics falls in the rejection
region, reject HO ; otherwise, fail to reject H0
Step 6 : interpret the results of the hypothesis test.
Example : table below show the pH levels for 15 lakes; test if
the lakes has pH greater than 6 at 5% significant level.( use the
critical value approach)
Solution :
Step 1: state the null and alternative hypotheses
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Dr. Nafez M. Barakat
H O :   6 ( mean PH Level is not greater than 6)
H a :   6 (mean PH Level is greater than 6)
Step 2 : decide on the significance level,   0.05
Step 3: compute the value of the test statistic
T
X  O
6.6  6

 3.458
S / n 0.672 / 15
Step 4: the critical value for a right-tailed test is
t  t 0.05  1.761 (from table) with df = 15-1 = 14
Step 5: the value of the test statistic, found in step 3 is T=3.458
fail in the rejection region. Consequently , we reject HO
 One sample t test for population Mean
(P-Value Approach)
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Dr. Nafez M. Barakat
Assumptions
3. Normal population or large sample
4.  unknown
Step 1: the null hypothesis is
H O :   O
and the alternative
hypothesis is
H a :    O (two  tailed ) H O :   O (left tailed )
H O :    O ( right tailed )
Step 2 : decide on the significance level, 
Step 3: compute the value of the test statistic
T
X  O
S/ n
Step 4: find the p-value by using table
with degrees of freedom (df= n-1)
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Dr. Nafez M. Barakat
Step 5 : if the P- value less than or equal  , ( p  value   ),
reject HO ; otherwise, fail to reject H0
Step 6 : interpret the results of the hypothesis test.
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Dr. Nafez M. Barakat
Example : table below show the pH levels for 15 lakes; test if
the lakes has pH greater than 6 at 5% significant level. ( use the
p-value Approach)
Solution :
Step 1: state the null and alternative hypotheses
H O :   6 ( mean PH Level is not greater than 6)
H a :   6 (mean PH Level is greater than 6)
Step 2 : decide on the significance level,   0.05
Step 3: compute the value of the test statistic
T
X  O
6.6  6

 3.458
S / n 0.672 / 15
Step 4: the p-value = p ( t>= 3.458) = 0.00192 (with df = 15-1 =
14 )
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Dr. Nafez M. Barakat
Step 5: p value < 0.05) so we reject HO
use SPSS program
use the SPSS program to perform the hypothesis in
previous example
STEP 1: Enter The Data As Shown Below
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Dr. Nafez M. Barakat
Step 3 : the result shown below
One-Sample Statistics
N
PH
15
Mean
6.600
Std. Deviation
.6719
Std. Error
Mean
.1735
One-Sample Test
Test Value = 6
PH
t
3.459
df
14
Sig. (2-tailed)
.004
13
Mean
Difference
.600
95% Confidenc e
Int erval of t he
Difference
Lower
Upper
.228
.972