Understandable Statistics
Seventh Edition
By Brase and Brase
Prepared by: Lynn Smith
Gloucester County College
Chapter Nine Part 5
(Section 9.7)
Hypothesis Testing
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1
Independent Sampling
Distributions
There is no relationship whatsoever
between specific values of the two
distributions.
Example: comparison of the
incomes of individuals from two
groups.
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2
Testing the Differences of
Means for Large, Independent
Samples
• Let x1 and x2 have normal distributions
with means 1 and 2 and standard
deviations 1 and 2 respectively.
• Take independent random samples of size
n1 and n2 from each distribution.
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3
The n the variablex1  x 2
hasthefollowingcharacte ri
stics:
1. A normal distribution.
2. Mean   1   2
3. Standard deviation 
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
2
1
n1


2
2
n2
4
If both n1 and n2 are 30 or
larger
The Central Limit Theorem can be
applied even if the original
distributions are not normal.
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5
When testing the difference of
means
The null hypothesis usually
indicates that there is no difference
between the means.
H0: 1– 2 = 0
or, equivalently
H0: 1= 2
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6
Left-Tailed Test
H0: 1= 2
H1: 1< 2
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7
Right-Tailed Test
H0: 1= 2
H1: 1 > 2
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8
Two-Tailed Test
H0: 1= 2
H1: 1  2
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9
Use the Critical Values for the
Normal Distribution
 = 0.05
 = 0.01
Critical z for
Left-Tailed Test
 1.645
 2.33
Critical z for
Right-Tailed Test
1.645
2.33
Critical z for
Two-Tailed Test
 1.96
 2.58
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10
To convert the sample x1  x 2 value
to a z value, use :
z
( x1  x 2 )  (  1   2 )

2
1
n1
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

2
2
n2
11
A college professor wishes to
determine (at 0.05 level of
significance) if there is a
difference in the final exam
results between the two groups
of students. One group studies
Calculus in the traditional
classroom setting; the other is
taught via distance learning.
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12
Sample exam results:
x
88
Distance
Learning
85

3.3
4.1
n
31
30
Traditional
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13
We wish to determine if the
observed difference in exam results
(88 versus 85) is significant.
H0: 1= 2
H1: 1  2
Using a two-tailed test, the critical z values
are  1.96 for  = 0.05.
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14
Compute the sample test
statistic
x1  x2  88  85  3
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15
Convert the test statistic to a z
value
z
( x1  x 2 )  (  1   2 )
 12
n1

 22
n2
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
30

3
.
14
3.3 2 4.12

31
30
16
Since 3.14 > 1.96
We reject the null hypothesis - the
statement that there was no
difference between the two groups.
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17
We conclude that the exam
results for the distance learning
class are significantly different
from those for the traditional
group at 5% level of
significance.
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18
P Value Approach
• To find the P value for the sample statistic
of the difference in the means, use the
corresponding z value, z = 3.14.
• Since we are working with a two-tailed
test, add the area to the right of z = 3.14
to the area to the left of z = – 3.14.
• P value = 2(0.0008) = 0.0016
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19
P value = 0.0016
We would reject H0
for any   0.0016.
We, therefore reject H0
for  = 0.05.
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20
Inferences about the
Differences of Two Means of
Small Independent Samples
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21
Assumptions
• Independent random samples are drawn
from two populations with means 1 and
2.
• The parent populations have normal
distributions.
• The standard deviations for the
populations (1 and 2) are equal.
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22
Best Estimate of the Common
or Pooled Standard Deviation
for Two Populations
( n1  1) s  ( n2  1) s
s
n1  n2  2
2
1
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2
2
23
When testing the difference of
means
The null hypothesis usually
indicates that there is no difference
between the means.
H0: 1– 2 = 0
or, equivalently
H0: 1= 2
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24
Left-Tailed Test
H0: 1= 2
H1: 1< 2
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25
Right-Tailed Test
H0: 1= 2
H1: 1 > 2
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26
Two-Tailed Test
H0: 1= 2
H1: 1  2
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27
To Test the Hypothesis Use the t
Statistic
x1  x 2
t
with d.f.  n1  n 2  2
1 1
s

n1 n2
where s  the pooled standard deviation
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28
The Pooled Standard Deviation
( n1  1) s  ( n2  1) s
s
n1  n2  2
2
1
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2
2
29
Test
H0: 1 = 2
Against H1: 1 > 2
Use a 1% Level of Significance.
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30
Sample Data
Population 1 Population 2
x
17.3
16.1
s
1.6
1.5
n
8
9
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31
Degrees of Freedom
d.f. = n1 + n2 – 2 =
8 + 9 – 2 = 15
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32
For a Right-Tailed Test at 1%
Level of Significance and d.f. = 15
Use the column headed by  = 0.010
The critical t value is
t = 2.602.
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33
Compute the Test Statistic x1  x2
x1  x2  17.3  16.1  1.2
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Pooled Standard Deviation
( n1  1) s  ( n2  1) s
s

n1  n2  2
2
1
2
2
(8  1)1.6  (9  1)1.5
 1.55
892
2
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2
35
Convert the Test Statistic x1  x 2
to a t Value
x1  x2  17.3  16.1  1.2
x1  x 2
1 .2
t

 1.593
1 1
1 1
s

1.55 
n1 n2
8 9
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36
Critical t Value = 2.602
2.602
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37
Test Statistic  1.593
1.593 2.602
Do not reject H0.
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P Value Approach
To test
H0: 1 = 2
Against H1: 1 > 2
t value = 1.593, d.f. = 15
• For a right-tailed test, use the column
headed by .
• For d.f. = 15, t = 1.593 falls between t =
1.517 and t = 1.753.
• We conclude that 0.050 < P < 0.075.
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39
Conclusion: 0.050 < P < 0.075
We reject H0 only for   P.
We could not reject H0 for  = 0.01.
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40
Tests for Differences of
Proportions
• We will draw independent random
samples of size n1 and n2 from two
binomial distributions.
• For large values of n1 and n2,, the
distribution of sample differences in
proportions of successes will be
approximately normally distributed.
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41
Symbols Used
r1 and r2 = number of successes from the
first and second samples
p1 and p2 = probability of success on each
trial from the first and second binomial
distributions
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42
For large values of n1 and n2:
The distribution of sample differences
r1 r2
pˆ 1  pˆ 2   is approximately normal with
n1 n2
mean    p1  p2
and standard deviation   
p1 q1
n1
p2 q2

n2
where q1  1  p1 and q 2  1  p2
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43
The Null Hypothesis
p1 = p2
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44
The proportions p1 and p2
are unknown and must be
estimated
Best (pooled)estimate  " p hat" 
r1  r2
pˆ 
n1  n2
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45
Critical Values for Tests
Involving a Difference of Two
Proportions (Large Samples)
 = 0.05
 = 0.01
Critical z for
Left-Tailed Test
 1.645
 2.33
Critical z for
Right-Tailed Test
1.645
2.33
Critical z for
Two-Tailed Test
 1.96
 2.58
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46
“Large” Samples
All of the following are larger than
five:
n1 pˆ , n1qˆ , n2 pˆ , n2 qˆ
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Sample Test Statistic
z
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pˆ 1  pˆ 2
pˆ qˆ pˆ qˆ

n1 n2
48
A college is attempting to
determine whether a reminder
phone call encourages students
to participate in early
registration. A group of 1200
students was divided into two
groups. One received a
reminder phone call; the other
did not.
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49
The data:
Reminder
No Reminder
n
600
600
Registered
early
475
452
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50
Use a 5% level of significance
to test the claim that the
reminder phone calls increased
participation in early
registration.
• Let p1 = proportion from the first group
(reminder) who registered early.
• Let p2 = proportion from the second
group (no reminder) who registered
early.
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The null hypothesis is that
there is no difference in
proportions.
H0: p1 = p2
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52
The alternate hypothesis is that
the reminders improve
participation rate.
H1: p1 > p2
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53
H1: p1 > p2
• Use a right-tailed test.
• For 5% level of significance, the critical z
value = 1.645.
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54
For the Reminder group,
r1 = 475, n1 = 600.
r1 475
pˆ 1  
 0.792
n1 600
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55
For the No Reminder group,
r1 = 452, n1 = 600.
r2 452
pˆ 2  
 0.753
n2 600
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56
The Sample Statistic
pˆ 1  pˆ 2  0.792  0.753  0.039
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57
The pooled estimates of
proportion
r1  r2 475  452
pˆ 

 0.7725
n1  n2 600  600
qˆ  1  pˆ  0.2275
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58
Convert the test statistic to a z
value:
z

pˆ 1  pˆ 2
pˆ qˆ pˆ qˆ

n1 n2
0.039
 1.61
.7725 (.2275) .7725 (.2275)

600
600
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59
Critical z Value = 1.645
1.645
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60
Test statistic = z = 1.61
1.61 1.645
Do not reject the null hypothesis.
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61
Conclusion
At 5% level of significance, we cannot
reject the claim that the proportions (of
students who register early) are equal.
We conclude that the reminder phone
calls do not make a significant difference
in the number of students who
participate in early registration.
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62
P Value Approach
• For a right-tailed test, the value of P is
the area in the right tail of the
distribution, larger than z = 1.61.
• From Table 5a, we find the P value.
• P = 1.000 – 0.9463= 0.0537
• We would reject H0 only for   0.0537.
• We cannot reject H0 for  = 0.05.
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63