Modular 17
Ch 10.2 Part II
and
Ch 10.3 Part II
Ch 10.2 Hypothesis Test for a Population Mean
Objective B : P – Value Approach
Ch 10.3 Hypothesis Test about a Population
Mean with unknown 
Objective B : P – Value Approach
Ch 10.2 Hypothesis Test for a Population Mean
Objective B : P – Value Approach
Z – Test for a Proportion
A hypothesis test involving a population proportion can be
considered as a binomial experiment. As we learned from Ch 8.2,
the best point estimate of p , the population proportion, is a
x
sample proportion, pˆ  , provided the sample is obtained by
n
1) simple random sampling ;
2) np0 (1  p0 )  10 to guarantee that a normal distribution can be
used to test hypotheses for H 0 : p  p 0 ;
3) the sampled values are independent of each other ( n  0.05N).
Testing Hypotheses Regarding a Population Proportion, p .
Use the following steps to perform a Z – Test for a Proportion.
Example 1: The percentage of physicians who are women is 27.9%. In a
survey of physicians employed by a large university health
system, 45 of 120 randomly selected physicians were
women. Use the P - Value Approach to determine whether
there is sufficient evidence at the 0.05 level of significance
to conclude that the proportion of women physicians at the
university health system exceeds 27.9%?
n  120 , x  45 ,   0.05
Step 1 :
H 0 : p  0.279
H1 : p  0.279
Step 2 :
  0.05
(right-tailed test)
Step 3 : Sample :
n  120 , x  45
x 45
pˆ  
 0.375
n 120
z0 
z0 
pˆ  p0

p0 (1  p0 )
n
0.375  0.279
(0.279)(1  0.279)
120
0.375  0.279
 2.34* (test statistic)
(0.279)(0.721)
120
P – value approach
-- > Let  alone.
-- > Convert z0  2.34 to a tailed area.
From Table V
0.04
0.9904
2.3
0.9904
z0  2.34
Step 4 :
P – value = 1 – 0.9904
= 0.0096
Is P – value <  ?
Is 0.0096 < 0.05 ? Yes, reject H 0 .
Step 5 :
There is enough evidence to support the claim that the
proportion of female physicians exceeds 27.9%.
Example 2: In 1997, 4% of mothers smoked more than 21 cigarettes
during their pregnancy. An obstetrician believes that the
percentage of mothers who smoke 21 cigarettes or more is
less than 4% today. She randomly selects 120 pregnant
mothers and finds that 3 of them smoked 21 or more
cigarettes during pregnancy. Does the sample data support
the obstetrician’s belief? Use the P - Value Approach with
  0.01 level of significant.
H 0 : p  0.04
H1 : p  0.04
(left-tailed test)
n  120 , x  3 ,   0.01
x
3
pˆ  
 0.025
n 120
z0 
pˆ  p0

p0 (1  p0 )
n
0.025  0.04
z0 
(0.04)(0.96)
120
0.025  0.04
(0.04)(1  0.04)
120
 0.84 * (test statistic)
P – value approach
-- > Let  alone.
-- > Convert z0  0.84 to a tailed area.
From Table V
0.04
0.2005
z0  0.84
P – value = 0.2005
 0.8
0.2005
Is P – value <  ?
Is 0.2005 < 0.01 ? No, do not reject H 0 .
There is not enough evidence to support the claim that the
percentages of mothers who smoke 21 cigarettes or more is less
than 4%.
Example 3: In 2000, 58% of females aged 15 years of age and older lived
alone, according to the U.S. Census Bureau. A sociologist
tests whether this percentage is different today by
conducting a random sample of 500 females aged 15 years
of age and older and finds that 285 are living alone. Use the
P -Value Approach to determine whether there is sufficient
evidence at the   0.1 level of significance to conclude
the proportion has changed since 2000?
H 0 : p  0.58
H1 : p  0.58
(Two-tailed test)
n  500 , x  285 ,   0.1
x 285
pˆ  
 0.57
n 500
pˆ  p0

p0 (1  p0 )
n
z0 
z0 
0.57  0.58
(0.58)(0.42)
500
0.57  0.58
(0.58)(1  0.58)
500
 0.45 * (test statistic)
P – value approach
-- > Let  alone.
-- > Convert z0  0.45 to a two-tailed area.
From Table V
0.05
0.3264
z0  0.45
0.3264
z0  0.45
P – value = Double (0.3264)
= 0.6528
 0.4
0.3264
Is P – value <  ?
Is 0.6528 < 0.1 ?
No, do not reject H 0 .
There is not enough evidence to support the claim.
Ch 10.2 Hypothesis Test for a Population Mean
Objective B : P – Value Approach
Ch 10.3 Hypothesis Test about a Population
Mean with unknown 
Objective B : P – Value Approach
Ch 10.3 Hypothesis Test about a Population
Mean with unknown 
Objective B :
P – Value Approach
The t – test procedure requires either that the sample was drawn
from a normally distributed population or the sample size was
greater than 30. Minor departures from normality will not
adversely affect the results of the test. However, if the data
include outliers, the t – test procedure should not be used. A
normality plot is used to test whether the sample was drawn
from a normally distributed population. A boxplot is used to
detect outliers.
Example 1: Find the P – value interval for each test value.
(a) t0  2.624 , n  15 , right-tailed
df  n  1  15  1  14
P  value  0.01
t0  2.624
df  14
From Table VI
df
0.01
14
2.624
(b) t0  1.703 , n  28 , right-tailed
df  n  1  28  1  27
P  value  0.05
t0  1.703
df  27
From Table VI
df
0.05
27
1.703
(c) t0  2.321, n  23 , left-tailed
From Table VI
df  n  1  23  1  22
0.01
df
0.02
22
2.321
2.183
2.508
0.01  P  value  0.02
t0  2.321
df  22
P – value is between 0.01 and 0.02.
0.01  P  value  0.02
(d) t0  1.562 , n  17 , two-tailed
df  n  1  17  1  16
From Table VI
df
0.10  P  value  0.20
t0  1.562
df  17
1
( P  value ) is between 0.05 and 0.10.
2
1
0.05  ( P  value )  0.10
2
0.10  P  value  0.20
16
0.05
0.10
1.562
1.337
1.746
Example 2: Use the P – value approach to determine whether to reject
or fail to reject H 0 .
(a) P – value  0.126 versus   0.05 for a right-tailed test
Since P – value is not smaller than  , fail to reject H 0 .
(0.05)
(0.126)
(b) P – value  0.001 versus   0.025 for a left-tailed test
Since P – value <  , reject H 0 .
(0.001)
(0.025)
Example 3: A survey of 15 large U.S. cities finds that the average
commute time one way is 25.4 minutes. A chamber of
commerce executive feels that the commute in his city is
less and wants to publicize this. He randomly selects 25
commuters and finds the average is 22.1 minutes with a
standard deviation of 5.3 minutes. At   0.01 , is there
enough evidence to support the claim? Use the P – value
approach.
Step 1 :
Setup :
H 0 :   25.4
H 0 :   25.4 (left-tailed test)
Step 2 :
  0.01
Step 3 :
0  25.4 , n  25 , x  22.1, s  5.3
x  0 22.1  25.4
  3.113 *
t0 

5.3 25
s n
P – value Approach
df  n  1  25  1  24
From Table VI
0.001
df
0.0025
24
3.113
3.091
3.467
0.001  P  value  0.0025
t0  3.113
df  24
P – value is between 0.001 and 0.0025.
0.001  P  value  0.0025
Step 4 :
Is P – value <  ? Yes , reject H 0 .
(Between 0.001 (0.01)
and 0.0025)
Step 5 :
There is sufficient evidence to support the executive’s claim that
the commute time in his city is less than 25.4 minutes.
As we have experienced from example 1 to 3, we were unable to
find exact P – values using the table because the t – table provides
t – values only for certain area. To find exact P – values, we use
statistical software.
Example 4: Use StatCrunch to calculate the following question.