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Notation - Comparing Two Means
Notation - Comparing Two Means
Chapter 11
Mean
Standard
Value Variance Deviation
Comparing Two Populations
or Treatments
1
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2
1
σx −x =
1
4
2
2
σ12 σ22
+
n1 n 2
Population or Treatment 1
n1
x1
s12
s1
Population or Treatment 2
µ2
σ22
σ2
Population or Treatment 2
n2
x2
s 22
s2
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5
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Hypothesis Test
Hypothesis Tests Comparing Two Means
against one of the usual alternate
hypotheses using the statistic
7
( x1 − x 2 ) − hypothesized value
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Sampling Distribution for
Comparing Two Means
If n1 and n2 are both large or the population
distributions are (at least approximately)
normal, then the distribution of
z=
8
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σ12 σ22
+
n1 n 2
6
Example
A sample of 35 cans from machine 1 gave a
mean of 16.031 and a sample of 31 cans from
machine 2 gave a mean of 16.009. State,
perform and interpret an appropriate
hypothesis test using the 0.05 level of
significance.
x1 − x 2 − (µ1 − µ 2 )
Is described (at least approximately) by the
standard normal (z) distribution.
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Hypothesis Test
Example
Comparing Two Means
µ1 = mean fill from machine 1
We would like to compare the mean fill of 32
ounce cans of beer from two adjacent filling
machines. Past experience has shown that the
population standard deviations of fills for the
two machines are known to be σ1 = 0.043 and
σ2 = 0.052 respectively.
H0: µ1 - µ2 = hypothesized value
σ12 σ 22
+
n1 n 2
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Comparing Two Means
Large size sample techniques allow
us to test the null hypothesis
z=
3
This implies that the sampling distribution of
x1 − x 2 is also normal or approximately
normal.
σ12 σ22
+
n1 n 2
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σ1
If n1 and n2 are both large or the population
distributions are (at least approximately)
normal then x1 and x 2 each have (at least
approximately) a normal distribution.
1. µ x −x = µ x − µ x = µ1 − µ 2
1
σ12
Sampling Distribution for Comparing
Two Means
If the random samples on which x1 and x 2
are based are selected independently of
one another, then
2. σ2x1 − x2 = σ2x1 + σ2x2 =
µ1
2
Sampling Distribution for
Comparing Two Means
Sample
Standard
Size
Mean Variance Deviation
Population or Treatment 1
µ2 = mean fill from machine 2
H0: µ1 - µ2 = 0
Ha: µ1 - µ2
Significance level: α = 0.05
Test statistic:
z=
9
x 1 − x 2 − hypothesized value
σ12 σ 22
+
n1 n 2
=
x1 − x 2 − 0
σ12 σ22
+
n1 n 2
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1
Hypothesis Test
Example
Hypothesis Test
Comparing Two Means
Since n1 and n2 are both large (> 30) we do not
have to make any assumptions about the nature of
the distributions of the fills.
z=
(16.031 − 16.009 )
0.0432 0.0522
+
35
31
The p-value of the test is 0.0628. There
is insufficient evidence to support a claim
that the two machines produce bottles
with different mean fills at a significance
level of 0.05.
= 1.86
Equivalently, we might say.
P-value = 2P(z > 1.86) = 2P(z < -1.86)
With a p-value of 0.063 we have been
unable to show the difference in mean
fills is statistically significant at the 0.05
significance level.
= 2(0.0314) = 0.0628
11
Two-Sample t Test for Comparing
Two Population Means
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12
df =
x1 − x 2 − (µ1 − µ 2 )
s12 s 22
+
n1 n 2
( V1 + V2 )
Alternate hypothesis and finding the P-value:
1. Ha: µ 1 - µ 2 > hypothesized value
P-value = Area under the
z curve to the right
of the calculated z
2
V12
V2
+ 2
n1 − 1 n 2 − 1
where V1 =
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Two-Sample t Tests for Difference
of Two Means
Two-Sample t Test for Comparing
Two Population Means
If n1 and n2 are both large or if the population
distributions are normal and when the two random
samples are independently selected, the
standardized variable
t=
σ12 σ 22
+
n1 n 2
=
P-value:
Accept this example for what it is, just a sample of
the calculation. Generally this statistic is used
when dealing with “what if” type of scenarios and
we will move on to another technique that is
somewhat more commonly used when σ1 and σ2
are not known.
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X1 − X 2 − 0
Example
Comparing Two Means
Calculation:
This example is a bit of a stretch, since knowing
the population standard deviations (without
knowing the population means) is very unusual.
10
Hypothesis Test
Example
Comparing Two Means
s12
s2
and V2 = 2
n1
n2
2. Ha: µ1 - µ2 < hypothesized value
P-value = Area under the
z curve to the left
of the calculated z
Has approximately a t distribution with
df should be truncated to an integer.
13
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14
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Hypothesis Test
Two-Sample t Tests for Difference
of Two Means
15
Hypothesis Test
Example
i. 2•(area to the right of z) if z is positive
ii. 2•(area to the left of z) if z is negative
Brand A
517, 495, 503, 491
503, 493, 505, 495
498, 481, 499, 494
H0: µ1 = µ2 (µ1 - µ2 = 0)
Ha: µ1 ≠ µ2 (µ1 - µ2 ≠ 0)
Significance level: α = 0.01
Brand B
493, 508, 513, 521
541, 533, 500, 515
536, 498, 515, 515
Test statistic:
t=
State and perform an appropriate hypothesis test.
16
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17
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Example
µ1 = the mean amount of acetaminophen in cold
tablet brand A
µ2 = the mean amount of acetaminophen in cold
tablet brand B
In an attempt to determine if two competing
brands of cold medicine contain, on the
average, the same amount of acetaminophen,
twelve different tablets from each of the two
competing brands were randomly selected and
tested for the amount of acetaminophen each
contains. The results (in milligrams) follow. Use
a significance level of 0.01.
3. Ha: µ 1 - µ 2 ≠ hypothesized value
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18
x1 − x 2 − hypothesized value
σ12 σ 22
+
n1 n 2
=
x1 − x 2 − 0
σ12 σ 22
+
n1 n 2
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2
Hypothesis Test
Example
Hypothesis Test
Example
Hypothesis Test
Example
Assumptions (continued):
Assumptions:The samples were selected
independently and randomly. Since the
samples are not large, we need to be able to
assume that the populations (of amounts of
acetaminophen are both normally distributed.
Calculation:
n1 = 12, x1 = 497.83, s1 = 8.8300
n 2 = 12, x 2 = 515.67, s 2 = 15.144
t=
As we can see from the normality plots and the
boxplots, the assumption that the underlying
distributions are normally distributed appears to be
quite reasonable.
19
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Hypothesis Test
20
Example
Hypothesis Test
Calculation:
s 2 8.83002
V1 = 1 =
= 6.4974
n1
121
V2 =
df =
2
1
2
2
2
V
V
+
n1 − 1 n 2 − 1
=
We truncate the degrees of freedom to give df = 17.
22
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Confidence Intervals
23
df =
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Confidence Intervals
Comparing Two Means
2
V12
V2
+ 2
n1 − 1 n 2 − 1
s12
s2
and V2 = 2
n1
n2
df should be truncated to an integer. The t
critical values are in the table of t critical
values (table III).
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8.83002 15.144 2
+
12
12
Comparing Two Means
The general formula for a confidence interval
for µ1 – µ2 when
1. The two samples are independently
chosen random samples, and
2. The sample sizes are both large
(generally n1 ≤ 30 and n2 ≤ 30) OR the
population distributions are
approximately normal is
(X
1
− X 2 ) ± (t critical value)
24
s12 s 22
+
n1 n 2
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Confidence Intervals
Example
Example
Comparing Two Means
s12 5.62 2
=
= 0.632
n1
50
Construct a 98% confidence interval for the
difference of the population means.
V2 =
s 22 6.312
=
= 0.796
n2
50
Sample
Sample Sample Standard
Size
mean Deviation
50
78.3
5.62
50
87.2
6.31
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= −3.52
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V1 =
Thread A
Thread B
26
497.83 − 515.67 − 0
Two kinds of thread are being compared for
strength. Fifty pieces of each type of thread
are tested under similar conditions. The
sample data is given in the following table.
where V1 =
25
=
Confidence Intervals
Comparing Two Means
The t critical value is based on
( V1 + V2 )
Example
Conclusion: Since P-value = 0.002 < 0.01 = α,
H0 is rejected. The data provides strong
evidence that the mean amount of
acetaminophen is not the same for both brands.
Specifically, there is strong evidence that the
average amount per tablet for brand A is less
than that for brand B.
(6.4974 + 19.112) 2
= 17.7
6.49742 19.1122
+
11
11
σ12 σ 22
+
n1 n 2
21
P-value: From the table of tail areas for t curve
(Table IV) we look up a t value of 3.5 with df = 17
to get 0.001. Since this is a two-tailed alternate
hypothesis, P-value = 2(0.001) = 0.002.
s 22 15.1442
=
=19.112
n2
12
( V1 + V2 )
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x1 − x 2 − 0
df =
( 0.632 + 0.796 )
2
0.632 0.796
+
49
49
2
2
= 96.7
Truncating, we have df = 96.
27
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3
Confidence Intervals
Confidence Intervals
Example
5.62 2 6.312
+
50
50
−8.9 ± 2.82
87 Octane
26.4, 27.6, 29.7
28.9, 29.3, 28.8
The 98% confidence interval estimate for the
difference of the means of the tensile strengths is
x1 =28.45, s1 = 1.228, x 2 =30.73, s 2 = 1.392
V1 =
90 Octane
30.5, 30.9, 29.2
31.7, 32.8, 29.3
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Confidence Intervals
29
Example
Looking on the table under 95% with 9 degrees of
freedom, the critical value of t is 2.26.
x1 - x 2 ± (t critical value)
= 28.45 − 30.73 ± 2.26
2
1
2
2
s
s
+
n1 n 2
The 95% confidence interval for the true difference of
the mean mileages is (-3.99, -0.57).
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35
2
= 9.8
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Comments: We had to assume that the samples
were independent and random and that the
underlying populations were normally
distributed since the sample sizes were small.
By looking at the following normality plots, we see
that the assumption of normality for each of the two
populations of mileages appears reasonable.
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33
1.
The samples are paired.
2.
The n sample differences can be viewed as a
random sample from a population of
differences.
3.
The number of sample differences is large
(generally at least 30) OR the population
distribution of differences is approximately
normal.
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Given the small sample sizes, the assumption of normality
is very important, so one would be a bit careful utilizing
this result.
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Comparing Two Population or
Treatment Means – Paired t Test
Alternate hypothesis and finding the P-value:
1. Ha: µd > hypothesized value
P-value = Area under the appropriate t
curve to the right of the calculated t
2. Ha: µd < hypothesized value
P-value = Area under the appropriate t
curve to the left of the calculated t
3. Ha: µd ≠ hypothesized value
i. 2•(area to the right of t) if t is positive
ii. 2•(area to the left of t) if t is negative
Assumptions:
Where n is the number of sample differences and x d
and sd are the sample mean and standard deviation of the
sample differences. This test is based on df = n-1
34
30
Comparing Two Population or
Treatment Means – Paired t Test
x d − hypothesized value
sd
n
( 0.2512 + 0.3231)
Confidence Intervals
Example
Comments about assumptions
Comparing Two Population or
Treatment Means – Paired t Test
t=
s 22 1.3922
=
= 0.3231
n2
6
Confidence Intervals
Example
Comments about assumptions
32
Null hypothesis: H0:µd = hypothesized value
Test statistic:
V2 =
0.2512 2 0.32312
+
5
5
Truncating, we have df = 9.
If we randomized the order of the tankfuls of
the two different types of gasoline we can
reasonably assume that the samples were
random and independent. By using all of the
observations from one car we are simply
controlling the effects of other variables such as
year, model, weight, etc.
1.2282 1.392 2
+
6
6
-2.28 ± 1.71
31
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s12 1.2282
=
= 0.2512
n1
6
df =
(-11.72, -6.08)
28
Example
Let the 87 octane fuel be the first group and the 90
octane fuel the second group, so we have
n1 = n2 = 6 and
A student recorded the mileage he obtained
while commuting to school in his car. He
kept track of the mileage for twelve different
tankfuls of fuel, involving gasoline of two
different octane ratings. Compute the 95%
confidence interval for the difference of
mean mileages. His data follow:
Looking on the table of t critical values (table
III) under 98% confidence level for
df = 96, (we take the closest value for df ,
specifically df = 120) and have the t critical
value = 2.36.
( 78.3 − 87.2 ) ± 2.36
Confidence Intervals
Example
Comparing Two Means
Comparing Two Means
36
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4
Paired Sample Example
Paired t Test Example
A weight reduction center advertises that
participants in its program lose an average of
at least 5 pounds during the first week of the
participation. Because of numerous
complaints, the state’s consumer protection
agency doubts this claim. To test the claim
at the 0.05 level of significance, 12
participants were randomly selected. Their
initial weights and their weights after 1 week
in the program appear on the next slide. Set
up and perform an appropriate hypothesis
test.
37
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Paired Sample Example
Member
Initial
Weight
One Week
Weight
1
195
195
2
153
151
3
174
170
4
125
123
5
149
144
6
152
149
7
135
131
8
143
147
9
139
138
10
198
192
11
215
211
12
153
152
38
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Paired Sample Example
continued
Paired Sample Example
continued
This is equivalent to the difference of means:
µd = µ1 – µ2 = µinitial weight - µ1 week weight
H0: µd = 5
Ha: µd < 5
Member
Initial
Weight
1
195
2
153
151
2
3
174
170
4
4
125
123
2
5
149
144
5
6
152
149
7
135
131
4
8
143
147
-4
9
139
138
1
10
198
192
6
11
215
211
4
12
153
152
1
39
40
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Paired Sample Example
3
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continued
Calculations: According to the statement of the
example, we can assume that the sampling is
random. The sample size (10) is small, so
n = 12, x d = 2.333, s d = 2.674
t=
Significance level: α = 0.05
Test statistic:
x − hypothesized value x d − 5
t= d
=
sd
sd
n
n
One Week Difference
Initial -1week
Weight
0
195
Paired Sample Example
continued
Assumptions: According to the statement of the
example, we can assume that the sampling is random.
The sample size (12) is small, so from the boxplot we
see that there is one outlier but never the less, the
distribution is reasonably symmetric and the normal
plot confirms that it is reasonable to assume that the
population of differences (weight losses) is normally
distributed.
µd = mean of the individual weight changes
(initial weight–weight after one week)
continued
x d − 5 2.333 − 5
=
= −3.45
sd
2.674
12
n
P-value: This is an lower tail test, so looking up the t value
of 3.0 under df = 11 in the table of tail areas for t curves
(table IV) we find that the P-value = 0.002.
41
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Paired Sample Example
continued
42
continued
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Large-Sample Inferences
Difference of Two Population (Treatment) Proportions
Minitab returns the following when asked to
perform this test.
Conclusions: Since P-value = 0.002 < 0.05 = α,
we reject H0.
Some notation:
This is substantially the same result.
Population
Sample
Sample Proportion of Proportion of
Size
Successes Successes
Paired T-Test and CI: Initial, One week
We draw the following conclusion.
There is strong evidence that the mean weight
loss for those who took the program for one
week is less than 5 pounds.
Paired T for Initial - One week
Initial
One week
Difference
N
12
12
12
Mean
160.92
158.58
2.333
StDev
28.19
27.49
2.674
95% upper bound for mean difference: 3.720
T-Test of mean difference = 5 (vs < 5): T-Value = -3.45
43
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44
Population or
treatment 1
Population or
treatment 2
SE Mean
8.14
7.93
0.772
n1
π1
p1
n2
π2
p2
P-Value = 0.003
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45
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5
Properties: Sampling Distribution of p1- p2
Large-Sample z Tests for π1 – π2 = 0
If two random samples are selected
independently of one another, the following
properties hold:
The combined estimate of the common
population proportion is
1. µ p − p = π1 − π2
1
2
z=
1
2
π1 (1 − π1 ) π 2 (1 − π2 )
+
and
n1
n2
2
=
3. If both n1 and n2 are large [n1 p1 ≥ 10,
n1(1- p1) ≥ 10, n2p2 ≥ 10, n2(1- p2) ≥ 10],
then p1 and p2 each have a sampling
distribution that is approximately normal
46
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total sample size
47
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Example - Student Retention
Gender
Male
Female
n1 = number of males surveyed
n2 = number of females surveyed
p1 = x1/n1 = sample proportion of males who plan on
returning
p2 = x2/n2 = sample proportion of females who plan on
returning
Null hypothesis: H0: π1 – π2 = 0
Alternate hypothesis: Ha: π1 – π2 ≠ 0
51
Example - Student Retention
Significance level: α = 0.05
Calculations:
Test statistic:
pc =
p1 − p 2
pc (1 − p c ) p c (1 − p c )
+
n1
n2
P-value:
+
182
Conclusion:
0.76727 − 0.77473
275
53
P-value = 2(0.4247) = 0.8494
p c (1 − p c )
0.77024(1 − 0.77024)
=
The P-value for this test is 2 times the area
under the z curve to the left of the computed
z = -0.19.
p1 − p 2
p c (1 − p c )
=
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Example - Student Retention
n1p1 + n 2 p 2 211 + 141 352
=
=
= 0.7702
n1 + n 2
275 + 182 457
z=
Assumptions: The two samples are independently
chosen random samples. Furthermore, the sample sizes
are large enough since
n1 p1 = 211 ≥ 10, n1(1- p1) = 64 ≥ 10
n2p2 = 141 ≥ 10, n2(1- p2) = 41 ≥ 10
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π2 = true proportion of females who plan on returning
Response
Yes
No Maybe
211
45
19
141
32
9
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275
52
Example - Student Retention
Test to see if the proportion of students planning on returning is
the same for both genders at the 0.05 level of significance?
50
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π1 = true proportion of males who plan on returning
The responses along with the gender of the person
responding are summarized in the following table.
Example - Student Retention
z=
48
A group of college students were asked what they
thought the “issue of the day”. Without a pause the
class almost to a person said “student retention”. The
class then went out and obtained a random sample
(questionable) and asked the question, “Do you plan
on returning next year?”
Alternate hypothesis and finding the P-value:
1. Ha: π1 - π2 > 0
P-value = Area under the z curve to the
right of the calculated z
2. Ha: π1 - π2 < 0
P-value = Area under the z curve to the
left of the calculated z
3. Ha: π1 - π2 ≠ 0
i. 2•(area to the right of z) if z is positive
ii. 2•(area to the left of z) if z is negative
p1 − p 2
p c (1 − p c ) p c (1 − p c )
+
n1
n2
Assumptions:
1. The samples are independently chosen random
samples OR treatments are assigned at random to
individuals or objects (or vice versa).
2. Both sample sizes are large:
n1 p1 ≥ 10, n1(1- p1) ≥ 10, n2p2 ≥ 10, n2(1- p2) ≥ 10
total number of successes in two samples
Large-Sample z Tests for π1 – π2 = 0
49
n1p1 + n 2p 2
n1 + n 2
pc =
π1 (1 − π1 ) π 2 (1 − π2 )
+
n1
n2
σ p −p =
1
Test statistic:
2
2
2
2
2. σ p −p = σ p + σ p =
1
Large-Sample z Tests for π1 – π2 = 0
Null hypothesis: H0: π1 – π2 = 0
+
Since P-value = 0.849 > 0.05 = α, the hypothesis H0 is
not rejected at significance level 0.05.
0.77024(1 − 0.77024)
182
-0.0074525
= -0.19
0.040198
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There is no evidence that the return rate is different for
males and females..
54
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6
Example
Example
A consumer agency spokesman stated that he
thought that the proportion of households having
a washing machine was higher for suburban
households then for urban households. To test to
see if that statement was correct at the 0.05 level
of significance, a reporter randomly selected a
number of households in both suburban and
urban environments and obtained the following
data.
Suburban
Urban
55
Number
surveyed
Number
having
washing
machines
Proportion
having
washing
machines
300
250
243
181
0.810
0.724
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Ha: π1 - π2 > 0
56
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Large-Sample Confidence
Interval for π1 – π2
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1. The samples are independently selected random
samples OR treatments that were assigned at
random to individuals or objects (or vice versa), and
The P-value = 1 - 0.9916 = 0.0084
Conclusion:
2. Both sample sizes are large:
Since P-value = 0.0084 < 0.05 = α, the hypothesis H0 is
rejected at significance level 0.05. There is sufficient
evidence at the 0.05 level of significance that the
proportion of suburban households that have washers is
more that the proportion of urban households that have
washers.
59
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Example
Example
A student assignment called for the students to survey
both male and female students (independently and
randomly chosen) to see if the proportions that
approve of the College’s new drug and alcohol policy.
A student went and randomly selected 200 male
students and 100 female students and obtained the
data summarized below.
For a 90% confidence interval the z value to use is
1.645. This value is obtained from the bottom row of
the table of t critical values (Table III).
n1 p1 ≥ 10, n1(1- p1) ≥ 10, n2p2 ≥ 10, n2(1- p2) ≥ 10
A large-sample confidence interval for π1 – π2 is
(p1 − p2 ) ± ( z critical value )
60
p1 (1 − p1 ) p 2 (1 − p 2 )
+
n1
n2
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
We use p1 to be the female’s sample approval
proportion and p2 as the male’s sample approval
proportion.
Number Number that Proportion
surveyed
approve
that approve
100
43
0.430
200
61
0.305
(0.430 − 0.305) ± 1.645
0.430(1 − 0.430) 0.305(1 − 0.305)
+
100
200
(0.125) ± 0.097
Use this data to obtain a 90% confidence interval estimate
for the difference of the proportions of female and male
students that approve of the new policy.
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
When
The P-value for this test is the area under the z
curve to the right of the computed z = 2.39.
0.810 − 0.742
1
1 
0.7709(1 − 0.7709) 
+

 300 250 
= 2.390
61
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
Example
=
Female
Male
57
P-value:
p1 − p2
pc (1 − pc ) pc (1 − pc )
+
n1
n2
p1 − p 2
p c (1 − p c ) p c (1 − p c )
+
n1
n2
Assumptions: The two samples are independently
chosen random samples. Furthermore, the sample sizes
are large enough since
n1 p1 = 243 ≥ 10, n1(1- p1) = 57 ≥ 10
n2p2 = 181 ≥ 10, n2(1- p2) = 69 ≥ 10
H0: π1 - π2 = 0
n1p1 + n 2 p 2 243 + 181 424
=
=
= 0.7709
n1 + n 2
300 + 250 550
58
z=
π1 - π2 is the difference between the proportions
of suburban households and urban
households that have washing machines.
Example
z=
Significance level: α = 0.05
Test statistic:
π2 = proportion of urban households having
washing machines
Calculations:
pc =
Example
π1 = proportion of suburban households having
washing machines
62
or
(0.028,0.222)
Based on the observed sample, we believe that the
proportion of females that approve of the policy exceeds the
proportion of males that approve of the policy by
somewhere between 0.028 and 0.222.
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
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