The Difference in Proportions
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Transcript The Difference in Proportions
Testing the Difference
between Proportions
Section 11.3
Assumptions
Randomly Selected Samples
Approximately normal since
𝑛1 𝑝1 ≥ 10
𝑛1 1 − 𝑝1 ≥ 10
and
𝑛2 𝑝2 ≥ 10
𝑛2 1 − 𝑝2 ≥ 10
Independent (at least 10n in population)
Normally the Standard Deviation of
Statistic is
p1q1 p2 q2
n1
n2
But, Since we claim p1 p2 in the
Ho
We can combine the values to form one
proportion:
x1 x2
pc
n1 n2
And the Standard Deviation of Statistic becomes
1 1
pc qc
n1 n2
Can you find this on the formula
Sheet?
They use a combined p.
A sample of 50 randomly selected men with high
triglyceride levels consumed 2 tablespoons of oat bran daily
for six weeks. After six weeks, 60% of the men had
lowered their triglyceride level. A sample of 80 men
consumed 2 tablespoons of wheat bran for six weeks. After
six weeks, 25% had lower triglyceride levels. Is there a
significant difference in the two proportions at the 0.01
significance level?
To calculate pc we need to find x1 and x2.
x1 0.60(50) 30
x2 0.25(80) 20
So…..
30 20 50
pc
0.385
50 80 130
Parameter:
p1 proportion of men with high triglyceride levels who ate oat bran.
p2 proportion of men with high triglyceride levels who ate wheat bran.
p p difference in prop. of men w/high triglycerides between those who ate oat & bran
2
1
Hypothesis:
H o : p1 p2
H A : p1 p2
Assumptions:
* Randomly Selected Samples
* Approximately Normal since
n1 p1 50(0.6) 30 5
n1q1 50(.4) 20 5
and
n2 p2 80(0.25) 20 5
n2 q2 80(0.75) 60 5
* Independent –
(at least 500 men eat oat and 800 eat wheat bran
Name of Test:
2-Proportion Z-Test
)
^ ^
p1 p 2 p1 p2
z
1 1
pc qc
n1 n2
z
.6 .25 0
1
1
.385 * .615 *
50 80
z 3.99
P Value P( z 3.99)* 2
P Value ncdf (3.99, )* 2
P Value 0
Reject the Ho since the P-Value(0) < (0.05)
There is sufficient evidence to support the claim that
there is a difference in the proportion of men who
lowered their triglycerides by eating oat bran and the
proportion who lowered their triglycerides by eating
wheat bran.
In a sample of 100 store customers, 43 used a Mastercard. In another sample
of 100, 58 used a Visa card. Is the proportion of customers who use
Mastercard less than those using Visa?
p1 prop using mastercard
p2 prop using visa
p1 p2 diff in prop using mastercard & visa
H o : p1 p2
H A : p1 p2
43 58
100 100
101
0.505
200
pc
Assumptions:
1. Randomly Selected Samples
2. Approx. Normal
n1 p1 100(.43) 43 5
n1q1 100(.57) 57 5
n2 p2 100(.58) 58 5
n q 100(.42) 42 5
2 2
3. Independent (at least 1000 of each)
^ ^
p1 p 2 p1 p2
.43 .58 0
z
2.12
1
1 1
1
.
505
*
.
495
*
pc qc
100 100
n1 n2
P Val P( z 2.12)
ncdf ( , 2.12) 0.017
Reject the Ho since the p-val(.017) < (0.05)
There is sufficient evidence to support the claim that the
proportion using mastercard is less than the proportion
using visa.
So how would we find a confidence
interval?
PANIC!
In a sample of 80 Americans, 55% wished that they were rich. In a sample
of 90 Europeans, 45% wished that they were rich. Is there a difference in
the proportions. Find and interpret the 95% confidence interval for the
difference of the two proportions.
p1 prop Americans who wish to be rich
p2 prop Europeans who wish to be rich
p1 p2 diff in prop Am & Eurp who wish to be rich
Assumptions:
1. Randomly Selected Samples
2. Approx Norm n1 p1 80(.55) 44 5
n1q1 80(.45) 36 5
n2 p2 90(.45) 41 5
n q 90(.55) 50 5
2 2
3. Independent (at least 800 Am and 900 Europeans.
^ ^
p1 p 2 z 2
p1q1 p2 q2
n1
n2
.55(.45) .45(.55)
.55 .45 1.96
80
90
.10 .150
.05, .25
We’re 95% confident that the difference in proportion of
Americans who wish to be rich and the proportion of
Europeans who wish to be rich is between -.05 and .25.
In fact, since this interval contains 0, there is no
significant difference.
Homework
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