Transcript Ch8-3

Section 8.3
Testing a claim about a Proportion
Objective
For a population with proportion p, use a
sample (with a sample proportion) to test
a claim about the proportion.
Testing a proportion uses the standard
normal distribution (z-distribution)
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Notation
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Requirements
(1) The sample used is a a simple random sample
(i.e. selected at random, no biases)
(2) Satisfies conditions for a Binomial distribution
(3) np0 ≥ 5 and nq0 ≥ 5
Note: 2 and 3 satisfy conditions for the normal
approximation to the binomial distribution
Note: p0 is the assumed proportion, not the sample proportion
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Test Statistic
Denoted z (as in z-score) since
the test uses the z-distribution.
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Traditional method:
If the test statistic falls within the
critical region, reject H0.
If the test statistic does not fall within
the critical region, fail to reject H0
(i.e. accept H0).
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Types of Hypothesis Tests:
Two-tailed, Left-tailed, Right-tailed
The tails in a distribution are the extreme
regions where values of the test statistic
agree with the alternative hypothesis
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Left-tailed Test “<”
H0: p = 0.5
 significance level
H1: p < 0.5
Area = 
-z
(Negative)
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Right-tailed Test “>”
H0: p = 0.5
 significance level
H1: p > 0.5
Area = 
z
(Positive)
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Two-tailed Test “≠”
H0: p = 0.5
H1: p ≠ 0.5
Area = /2
-z/2
 significance level
Area = /2
z/2
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Example 1
The XSORT method of gender selection is believed to
increases the likelihood of birthing a girl.
14 couples used the XSORT method and resulted in the
birth of 13 girls and 1 boy.
Using a 0.05 significance level, test the claim that the
XSORT method increases the birth rate of girls.
(Assume the normal birthrate of girls is 0.5)
What we know:
p0 = 0.5
n = 14
Claim: p > 0.5
x = 13
using
p = 0.9286
α = 0.05
n p0 = 14*0.5 = 7
n q0 = 14*0.5 = 7
Since n p0 > 5 and n q0 > 5, we can perform a hypothesis test.
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Example 1
What we know:
p0 = 0.5
n = 14
Claim: p > 0.5
x = 13
using
p = 0.9286
α = 0.01
H0 : p = 0.5
H1 : p > 0.5
Right-tailed
Test statistic:
Critical value:
zα = 1.645
z = 3.207
z in critical region
Initial Conclusion: Since z is in the critical region, reject H0
Final Conclusion: We Accept the claim that the XSORT
method increases the birth rate of girls
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P-Value
The P-value is the probability of getting a
value of the test statistic that is at least as
extreme as the one representing the
sample data, assuming that the null
hypothesis is true.
Example
z Test statistic
zα Critical value
P-value = P(Z > z)
p-value
(area)
z zα
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P-Value
Critical region
in the right tail:
P-value = area to the right of
the test statistic
Critical region
in the left tail:
P-value = area to the left of the
test statistic
Critical region
in two tails:
P-value = twice the area in the tail
beyond the test statistic
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P-Value method:
If P-value   , reject H0.
If P-value >  , fail to reject H0.
If the P is low, the null must go.
If the P is high, the null will fly.
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Caution
Don’t confuse a P-value with a proportion p.
Know this distinction:
P-value = probability of getting a test statistic
at least as extreme as the one
representing sample data
p = population proportion
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Calculating P-value for a Proportion
Stat → Proportions → One sample → with summary
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Calculating P-value for a Proportion
Enter the number of successes (x) and the
number of observations (n)
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Calculating P-value for a Proportion
Enter the Null proportion (p0) and select the
alternative hypothesis (≠, <, or >)
Then hit Calculate
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Calculating P-value for a Proportion
The resulting table shows both the
test statistic (z) and the P-value
P-value
Test statistic
P-value = 0.0007
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Example 1 Using P-value
What we know:
p0 = 0.5
n = 14
Claim: p > 0.5
H0 : p = 0.5
H1 : p > 0.5
x = 13
using
p = 0.9286
α = 0.01
Stat → Proportions→ One sample → With summary
Number of successes:
13
Number of observations:
14
● Hypothesis Test
Null: proportion=
Alternative
0.5
>
P-value = 0.0007
Initial Conclusion: Since p-value < α (α = 0.05), reject H0
Final Conclusion: We Accept the claim that the XSORT
method increases the birth rate of girls
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Do we prove a claim?
A statistical test cannot definitely prove
a hypothesis or a claim.
Our conclusion can be only stated like this:
The available evidence is not strong enough
to warrant rejection of a hypothesis or a
claim
We can say we are 95% confident it holds.
“The only definite is that there are no definites” -Unknown
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Example 2
Problem 32, pg 424
Mendel’s Genetics Experiments
When Gregor Mendel conducted his famous
hybridization experiments with peas, one such
experiment resulted in 580 offspring peas, with 26.2%
of them having yellow pods. According to Mendel’s
theory, ¼ of the offspring peas should have yellow
pods. Use a 0.05 significance level to test the claim that
the proportion of peas with yellow pods is equal to ¼.
What we know:
p0 = 0.25
n = 580
Claim: p = 0.25
p = 0.262
using
α = 0.05
n p0 = 580*0.25 = 145
n q0 = 580*0.75 = 435
Since n p0 > 5 and n q0 > 5, we can perform a hypothesis test.
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Example 2
What we know:
p0 = 0.25
n = 580
Claim: p = 0.25
p = 0.262
using
α = 0.05
H0 : p = 0.25
H1 : p ≠ 0.25 Two-tailed
Test statistic:
zα = 1.960
zα = -1.960
z = 0.667
Critical value:
z not in critical region
Initial Conclusion: Since z is not in the critical region, accept H0
Final Conclusion: We Accept the claim that the proportion
of peas with yellow pods is equal to ¼
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Example 2 Using P-value
What we know:
p0 = 0.25
n = 580
Claim: p = 0.25
H0 : p = 0.25
H1 : p ≠ 0.25
p = 0.262
using
α = 0.05
Stat → Proportions→ One sample → With summary
Number of successes: 152
Number of observations: 580
● Hypothesis Test
Null: proportion=
0.25
Alternative
≠
x = np = 580*0.262 ≈ 152
P-value = 0.5021
Initial Conclusion: Since P-value > α, accept H0
Final Conclusion: We Accept the claim that the proportion
of peas with yellow pods is equal to ¼
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