1 section 8.3 testing a claim about a proportion objective for a population with proportion p, use a...
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Section 8.3Testing 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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(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) n p0 ≥ 5 and n q0 ≥ 5
Requirements
Note: p0 is the assumed proportion, not the sample proportion
Note: 2 and 3 satisfy conditions for the normal approximation to the binomial distribution
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Test StatisticDenoted z (as in z-score) since the test uses the z-distribution.
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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).
Traditional method:
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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
H1: p < 0.5 significance level
Area =
-z
(Negative)
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Right-tailed Test “>”
H0: p = 0.5
H1: p > 0.5 significance level
Area =
z
(Positive)
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Two-tailed Test “≠”
H0: p = 0.5
H1: p ≠ 0.5 significance level
z
Area = Area =
-z
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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 x = 13 p = 0.9286
Claim: p > 0.5 using α = 0.05
Example 1
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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H0 : p = 0.5
H1 : p > 0.5
Example 1
Right-tailed
What we know: p0 = 0.5 n = 14 x = 13 p = 0.9286
Claim: p > 0.5 using α = 0.01
z in critical region
z = 3.207zα = 1.645
Test statistic:
Critical value:
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.
z Test statistic
zα Critical value
z zαP-value = P(Z > z)
p-value(area)
Example
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P-Value
Critical region in the left tail:
Critical region in the right tail:
Critical region in two tails:
P-value = area to the left of the test statistic
P-value = area to the right of the test statistic
P-value = twice the area in the tail beyond the test statistic
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P-Value method:
If the P is low, the null must go.If the P is high, the null will fly.
If P-value , reject H0.
If P-value > , fail to reject H0.
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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
Test statistic
P-value
P-value = 0.0007
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Using P-value
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
P-value = 0.0007
Stat → Proportions→ One sample → With summary
Null: proportion=
Alternative
Number of successes:
Number of observations:
H0 : p = 0.5
H1 : p > 0.5
Example 1What we know: p0 = 0.5 n = 14 x = 13 p = 0.9286
Claim: p > 0.5 using α = 0.01
13
14
0.5
>
● Hypothesis Test
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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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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 p = 0.262
Claim: p = 0.25 using α = 0.05
Example 2
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.
Problem 32, pg 424
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H0 : p = 0.25
H1 : p ≠ 0.25
Example 2
Two-tailed
z not in critical region
z = 0.667
zα = -1.960
Test statistic:
Critical value:
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
What we know: p0 = 0.25 n = 580 p = 0.262
Claim: p = 0.25 using α = 0.05
zα = 1.960
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Using P-valueExample 2
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
H0 : p = 0.25
H1 : p ≠ 0.25
What we know: p0 = 0.25 n = 580 p = 0.262
Claim: p = 0.25 using α = 0.05
x = np = 580*0.262 ≈ 152 P-value = 0.5021
Stat → Proportions→ One sample → With summary
Null: proportion=
Alternative
Number of successes:
Number of observations:
152
580
0.25
≠
● Hypothesis Test