Download 10.1: 2-Proportion Situations

10.1: 2-Proportion Situations
3.9.2017
• 2-proportion confidence intervals
• 2-proportion significance tests
• WE ARE LOOKING AT THE DIFFERENCE
BETWEEN TWO PROPORTIONS
– What is the plausible range for the difference
between the two? (confidence interval)
– Are the two different from each other?
(significance test(
2 Proportions
• So now we have 2 samples
– 1 from one population
– 1 from another population
• We want to know what the difference between
the two populations is
– NOT what the difference between the samples is
– We are just using the samples to draw conclusions
about the populations
–
Shape
• Difference between two proportions (p1-p2)
• Normality conditions
Center
• If the new distribution is defined as the
difference between the two proportions (p1p2), then the mean of the distribution is p1-p2
• Since we don’t know p1 and p2, we estimate
the center to be
Spread
• The standard deviation is our usual measure
of spread
• This is on the AP formula sheet
Clarification
• We have these two proportions
• To make conclusions about the difference between
them, we are effectively creating a new
variable/distribution that is defined by one minus the
other
• So once we do that, we now have a single sampling
distribution that allows us to draw conclusions
• So it behaves much the same way as chapters 8 and 9
Scenario 1
• We know (or think we know) the true
population proportions
• The question may then ask us, assuming that
those values are true, to find the probability
of getting a certain result
– See next slide for example
• Find the probability of getting a difference of
.1 or less from the two samples
• Find the probability of getting a difference of
.1 or less from the two samples
– Z=(value-mean)/(st. dev)
– Z=(.1-.2)/st. dev
• St dev. =.0579
– (-.1)/(.0579)=-1.727
– Normalcdf(-555555,-1.727,0,1)=.0427
– .0427 or about a 4% chance
• Find the probability of getting a difference of
.1 or less from the two samples
– .0427 or about a 4% chance
• Does this give us reason to doubt the
counselors’ reports?
Scenario 2
• We want a confidence interval for the difference
between the two proportions
• We don’t know the true proportion—we use
sample data to create the confidence interval
• Remember:
• (Point estimate) ± (Critical value)(St dev)
•
± (critical value)
• First, let’s check our conditions:
– Random? YES
– Normal? YES
• Independent? YES—one person’s response
doesn’t affect another’s
• Now, let’s construct our 95% confidence
interval
– “difference between teens and adults”
– Teens minus adults
• Point estimate:
• Critical value:
• St. dev:
• Point estimate: (.73-.47)=.26
• Critical value: 1.96 (or 2)
• St. dev:
– .0189
• .26 ± (1.96)(.0189)
• .26 ± .037 or .223--.297
You try
• In a random sample of 50 at-bats, one major
league baseball player reached base 19 times.
Another player, (also in a sample of 50 at-bats)
reached base 14 times. Find a 95% confidence
interval to represent the difference in their
on-base rates
• Point estimate: (.38-.28)=.1
• Critical value: 1.96
• St. Dev: .0935
• .1 ± (1.96)(.0935)
• .1 ± .183
• -.083 -- .283
Now different confidence level
• In a random sample of 50 at-bats, one major
league baseball player reached base 19 times.
Another player, (also in a sample of 50 at-bats)
reached base 14 times. Find an 84%
confidence interval to represent the difference
in their on-base rates
• Point estimate: (.38-.28)=.1
• Critical value: 1.405 (use invnorm
• St. Dev: .0935
• .1 ± (1.405)(.0935)
• .1 ± .131
• -.031 -- .231
Scenario 3
• We want to perform a significance test to see
if the two proportions are different from each
other
– We don’t know the population proportions—if we
did, no need for significance test
• Null hypothesis is that there is no difference
– Or that the difference equals 0
• Alternative hypothesis will be 1 of 3 choices
– The difference is >0
– The difference is <0
– The difference ≠ 0
• We are going to use
standard deviation
when we calculate the
• For our test statistics (Z) we use the same
procedure as always: (value-mean)/(st dev)
• For the standard deviation, we use the same
formula as before, but we plug in for both p1
and p2:
•
• H0:
• Ha:
• Z=(.2375-.1733)/(st. dev)
• Pooled prob: (45/230)=.1957
• St dev= .0549
•
•
•
•
•
Z=(.0642)/(.0549)=1.169
Normalcdf(1.169,BIG,0,1)=.121
.121 x 2 = .242
Fail to reject at a .05 significance level
Cannot conclude that there is a difference
You try
•
•
•
•
•
•
•
H0: p1-p2=0
Ha: p1-p2<0
P-hat1: .027
P-hat2: .041
N1: 2051
N2: 2030
Pc: .034
•
•
•
•
•
•
•
H0: p1-p2=0
Ha: p1-p2<0
P-hat1: .027
P-hat2: .041
N1: 2051
N2: 2030
Pc: .034
•
•
•
•
Z= (.027-.041)/(st dev)
St dev: .0057
Z=(-.014)/.0057=-2.456
Normalcdf(SMALL,2.456,0,1)=.007
• Reject null
• Conclude that the heart
attack rate is lower for
those who take the drug
compared to the placebo
•
•
•
•
•
•
•
H0: p1-p2=0
Ha: p1-p2<0
P-hat1: .027
P-hat2: .041
N1: 2051
N2: 2030
Pc: .034
• We can also do this on
our calculator
• 2-PropZTest
• X1:56
• N1: 2051
• X2: 84
• N2: 2030
You try
• Out of a sample of 3000 people in city #1,
1567 say that they will vote for the Republican
candidate. Out of a sample of 3700 people in
city 2, 1801 say that they will vote for the
Republican candidate. Is there a statistically
significant difference between the preferences
of the two cities at the .01 significance level?