AP Statistics
I.
Inference Part 2
10-2: Inference for Two Proportions
Unit 9 Day 3
Confidence Intervals for the difference of two proportions(2-prop Z-Int)
P – define parameter, population, and process
A – check Assumptions & Conditions
-Random – both samples must be random
-Normal – check that the number of successes & failures for each sample is at least 10 (there should
be 4 numbers to verify).
-Independent – for experiments, state independence if experiment was done properly; for samples,
check the 10% condition (add a 0 to each sample size and check population size).
S – Statistical Work (2-Prop Z-Int)
𝑝̂1 (1−𝑝̂1 )
-by hand: Formula: (𝑝̂1 − 𝑝̂2 ) ± 𝑧 ∗ √
𝑛1
+
𝑝̂2 (1−𝑝̂2 )
Calculator:
𝑛2
*point estimate = (𝑝̂1 − 𝑝̂2 ); critical value = 𝑧 ∗ ; standard error = √
𝑝̂1 (1−𝑝̂1 )
𝑛1
+
𝑝̂2 (1−𝑝̂2 )
𝑛2
*margin of error = (critical value)(standard error)
*the critical value, z*, comes from the Normal curve. The Confidence Level is the middle percent, so you
subtract from 1 and divide by 2 to put the “area to the left into 2 ndVars3: InvNorm(area to left)
S – Conclusion: “We are ____% confident that the interval between ____________ and ____________ will
capture the true difference in proportion of parameter 1 and parameter 2 (in context).
Example 1) Just before the presidential election in November 2008, a local newspaper conducted a poll of
residents in a medium-sized city and found that 120 out of a simple random sample of 250 men intended to
vote for Barack Obama and 132 out of an SRS of 240 women intended to vote for Obama. Construct and
interpret a 95% confidence interval for the difference in proportion of women and men who supported
Obama in this city.
P – 2-proportion Z-interval
𝑝1 = true proportion of male voters who supported Obama in this city
𝑝2 = true proportion of female voters who supported Obama in this city
A -Random – both samples are an SRS
-Normal – men: 120 successes (votes for Obama) and 130 failures (not voting for Obama) are both at
least 10; women: 132 successes, 108 failures are both at least 10
N -Independent – there are at least 2500 men and 2400 women in this medium-sized city
S–
**x1 and x2 must be whole numbers, so if given the
percentages of the samples, you have to find the
actual numbers for x1 and x2.
(-.1583, .0183)  only need to report the confidence
interval for proportions (df doesn’t apply here because
it’s only for the t-distribution).
S – We are 95% confident the interval between -.1583 and .0183 will capture the true difference in
proportions of men and women who support Obama in this medium-sized city.
**formulas for confidence intervals use the sample statistic, 𝑝̂ , in the calculations because the goal is to
estimate the population proportion p. That’s why the normal condition counts successes and failures instead
of np and n(1-p), because we don’t know the value of p.
II.
Significance Tests for the difference of two proportions (2-proportion Z-Test)
P – state hypothesis, define parameters & populations, and identify procedure
Null: H0 : p1 = p2
*the null will ALWAYS be that there is no difference between proportions
Alternative: Ha : p1 >/</≠ p2
*choose the appropriate inequality symbol based on the problem
a
and always state p1 first.
A – check Assumptions & Conditions & state alpha level
-all conditions for significance tests for two-proportions are the same as confidence intervals for two
proportions
S – Statistical Work (2-Prop Z-Test)
-by hand – Formula: 𝑡 =
(𝑝̂1 −𝑝̂2 )−0
,
̂ 𝐶 (1−𝑝
̂ 𝐶) 𝑝
̂ (1−𝑝
̂ 𝐶)
𝑝
√
+ 𝐶
𝑛1
𝑛2
then use tcdf to find p-value -Calculator:
𝑝̂𝐶 = pooled sample proportion; the combined sample proportion of the two samples
𝑋1 + 𝑋2
𝑝̂𝐶 =
𝑛1 + 𝑛2
𝑝̂𝐶 is used in the formula for standard deviation of the test statistic
S – Conclusion:
Because our p-value of _______ is less than [greater than] α = 0.05, we [fail to] reject the null
hypothesis and [cannot] conclude ______________________________________________ (alternative in context).
**if the p-value is greater than alpha, then insert the words in brackets in the conclusion**
Example 2) Is there convincing evidence that there was a gender difference in Obama’s support in this
city? (scenario from example 1) Support your conclusion with a test of significance using α = 0.05
P
2-proportion Z-test
H0 : p1 = p2
Ha : p1 ≠ p2
𝑝1 = true proportion of male voters who supported Obama in this city
𝑝2 = true proportion of female voters who supported Obama in this city
A – α=0.05
Random, Normal, & Independent were checked & verified in Example 1.
S–
z = -1.550
p-value = .1212
̂𝑪 =.5143
𝒑
S – Because our p-value of 0.1212 is greater than α = 0.05, we fail to reject the null hypothesis and cannot
conclude that there is a difference in proportion of men and women supporters for Obama in this city.
**Notice how the confidence interval captured 0 as a possible parameter value for the true difference in
proportions for men and women supporters of Obama in that city and we failed to reject the null hypothesis &
could not conclude there was a difference***
*confidence intervals & significance tests go hand-in-hand