1. 10.1: 2-Proportion Situations

2. A Choice Combine Chapters 10 and 11 together
Leave Chapters 10 and 11 separate
Would give us more time to review for AP exam in April/May
Would mean fewer tests to stress about and study for
Would also mean fewer opportunities to raise your grade
And more content to remember for the test
Would give less time to study for AP exam in April/May
More tests to stress about and study for
More opportunities to raise your grade
Less content to remember for each individual test

3. 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)

4. 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

5. Shape
Difference between two proportions (p1-p2)
Normality conditions

6. Center
If the new distribution is defined as the difference between the two proportions (p1-p2), then the mean of the distribution is p1-p2
Since we don’t know p1 and p2, we estimate the center to be

7. Spread The standard deviation is our usual measure of spread
This is on the AP formula sheet

8. 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

9. First Type of Problem (Similar to Chapter 7)
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 sample
See next slide for example
This is very similar to what we did in chapter 7

10. Find the probability of getting a difference of
Find the probability of getting a difference of .1 or less from the two samples
Given the true values in the question

11. Find the probability of getting a difference of
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( ,-1.727,0,1)=.0427
.0427 or about a 4% chance

12. Does this give us reason to doubt the counselors’ reports?
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?

13. Second Type of Problem (Similar to Chapter 8)
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)

14. First, let’s check our conditions:
Random? YES
Normal? YES
Independent? YES—10% condition is met

15. Now, let’s construct our 95% confidence interval
“difference between teens and adults”
Teens minus adults
Point estimate:
Critical value:
St. dev:

16. Point estimate: (.73-.47)=.26 Critical value: 1.96 (or 2) St. dev:
.0189
.26 ± (1.96)(.0189)
.26 ± or (.223, .297)

17. Phrasing our conclusion
Our interval was (.223, .297)
“We are 95% confident that the interval from .223 to .297 captures the true difference in the proportion of US teens and US adults that use social networking sites”
“We are 95% confident that the proportion of US teens that use social media is between 22.3% and 29.7% higher than the proportion of US adults that use social media”

18. 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

19. Point estimate: ( )=.1
Critical value: 1.96
St. Dev: .0935
.1 ± (1.96)(.0935)
.1 ± .183

20. 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

21. Point estimate: ( )=.1
Critical value: (use invnorm)
St. Dev: .0935
.1 ± (1.405)(.0935)
.1 ± .131

22. If we wanted to say how much better the first player is than the second player, neither of those intervals is particularly useful
They are both quite wide
And they both include zero
What could we do that would make our intervals narrower (without decreasing our confidence level)?

23. Third Type of Problem (Similar to Chapter 9)
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

24. Steps (Same as Chapter 9)
Hypotheses and alpha level
Random sample from population (usually done for us already)
Make sure conditions are met
Calculate test statistic (Z-score)
Calculate p-value
Draw conclusions in context (reject or fail to reject the null hypothesis)

25. 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

26. More Complicated Hypotheses
As the previous slide indicated, we will always use a difference of 0 as our null hypothesis
Our alternative will then always be: A. that the difference is greater than 0, B. that the difference is less than 0, or C. that the difference is not equal to zero
You CAN do more complicated hypotheses
For example, a null hypothesis that the difference is .12, with an alternative that it is less than .12
However, this is NOT something you need to know for AP statistics
You may see it if you take more advanced statistics/econometrics/business modeling/quantitative methods classes

28. We are going to use when we calculate the standard deviation
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:

30. Step 3 conditions (many conditions, unfortunately)
H0: 𝑝 1 − 𝑝 2 = 𝑜𝑟 𝑝 1 = 𝑝 2
Ha: 𝑝 1 − 𝑝 2 ≠ 𝑜𝑟 𝑝 1 ≠ 𝑝 2
Step 2: random sample?
Step 3 conditions (many conditions, unfortunately)
Sample 1: (80)(.2375)=19 (80)(.7625)=61
Sample 2: (150)(.1733)=26 (150)(.8267)=124
10% condition met (both schools have more than 1500 students)

31. Z=( )/(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 = p-value=.242
Fail to reject at a .05 significance level
Cannot conclude that there is a difference between the two schools

32. You try

33. H0: p1-p2=0
Ha: p1-p2<0
P-hat1: .027
P-hat2: .041
N1: 2051
N2: 2030
Pc: .034

34. H0: p1-p2=0
Ha: p1-p2<0
P-hat1: .027
P-hat2: .041
N1: 2051
N2: 2030
Pc: .034
Z= ( )/(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

35. 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

36. You try
Out of a sample of 3000 people in a city, 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. Assume that all conditions for inference are met. Is there a statistically significant difference between the preferences of the two cities at the .01 significance level?

37. Calculator Tests (make sure you’re showing enough work)
Hypothesis Test
Confidence Interval
2-PropZTest
State Hypotheses
Check Conditions
State what test you’re doing
Report 𝑝 1 and 𝑝 2
Report Test Statistic (Z)
Report p-value
Draw conclusions in context
2-PropZInt
They are much more lenient on this one
Check conditions
State what type of interval (i.e. “2 proportion Z interval”)

38. 4th Type of Problem

39. ME=(critical value)(Standard deviation) .02= 1.645 .5 .5 𝑛 + .5 .5) 𝑛
.02= 𝑛 ) 𝑛
.02= 𝑛 𝑛
.02= 𝑛
.01216= .5 𝑛
= .5 𝑛
𝑛 =.5
𝑛=3381
Book got 3383 because of different rounding

40. Homework
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