1. HUDM4122 Probability and Statistical Inference
April 1, 2015

2. Questions from last class?

3. Continuing from last class…

4. Another way to think about Confidence Intervals
For a 95% Confidence Interval
5% of the time , the true value will be outside the confidence interval

5. Another way to think about Confidence Intervals
For a 95% Confidence Interval
5% of the time , the true value will be outside the confidence interval
We can write this proportion of 5% as a
If a is 0.05, then we have a 95% Confidence Interval

6. So, for a 95% Confidence Interval
a is 0.05,
Cumulative Normal Probability is from to 0.975
Meaning that Confidence Intervals bounds are SE and SE
These values are called − 𝒁 ∝/𝟐 and + 𝒁 ∝/𝟐
− 𝒁 ∝/𝟐 =−𝟏.𝟗𝟔
+ 𝒁 ∝/𝟐 =

7. Why is it 𝑍 ∝/2 ? Because to get a 95% confidence interval ∝ = 0.05
0.975 – = 0.95
0.025 = ∝/2
0.975 = 1-(∝/2)

8. So, formally For given a The confidence intervals are 𝑥 ± 𝑍 ∝/2 𝜎 𝑛
Where 𝜎 𝑛 is just your standard error
And we can use s (sample standard deviation) for 𝜎 whenever the sample is sufficiently large

9. Regarding this whole “sufficiently large” thing
MBB say “sufficiently large” is when N>30
That’s probably reasonable
We’ll come back to this issue in a couple weeks when we discuss the difference between Z statistical tests and t statistical tests

10. Questions? Comments?

11. For a 99% Confidence Interval
a is 0.01,
Cumulative Normal Probability is from to 0.995
Look in your table to get and 0.995
That’s and +2.57
Meaning that Confidence Intervals bounds are SE and SE
− 𝑍 ∝/2 =−2.57
+ 𝑍 ∝/2 =

12. You try it: 90% Confidence Interval
What are the bounds?

13. You try it: 90% Confidence Interval
a is 0.10,
Cumulative Normal Probability is from 0.05 to 0.95
Look in your table to get 0.05 and 0.95
That’s and +1.64
Meaning that Confidence Intervals bounds are SE and SE

14. You try it: 90% Confidence Interval
a is 0.10,
Cumulative Normal Probability is from 0.05 to 0.95
Look in your table to get 0.05 and 0.95
That’s and +1.64
− 𝑍 ∝/2 =−1.64
+ 𝑍 ∝/2 =

15. Comments? Questions?

16. Note The smaller your a The larger your % Confidence Interval
In other words, the bigger your interval
The more certain you are
And the smaller your interval
The less certain you are

17. What you can adjust You can adjust You can’t adjust
Your level of certainty
Your sample size
You can’t adjust
The population mean
The population standard deviation

18. So if you want to be more certain
Get a bigger sample size

19. So if you want to be more certain
Get a bigger sample size
Which is not always easy 

20. On to today’s class…

21. Today Chapter 8.6-8.7 in Mendenhall, Beaver, & Beaver
Estimating Differences Between Means and Proportions

22. Often… We don’t just want to estimate one mean or proportion
We want to estimate the difference between two means or proportions

23. Examples
You conduct a randomized experiment on the effectiveness of ASSISTments versus Dreambox. Which one has higher learning gains?
You test out two medicines in a randomized experiment. Which one leads to a higher proportion of patients surviving?
You test the SAT scores of students who take honors classes versus regular classes. Which group has higher SAT scores?

24. Formally
Once assigned and treated differently, these become two populations
Population 1
Population 2

25. Each has their own set of statistics

26. What is the difference between means?

27. What is the difference between means?
We know already that each group’s mean can be treated as a single value, a point estimate
Or as a distribution

28. What is the difference between means?
So it stands to reason that the difference between group means
Can be treated as a single value, a point estimate
Or as a distribution

29. What is the difference between means?
Point estimate of the difference
Is the difference between the two point estimates of the mean
E.g. our best estimate of ( 𝜇 1 − 𝜇 2 ) is ( 𝑥 1 − 𝑥 2 )

30. We can call this value The mean difference between groups
Or the most likely value for the difference between groups

31. Example
Students using ASSISTments gain 30 points pre-post. Students using Dreambox gain 5 points pre-post. What is the mean difference in learning gains?
30-5 = 25 points

32. You try it
You test the SAT scores of students who take honors classes versus regular classes. Honors students average 650. Regular students average 480. What is the mean difference in SAT?

33. You try it
You test the SAT scores of students who take honors classes versus regular classes. Honors students average 650. Regular students average 480. What is the mean difference in SAT?
170 points

34. The Standard Error For the difference between groups
For sufficiently large samples
𝑆𝐸= 𝑠 𝑛 𝑠 𝑛 2

35. Example
Students using ASSISTments gain 30 points pre-post, with a standard deviation of 10 points. Students using Dreambox gain 5 points pre-post, with a standard deviation of 8 points. There are 30 students in each condition. What is the standard error for the mean difference in learning gains?
𝑆𝐸= 𝑠 𝑛 𝑠 𝑛 2 =

36. Example
Students using ASSISTments gain 30 points pre-post, with a standard deviation of 10 points. Students using Dreambox gain 5 points pre-post, with a standard deviation of 8 points. There are 30 students in each condition. What is the standard error for the mean difference in learning gains?
= = 2.34

37. You try it
You test the SAT scores of students who take honors classes versus regular classes. In your school, honors students average 650, with standard deviation 40. Regular students average 480, with standard deviation 20. Your school has 40 honors students and 200 regular students. What is the standard error for the mean difference in SAT?

38. 𝑆𝐸= 𝑠 𝑛 𝑠 𝑛 2 =
You test the SAT scores of students who take honors classes versus regular classes. In your school, honors students average 650, with standard deviation 40. Regular students average 480, with standard deviation 20. Your school has 40 honors students and 200 regular students. What is the standard error for the mean difference in SAT?

39. = =
You test the SAT scores of students who take honors classes versus regular classes. In your school, honors students average 650, with standard deviation 40. Regular students average 480, with standard deviation 20. Your school has 40 honors students and 200 regular students. What is the standard error for the mean difference in SAT?

40. 40+2 = = 6.5
You test the SAT scores of students who take honors classes versus regular classes. In your school, honors students average 650, with standard deviation 40. Regular students average 480, with standard deviation 20. Your school has 40 honors students and 200 regular students. What is the standard error for the mean difference in SAT?

41. By Central Limit Theorem
The sampling distribution of 𝑥 1 − 𝑥 2
Is approximately normal when
Both n1 and n2 are > 30
𝑆𝐸= 𝑠 𝑛 𝑠 𝑛 2

42. We’ll talk about cases with smaller data sets
Later in the semester

43. Questions? Comments?

44. Since it is normally distributed
You can compute the 95% Confidence Interval as we discussed in the last class

45. Example
Students using ASSISTments gain 30 points pre-post, with a standard deviation of 10 points. Students using Dreambox gain 5 points pre-post, with a standard deviation of 8 points. There are 30 students in each condition. What is the standard error for the mean difference in learning gains?
𝑥 1 − 𝑥 2 =25, 𝑆𝐸= 2.34

46. Example
Students using ASSISTments gain 30 points pre-post, with a standard deviation of 10 points. Students using Dreambox gain 5 points pre-post, with a standard deviation of 8 points. There are 30 students in each condition. What is the standard error for the mean difference in learning gains?
𝑥 1 − 𝑥 2 =25, 𝑆𝐸= 2.34
95% CI = [25-(1.96)(2.34), 25+(1.96)(2.34)]

47. Example
Students using ASSISTments gain 30 points pre-post, with a standard deviation of 10 points. Students using Dreambox gain 5 points pre-post, with a standard deviation of 8 points. There are 30 students in each condition. What is the standard error for the mean difference in learning gains?
𝑥 1 − 𝑥 2 =25, 𝑆𝐸= 2.34
95% CI = 20.41, 29.59

48. You try it
You test the SAT scores of students who take honors classes versus regular classes. In your school, honors students average 650, with standard deviation 40. Regular students average 480, with standard deviation 20. Your school has 40 honors students and 200 regular students. What is the standard error for the mean difference in SAT?
𝑥 1 − 𝑥 2 =170, 𝑆𝐸=6.5

49. Estimating the Difference Between Two Proportions
Coming from a binomial distribution

50. What is the difference between means?

51. What is the difference between means?
We know already that each group’s mean proportion can be treated as a single value, a point estimate
Or as a distribution

52. What is the difference between means?
So it stands to reason that the difference between group mean proportions
Can be treated as a single value, a point estimate
Or as a distribution

53. What is the difference between means?
Point estimate of the difference
Is the difference between the two point estimates of the mean proportions
E.g. our best estimate of ( 𝑝 1 − 𝑝 2 ) is the difference between the sample proportions ( 𝑝 1 − 𝑝 2 )

54. Example
You test out two medicines in a randomized experiment. Fworplomycin leads to 80% of patients surviving, while Penicillin leads to 30% of patients surviving. What is the mean difference in proportion of survival?
80%-30%=50%

55. Example
You compare students in two sections of HUDM4122. If they later take HUDM5122, 85% of students in Section A pass HUDM5122, while only 80% of students in Section B pass HUDM5122. What is the mean difference in pass rates?
85%-80%=5%

56. The Standard Error For the difference between proportions
For sufficiently large samples
𝑆𝐸= 𝑝 1 𝑞 1 𝑛 𝑝 2 𝑞 2 𝑛 2

57. By Central Limit Theoreom
The sampling distribution of ( 𝑝 1 − 𝑝 2 )
Is approximately normal when samples are sufficiently large
𝑛 1 𝑝 1 > 5 AND 𝑛 1 𝑞 1 > 5 AND 𝑛 2 𝑝 2 > 5 AND 𝑛 2 𝑞 2 > 5
𝑆𝐸= 𝑝 1 𝑞 1 𝑛 𝑝 2 𝑞 2 𝑛 2

58. Example
You test out two medicines in a randomized experiment, with 50 subjects in each group. Fworplomycin leads to 80% of patients surviving, while Penicillin leads to 30% of patients surviving. What is the standard error of the mean difference between proportions?

59. 𝑆𝐸= 𝑝 1 𝑞 1 𝑛 1 + 𝑝 2 𝑞 2 𝑛 2 = (0.8)(0.2) 50 + (0.3)(0.7) 50
𝑆𝐸= 𝑝 1 𝑞 1 𝑛 𝑝 2 𝑞 2 𝑛 2 = (0.8)(0.2) (0.3)(0.7) 50
You test out two medicines in a randomized experiment , with 50 subjects in each group. Fworplomycin leads to 80% of patients surviving, while Penicillin leads to 30% of patients surviving. What is the standard error of the mean difference between proportions?

60. (0.8)(0.2) (0.3)(0.7) 50 = 0.086
You test out two medicines in a randomized experiment , with 50 subjects in each group. Fworplomycin leads to 80% of patients surviving, while Penicillin leads to 30% of patients surviving. What is the standard error of the mean difference between proportions?

61. Example
You test out two medicines in a randomized experiment, with 50 subjects in each group. Fworplomycin leads to 80% of patients surviving, while Penicillin leads to 30% of patients surviving. What is the 95% Confidence Interval of the mean difference between proportions?

62. Example
You test out two medicines in a randomized experiment, with 50 subjects in each group. Fworplomycin leads to 80% of patients surviving, while Penicillin leads to 30% of patients surviving. What is the 95% Confidence Interval of the mean difference between proportions?
( 𝑝 1 − 𝑝 2 )= 0.5, SE = 0.086

63. 95% CI = [0.5-(1.96)(0.086), 0.5+(1.96)(0.086)]
You test out two medicines in a randomized experiment , with 50 subjects in each group. Fworplomycin leads to 80% of patients surviving, while Penicillin leads to 30% of patients surviving. What is the 95% Confidence Interval of the mean difference between proportions?
( 𝑝 1 − 𝑝 2 )= 0.5, SE = 0.086

64. 95% CI = [0.33,0.67]
You test out two medicines in a randomized experiment , with 50 subjects in each group. Fworplomycin leads to 80% of patients surviving, while Penicillin leads to 30% of patients surviving. What is the 95% Confidence Interval of the mean difference between proportions?
( 𝑝 1 − 𝑝 2 )= 0.5, SE = 0.086

65. You try it
You compare students in two sections of HUDM4122. If they later take HUDM5122, 85% of students in Section A (40 students) pass HUDM5122, while only 80% of students in Section B (30 students) pass HUDM5122. What is the standard error of the mean difference in pass rates, and what is the 95% Confidence Interval?

66. Note that
Sample sizes can be different between groups

67. Comments? Questions?

68. Final questions or comments for the day?

69. Upcoming Classes 4/6 Z tests 4/8 No class 4/13 Types of Errors HW7 due