HUDM4122 Probability and Statistical Inference April 1, 2015

Questions from last class?

Continuing from last class…

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

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

So, for a 95% Confidence Interval a is 0.05, Cumulative Normal Probability is from 0.025 to 0.975 Meaning that Confidence Intervals bounds are -1.96 SE and +1.96 SE These values are called − 𝒁 ∝/𝟐 and + 𝒁 ∝/𝟐 − 𝒁 ∝/𝟐 =−𝟏.𝟗𝟔 + 𝒁 ∝/𝟐 = + 1.96

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

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

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

Questions? Comments?

For a 99% Confidence Interval a is 0.01, Cumulative Normal Probability is from 0.005 to 0.995 Look in your table to get 0.005 and 0.995 That’s -2.57 and +2.57 Meaning that Confidence Intervals bounds are -2.57 SE and +2.57 SE − 𝑍 ∝/2 =−2.57 + 𝑍 ∝/2 = + 2.57

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

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 -1.64 and +1.64 Meaning that Confidence Intervals bounds are -1.64 SE and +1.64 SE

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 -1.64 and +1.64 − 𝑍 ∝/2 =−1.64 + 𝑍 ∝/2 = + 1.64

Comments? Questions?

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

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

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

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

On to today’s class…

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

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

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?

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

Each has their own set of statistics

What is the difference between means?

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

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

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 )

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

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

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?

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

The Standard Error For the difference between groups For sufficiently large samples 𝑆𝐸= 𝑠 1 2 𝑛 1 + 𝑠 2 2 𝑛 2

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 1 2 𝑛 1 + 𝑠 2 2 𝑛 2 = 10 2 30 + 8 2 30

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? 10 2 30 + 8 2 30 = 100 30 + 64 30 = 2.34

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?

𝑆𝐸= 𝑠 1 2 𝑛 1 + 𝑠 2 2 𝑛 2 = 40 2 40 + 20 2 200 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 2 40 + 20 2 200 = 1600 40 + 400 200 = 40+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?

40+2 = 42 = 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?

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

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

Questions? Comments?

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

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

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)]

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

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

Estimating the Difference Between Two Proportions Coming from a binomial distribution

What is the difference between means?

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

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

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 )

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%

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%

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

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 1 + 𝑝 2 𝑞 2 𝑛 2

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?

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 1 + 𝑝 2 𝑞 2 𝑛 2 = (0.8)(0.2) 50 + (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?

(0.8)(0.2) 50 + (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?

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?

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

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

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

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?

Note that Sample sizes can be different between groups

Comments? Questions?

Final questions or comments for the day?

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