+
Proportions
Estimating population proportions
Difference of proportions
+
Review: Setting up a c.i. around a
proportion
1. estimate the proportion
2. Take the SD with this formula:
s= sqrt(p * (1-p))
3. Find the s.e. with this formula:
s / sqrt(n)
4. Set up the confidence interval with this
formula:
proportion plus or minus t * s.e.
+
Example
A teacher offers extra tutoring in math. He
takes a sample of 100 students who went
through his tutoring sessions, and found that
73 started to get higher marks. He wants to
figure out the 95% confidence limits in his
survey before he goes to the principal and
tells her the program was successful.
+
Step 1-2
Step 1: estimate the population proportion
=0.73 start to perform better in math
Step 2: get the sample standard deviation using
this formula:
s= sqrt(p * (1-p))
=sqrt(0.73 * 0.27)
=0.444
+
Step 3
Step 3: Use this in order to find the standard
error:
= s / sqrt(n)
=0.44/ sqrt(100) = 0.044
+
Build confidence interval
Step 4: What are the 95% confidence limits of the
proportion? Since n is bigger than 30, the normal
curve can be used. Set up a confidence interval
using this formula: proportion plus or minus t *
s.e.
=0.73 + or - 1.96 * 0.044
=0.73 + or - 0.087
=0.63 to 0.81
+
Testing the difference
between 2 groups
Chapter 14
+
Difference of means test
Used if someone wants to know if two sample
means or proportions are different (statistically)
-could both sample means have been drawn
from the same population (and the
difference is attributed to chance alone)
-or are they so different that there is no way they
could have been drawn from the same
population
+
Vs. what we’ve already done


-So far, we have used single samples, meaning we wanted to
see if a single sample of a particular mean could be drawn
from a population with a known or hypothesized mean
now we use 2 samples
+
A sample problem to walk through
the logic of these types of
problems
A veterans support agency
offers continuing education
seminars for veterans. They
want to evaluate the effect it has
on job placement. The agency
has half the veterans take the
seminars, and the other half
does not. They randomly select
50 who have done the seminars
and 50 who have not. They want
to evaluate whether the job
placement rates are different.
+
Formulate a research and a null
hypothesis
The research hypothesis: tests whether one of the
sample means is larger or smaller than the other
sample mean (difference of means)
The null hypothesis: when you fail to reject it, you’re
saying that the population means in questions are not
different (i.e. an after school reading program didn’t
raise mean scores)
+
For this example:
H_A: Employees with the seminar will have
higher job placement rates
H_0: Employees who attend the seminar and
those who do not attend the seminar will show
no difference in job placement rates
+
What does it mean to reject the
null?
-If we fail to reject the null, this is like saying the mean
scores of the two populations show no difference
-i.e. the population mean is the same and the
seminars don’t lead to higher job placement
-If we reject the null, the conclusion there is that the
test scores were different and job placement rates are
in fact different
+
Practice Problem

The career development center wants to see if its new ad
campaign is working, so they can decide whether or not to
fire their intern and use the money elsewhere. They
randomly sample 9 MIT courses and send them weekly
reminders about the career center. They randomly sample
another 9 courses and send no emails. They then record the
mean student visits per sample.

The average visits in courses with no ads is 135 (SD=110)
per term

The average visits in courses with ads is 405 (SD=135) per
term
+
Thinking through the problem


The first step is to state the null and alternative hypothesis:

Even if you aren’t asked for them, it helps you think

Ho: the ads have had no effect on visitations to the career center

Ha: the ads have increased visitations to the career center
What you’re conceptualizing here is the difference between
the average visitations:

Mu(experiment) – Mu(control) = d (difference)
+
Perform calculations (that we
already know how to do)


The mean and SD were already given:

The average visits in courses with no ads is 135 (SD=110) per
term

The average visits in courses with ads is 405 (SD=135) per term
Get the Standard error: SD / sqrt(n)

Pre advertisements: 110 / sqrt (9) = 36.667

Post advertisements: 135 / sqrt (9) = 45
+
New step for difference of means

Calculate the pooled standard error with the following
formula:

=58.04
+
New step 2 for difference of means

Get the t score using the pooled standard error s.e._d.

Why? Because the point of the difference of means test is to see
the probability that the groups could have been drawn by the
same population and the difference is just by chance

The formula for the t score is:

(135-405)/58.04= -4.65
+
Look this t score up in the t table

Find the degrees of freedom (n1+n2-2)=18-2=16

P=.0001

Reject null
+
Can also use a calculator

http://stattrek.com/online-calculator/t-distribution.aspx
+
Types of difference tests
+
Before we start

We see the word “variance” a lot in this chapter

The variance is just the standard deviation squared
+
General types of difference of
means tests


Independent samples: you have two samples that are not
paired or matched in any way

These samples were obtained using random sampling methods

Example: someone at the IRS picks 2 samples from a database of
250 tax returns

These samples could have equal or unequal variances (more on
that later)
Dependent Samples: “before and after test” where each item
in one sample is paired with an item in the second sample

Example: an agency selects 20 people with low performance
scores and has them do a workshop for a month, the same 20
employees are then tested again after the workshop to see if they
improve
+

Difference of means: Independent
samples, unequal variances
If you don’t know what type of difference
test you’re doing assume it is this one

This is the most conservative test and the
one you see the most in real life studies

Conservative means it is hard to reject the
null hypothesis

Why? The standard error calculations take
large differences in sample variances (s^2)
into account

Sampling error is to blame for unequal
variances
+
Calculating the degrees of
freedom

Use this formula (it produces smaller df which makes the test
more conservative versus the n1+n2-2 formula:

Remember, the lower the df, the bigger the test statistic
needs to be when deciding to reject the null or not
+
Calculating degrees of freedom

The formula is useful when the number of cases in each
sample is different



Or if the number of cases in each sample is small (less than 30)
Example: if sample one has 150 cases and sample 2 only has 20
 The variances will be different
More conservative tests make it harder to commit a type I
error
+ Practice Problem: difference of
means, independent samples,
unequal variances

The president of MIT wants to know if a new technology
program for professors has made them use interactive visual
aides more in the classroom. He randomly selects 10 courses
where the professors in them received the training, and 8
courses where they have not yet taken the course.
Mean use of
visual aides per
term
s
No course
32.7
6.4
Tech course
37.6
6.3
+
State the null and alternative
hypotheses

Ho: the use of visual aides with course = use without course

Ha: the use of visual aides with course > use without course
+
Calculate the standard error
Mean use of
visual aides per
term
s
No course
32.7
6.4
Tech course
37.6
6.3
• Use s/sqrt(n)
• No course: 6.4 / sqrt 8= 2.262
• Course: 6.3 / sqrt 10=1.99
+
Get the pooled standard error

Use this formula:

=3.015
+
Get the t score


Use this formula:
t= (32.7-37.6) / 3.015=-1.625
+
Calculate df

Use this formula:

Numerator: [(6.4 ^2 / 8) + (6.3 ^2 /10)] ^2 =82

Denominator: 3.74 + 1.75

82/(3.74+1.75) = 14.93

Round to 15

(compare to 10+8-2)=16
+
Look up in t table

df=15

T=-1.625

=0.2012

~20% chance these samples were taken from the same
population

Between .10 and .05
+
Visualized
+
Difference of means, independent
samples, equal variances

Less conservative than the test for unequal variances


You determine if two sample variances are equal by using
the Levene test



Because the former makes for larger standard errors and higher t
scores
We will go over this in our next Stata lab
This is a super common task for students to use Stata for
The Levene gets interpreted as follows:



Null: two sample variances are equal
Research: two sample variances are unequal
The test statistic here is the F statistic
 Example: F=87.4 at significance .00 -> reject the null since the
probability the two variances are equal is quite small
+
Steps to solve these types of
problems

Once you have the mean and s1 and s2, calculate a new
“pooled” standard deviation using this formula:

This is nothing but a weighted average of the two sample
standard deviations
+
Calculate standard error

Convert the standard deviation to standard error using this
formula:

Then get the t statistic using (x bar 1 – x bar 2)/s.e.

Then look it up in the t chart
+

Example
NOAA scientists sampling fish larvae in New England
fisheries have received a bigger budget to buy finer mesh
nets for their sampling trips in the spring and in the fall. They
believe this net will help them to better survey for fish larvae.
Better data means less angry stakeholder assessments. They
randomly select 10 boats with old nets and 10 boats with new
nets and collect the following information. They want to show
that this was money well spent.
Mean number
of larvae
s
Old net
326
64
New Net
526
64
+
State the null and the research
hypothesis

Ho: the new net’s catch yield = the old net’s

Ha: the new net’s catch > old net’s catch
+
Get the pooled standard deviation

Use this formula:

Numerator= 73728


Denominator= 4096
S_d= 64
Mean number
of larvae
s
Old net
326
64
New Net
526
64
+
Get the standard error & t statistic


Use this formula:
=28.622

(326-526)/28.622

= -6.98
Mean number
of larvae
s
Old net
326
64
New Net
526
64
+
Look it up in the chart at 18 df

p<.0005

This means we can reject the null and say with high certainty
that the new nets are working.

Let’s extrapolate these results to make claims on government
spending in general.

Just kidding.
+
Difference of means tests:
Dependent Samples

This is the “before and after” where the befores are paired
with the afters
Example: The IRS implemented a training program to reduce
the time it takes to process an organization’s tax exempt status.
They took the following data from 10 different regional offices
before and after the program:
+
The remaining steps to solve

ARE ALL PERFORMED ON THE D COLUMN

Your results and the statistical inference you make are on the
difference
+
Step 1: get the standard error

Mean=4.07

S=4.56

s.e.= 4.56/ sqrt(10) = 1.44
+
Get the t score

(4.07-0) / 1.44

Use 0 because you are seeing if there is a difference between
the difference you found (d) and no difference at all 0

=2.83

Look this up in the t table at df=9, or use stat calculator:
+
Difference of proportions
• The t test can be used for the
difference of 2 sample proportions in
the same way that it can be used for
differences between sample means
+
Practice Problem

DUSP wants to know if math camp is working for its new MCP
admits. 65% of the incoming class was put through the
program. 80 Students are sampled from the math camp group
65 passed quant. 40 MCPs were sampled from the group that
was exempted from math camp. From those, 29 passed quant.
Does math camp work?

Calculate the proportions:

Math camp: 81.2% pass

No math camp: 72.5% pass
+
Calculate the s

S=sqrt(p*(1-p))

Those who did math camp: sqrt (.81*.19)=.39

Those who did not do math camp: sqrt(.72*.25)=.42
+
Get the s.e.

s / sqrt(n)

Those who did math camp: .39 / sqrt (80) = .044

Those who did not do math camp: .42 / sqrt (40)= .067
+
Get the pooled s.e.

Use this formula:

=sqrt(.0438^2+.067^2)

=0.08
+
Get the t score

Use this formula: (proportion 1 – proportion 2) / pooled s.e.

(0.81-0.72) / 0.08 = 1.125

Look it up in the infinity df in the t table because the sample
sizes are both over 30

~.13

There is a .13 chance that these two samples were drawn
from the same population