Thus far…
Inferences When Comparing
Two Means
Dr. Tom Ilvento
STAT 200
„
„
Testing differences between two
means or proportions
„
„
„
The same strategy will apply for testing differences
between two means or proportions
These will be two independent, random samples
We will have
„ Estimate (of the difference)
„ Hypothesized value
„ Standard error
„ Knowledge of a sampling distribution
We have made an inference from a single
sample mean and proportion to a population,
using
„ The sample mean (or proportion)
„ The sample standard deviation
„ Knowledge of the sampling distribution for
the mean (proportion)
And it matters if sigma is known or unknown
Testing differences between two
means or proportions
„
„
„
We can make a point estimate and a hypothesis of
the difference of the two means
Or a Confidence Interval around the difference of
the two means
With a few twists
„
Mean
„
„
„
Proportions
„
Testing differences between two
means or proportions
„
„
„
We will also need to come up with:
„ An estimator of the difference of two
means/proportions
„ The standard error of the sampling
distribution for our estimator
With two sample problems we have two
sources of variability and sampling error
We also must assume the samples are
independent random samples
Sigma known or unknown
Should we pool the variance
When testing Ho: we need to check if p1
= p2
What are independent, random
samples?
„
„
Independent samples means that each sample and
the resulting variables do not influence the other
sample
„ If we sampled the same subjects at two
different times we would not have independent
samples
„ If we sampled husband and wife, they would
not be independent
However, we have a strategy to assess change
over time of the same subject – paired difference
test
1
Decision Tree for two
Proportions
Decision Tree for Two Means
Target
H0:
µ1-µ2=D
Assumptions
Independent random samples
Sigma Known
Test Statistic
Testing
z, using
pop. variances
H0:p1- p2 = 0 Independent random samples
Large sample size (n1, n2>30)
Known that p1 = p2 under H0
z, using
pooled sample
proportion Pp
H0:p1- p2 = D0 Independent random samples
Large sample size
When D0 ≠ 0
z
Independent random samples
Sigma Unknown
t, using sample
Assumptions
Test Statistic
variances
If we can assume
Equal variances
use pooled
variance Sp2
Example Problem: Manufacturer of
Roof Shingles
„
„
„
„
„
„
Descriptive Statistics
Studies have shown the weight is an important
customer perception of quality, as well as a company
cost consideration.
The last stage of the assembly line packages the
shingles before placement on wooden pallets.
The data below are the weight (in pounds) of pallets
of Boston and Vermont variety of shingles.
Boston
Mean
Standard Error
Median
Mode
Standard Deviation
Sample Variance
Kurtosis
Skewness
Range
Minimum
Maximum
Sum
Count
Confidence Level(95.0%)
Open up PALLET.xls
Run Descriptives
Produce a Box and Whisker for both groups
Vermont
3124.21
3704.04
1.81
2.57
3122.00
3704.00
3098.00
3728.00
34.71
46.74
1204.99
2185.03
0.77
0.21
0.53
0.29
222.00
290.00
3044.00
3566.00
3266.00
3856.00
1149711.00 1222334.00
368
330
3.558
5.062
Lecture Notes problem
„
B o x- and - w hi sker Pl o t o f Pal l et W eig ht i n p o und s
„
„
„
V e r mont
Bos t on
3040
3240
3440
3640
3840
4040
4240
Class with Lecture
Notes
n1= 86
mean1 = 8.48
s12= .94
„
„
„
„
Class without Lecture
Notes
n2= 35
mean2 = 7.80
s22= 2.99
Do the samples provide sufficient evidence to
conclude that there is a difference in mean
responses of the two groups?
Use α = .01
2
We need to figure out the sampling
distribution (mean1-mean2)
Roof Shingles problem
Null hypothesis H0: (µ1-µ2) = ?
Alternative
Assumptions
Test Statistic
Rejection Region
Calculation
„
Ha: (µ1-µ2) … ? two-tailed test
Two independent samples, n is Large
„
t* =
„
tα=.05/2 = ?
t* =
t* ? tα=.025
Conclusion ?
The mean of the sampling distribution for
(mean1-mean2)
Will equal = (µ1-µ2)
= D0
„ We usually designate the expected
difference as D0 under the the null
hypothesis
„ Most often we think of D0 = 0; no difference
?? H0: (µ1-µ2) = ?
Standard Error of the difference
of two means
„
The Standard Error of the sampling
distribution difference of two means is given
as:
2
2
1
2
( x1 − x2 )
1
2
σ
=
σ
n
+
σ
n
The sampling distribution of (mean1-mean2) is
approximately normal for large samples under the
Central Limit Theorem
The Standard Error for the
difference of two means
„
„
Is based on two independent random
samples
We typically use the sample estimates of σ1
and σ2
„ Which are s1 and s2
σ (x −x ) =
1
2
s12 s22
+
n1 n2
And then use the t-distribution
The Test Statistic for our
problem
t* =
(3124.2 − 3704.0) − 0
1204.99 2185.03
+
368
330
Comparison of two means
Null hypothesis
Alternative
Assumptions
Two independent samples, sigma is unknown
Test Statistic
t* = (3124.2-3704.0- 0)/[(1204.99/368)+(2185.03/330)].5
Rejection Region
Calculation
t* = (-579.8 – 0)/3.15 = -184.31
H0: (µ1-µ2) = 0
Ha: (µ1-µ2) … 0 two-tailed test
Conclusion
tα=.05/2, 603 d.f. = -1.964 or 1.964
t* = -184
t* < tα=.05/2, 603 d.f.
-184 < -1.964
Reject H0: (µ1-µ2) = 0
3
What is the p-value for our
statistic?
„
„
„
„
„
t* = -184
IT IS HUGE!!!!
Do this with EXCEL Data Analysis
„
This is smaller than α = .05/2
P < .001
Therefore, we reject
„ H0: (µ1-µ2) = 0
95% Confidence Interval for the
Difference of Two Means
Excel output
t-Test: Two-Sample Assuming Unequal Variances
Mean
Variance
Observations
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
„
Boston Vermont
3124.215 3704.042
1204.992 2185.032
368
330
0
603
-184.321
0
1.647385
0
1.963906
„
„
„
„
There may be times when we think the
difference between the two samples is
primarily the means
But the variances are similar
In this case we ought to use information from
both samples to estimate sigmas
We will use a t-test and the t distribution
Assumptions
„ The population variances are equal
„ Random samples selected independently
of each other
( x1 − x2 ) ± zα / 2σ ( x1 − x2 )
( x1 − x2 ) ± tα / 2 s( x1 − x2 )
What about if the variances are equal
to each other?
„
Tools
„ Data Analysis
„ t-test: two sample assuming unequal
variances
„ Pick both variables
„ Note labels or not
„ Tell Excel where to put the output
„ Dress it up
The standard error for a small sample
difference of means
„
„
„
„
Since we assume σ1 = σ2
thus (s1 = s2)
We should pool our estimate of the standard
error of the sampling distribution
Using information from both sample estimates
4
Step 2: Use the Pooled Estimate of the
Variance to calculate the estimate of the
Standard Error
POOLED ESTIMATE OF THE VARIANCE
„
Step 1: Then our formula will be a weighted
average of s1 and s2
σˆ ( x − x ) =
1
2
(n − 1) s12 + (n2 − 1) s22
s = 1
(n1 − 1) + (n2 − 1)
2
p
s 2p
n1
+
⎛1 1⎞
= s 2p ⎜⎜ + ⎟⎟
n2
⎝ n1 n2 ⎠
s 2p
alternatively
σˆ ( x − x ) = s p
Note: the denominator reduces to (n1 +n2 –2)
which is the d.f. for the t distribution
1
2
1 1
+
n1 n2
Do this with Excel – two sample
assuming equal variances
What does pooling do for us?
• Pooling generates a weighted average as the
estimate of the variance
t-Test: Two-Sample Assuming Equal Variances
• The weights are the sample sizes for each
sample
Mean
Variance
Observations
Pooled Variance
Hypothesized Mean Difference
df
t Stat
P(T<=t) one-tail
t Critical one-tail
P(T<=t) two-tail
t Critical two-tail
• A pooled estimate is thought to be a better
estimate if we can assume the variances are
equal
• And our degrees of freedom are larger d.f.= n1 + n2 – 2
• Which means the t-value will be smaller
What about the difference in
proportions?
„
„
Based on large sample only
Same strategy as for the mean
„ We calculate the difference in the two
sample proportions
„ Establish the sampling distribution for our
estimator
„ Calculate a standard error of this sampling
distribution
„ Conduct a test
Boston
Vermont
3124.215 3704.042
1204.992 2185.032
368
330
1668.258
0
696
-187.2495
0
1.647046
0
1.963378
Decision Tree for two
Proportions
Testing
Assumptions
Test Statistic
H0:p1- p2 = 0 Independent random samples
Large sample size (n1, n2>30)
Known that p1 = p2 under H0
z, using
pooled sample
proportion Pp
H0:p1- p2 = D0 Independent random samples
Large sample size
When D0 ≠ 0
z
5
Differences of Proportions
„
„
For the null hypothesis
„ H : (p -p ) = D
0
1 2
0
For Proportions
„
The Standard Error for ( ps1 - ps2)
s( p s 1 − p s 2 ) =
ps1qs1 ps 2 qs 2
+
n1
n2
100(1 - α )% C.I. = ( ps1 − ps 2 ) ± zα / 2 s( ps1 − ps 2 )
For Proportions
„
„
So for Proportions
Special Note: Under
the Null Hypothesis
where p1 – p2 =0
We use a pooled
average, pp, based on
adding the total number
of successes and divide
by the sum of the two
sample sizes
„
„
p=
( x1 + x2 )
(n1 + n2 )
For a confidence interval, use the sample
estimates in this manner to generate the
standard error – since there is no
assumption that the variances are equal
s( ps1 − ps 2 ) =
p s1 qs1 ps 2 qs 2
+
n1
n2
For a Hypothesis Test where we specify that
the proportions are equal, (p1 – p2 = 0) we
should pool the information and use this
approach
s( p s 1 − p s 2 ) =
where x = # of
successes for each
group
⎛1 1⎞
pq ⎜⎜ + ⎟⎟
⎝ n1 n2 ⎠
Gender Differences in Shopping for
Clothing
So for Proportions
„
Since this is a large sample problem, we
could use the sample estimates of ps1 and ps2
to estimate the standard error for a Confidence
Interval
„
„
„
„
A survey was taken of 500 people to determine
various information concerning consumer
behavior. One question asked, “Do you enjoy
shopping for clothing?” The data are shown
below by sex.
For women 224 out of 260 indicated Yes
For men, 136 out of 240 indicated yes
Conduct a test that a higher proportion of
women favor shopping using α=.01
Use PHSTAT – two sample Test, z-test
for difference of two proportions
6
Use PHStat, z-test for difference in
two proportions
Z Test for Differences in Two Proportions
Data
Hypothesized Difference
Level of Significance
Group 1
Number of Successes
Sample Size
Group 2
Number of Successes
Sample Size
Intermediate Calculations
Group 1 Proportion
Group 2 Proportion
Difference in Two Proportions
Average Proportion
Z Test Statistic
Upper-Tail Test
Upper Critical Value
p -Value
Reject the null hypothesis
0
0.01
224
260
136
240
0.862
0.567
0.295
0.7200
7.34
2.326
0.000
I conducted a test to determine if women
indicated they enjoyed shopping for clothing
more than men did.
Because the Null Hypothesis was that the
two groups are equal, I used a pooled
estimate of the proportion who enjoyed
shopping for clothing.
Paired Difference Test
„
„
„
For women, the proportion was .862 and for
men it was .567
The results of my test indicate a highly
significant difference between women and
men, p < .001. I have evidence to suggest
that women enjoy shopping for clothing
more than men.
Paired Difference Test
„
„
„
„
Paired Difference Test
The strategy is relatively simple
We simply create a new variable which is the
difference of the pre-test from the post test
This new variable can be thought as a single
random sample
For this new variable we calculate sample
estimates of the mean and standard deviation
„
And then calculate C.I. or conduct a hypothesis test
on this new variable
Often times the mean difference is referred to as
„
And the hypothesis is often:
„
„
„
„
Example of paired difference
data
Patient
When you have a situation where we record a
pre and post test for the same individual
We cannot treat the samples as independent
In these cases we can do a Paired
Difference Test.
„ It’s called “Matched Pairs Test” in some
books.
„ Or Mean Difference in your book
Difference
2-1
Time 1
Time 2
1
5
5
0
2
1
3
-2
3
0
0
0
4
1
1
0
5
0
1
-1
6
2
1
1
meanD
H 0 : µD = 0
This is no different than any single mean test, large
sample or small sample
Alzheimer’s study: Paired Difference
Test
„
„
„
Twenty Alzheimer patients were asked to spell 24
homophone pairs given in random order
„ Homophones are words that have the same
pronunciation as another word with a different
meaning and different spelling
„ Examples: nun and none; doe and dough
The number of confusions were recorded
The test was repeated one year later
7
Alzheimer’s study: Paired Difference
Test
„
The researchers posed the following question:
Do Alzheimer’s patients show a significant
increase in mean homophone confusion
over time?
„
Use an alpha value of .05
Alzheimer’s study: Paired Difference
Test
Statistics
Mean
Standard Error
Median
Mode
Standard Deviation
Sample Variance
Kurtosis
Skewness
Range
Minimum
Maximum
Sum
Count
Time 1
4.15
0.78
5.00
5.00
3.50
12.24
-0.85
0.41
11
0
11
83
20
Time 2 Difference
5.80
0.94
5.50
3.00
4.21
17.75
0.13
0.64
16
0
16
116
20
1.65
0.72
1.00
0.00
3.20
10.24
0.08
0.48
12
-3
9
33
20
Alzheimer’s study: Paired Difference
Test
Null hypothesis
Alternative
Assumptions
Test Statistic
Rejection Region
H 0 : µD = 0
Ha: µD > 0 one-tailed test, upper
small sample, normal
t* = (1.65 – 0)/(3.2/20.5)
tα=.05, 19 d.f. = 1.729
Calculation
t* = 2.31
Conclusion
t* > tα=.05, 19 d.f.
2.31 > 1.729
Reject H0: µD = 0
8