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Testing means, part III
The two-sample t-test
One-sample t-test
Null hypothesis
The population mean
is equal to o
Sample
Test statistic
Y  o
t
s/ n
compare
Null distribution
t with n-1 df
How unusual is this test statistic?

P < 0.05
Reject Ho
P > 0.05
Fail to reject Ho

Paired t-test
Null hypothesis
The mean difference
is equal to o
Sample
Test statistic
d  do
t
SE d
compare
Null distribution
t with n-1 df
*n is the number of pairs
How unusual is this test statistic?
P < 0.05
Reject Ho
P > 0.05
Fail to reject Ho
Comparing means
• Tests with one categorical and one
numerical variable
• Goal: to compare the mean of a numerical
variable for different groups.
4
Paired vs. 2 sample comparisons
5
2 Sample Design
• Each of the two samples is a random sample
from its population
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2 Sample Design
• Each of the two samples is a random sample
from its population
• The data cannot be paired
7
2 Sample Design - assumptions
• Each of the two samples is a random sample
• In each population, the numerical variable
being studied is normally distributed
• The standard deviation of the numerical
variable in the first population is equal to
the standard deviation in the second
population
8
Estimation:
Difference between two means
Y1 Y2
Normal distribution
Standard deviation s1=s2=s

Since both Y1 and Y2 are normally distributed,
their difference will also follow a normal
distribution
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Estimation:
Difference between two means
Y1 Y2
Confidence interval:

Y1  Y2  SEY Y
1
t
 2,df
2
10

Standard error of difference in
means


1
1
2
SEY1 Y2  sp   
n1 n 2 
s2p
= pooled sample variance
n1 = size of sample 1
n 2 = size of sample 2

11

Standard error of difference in
means


1
1
2
SEY1 Y2  sp   
n1 n 2 
Pooled variance:
df s  df s
s 
df1  df2
2
p
2
11
2
2 2
12
Standard error of difference in
means
Pooled variance:
df s  df s
s 
df1  df2
2
p
2
11
2
2 2
df1 = degrees of freedom for sample 1 = n1 -1
df2 = degrees of freedom for sample 2 = n2-1
s12 = sample variance of sample 1
s22 = sample variance of sample 2

13
Estimation:
Difference between two means
Y1 Y2
Confidence interval:

Y1  Y2  SEY Y
1
t
 2,df
2
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Estimation:
Difference between two means
Y1 Y2
Confidence interval:

Y1  Y2  SEY Y
1
t
 2,df
2
df = df1 + df2 = n1+n2-2
15
Costs of resistance to disease
2 genotypes of lettuce: Susceptible and Resistant
Do these differ in fitness in the absence of disease?
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
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Data, summarized
Susceptible
Resistant
Mean number
of buds
720
582
SD of number
of buds
223.6
277.3
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16
Sample size
Both distributions are approximately normal.
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Calculating the standard error
df1 =15 -1=14;
df2 = 16-1=15
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Calculating the standard error
df1 =15 -1=14;
df2 = 16-1=15
df s  df s 14 223.6  15277.3
s 

 63909.9
df1  df 2
14  15
2
p
2
1 1
2
2 2
2
2
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Calculating the standard error
df1 =15 -1=14;
df2 = 16-1=15
df s  df s 14 223.6  15277.3
s 

 63909.9
df1  df 2
14  15
2
p
2
1 1
SE x1 x2
2
2 2
2
2


 1 1 
1
1
2
 sp     63909.9    90.86
15 16 
n1 n 2 
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Finding t
df = df1 + df2= n1+n2-2
= 15+16-2
=29
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Finding t
df = df1 + df2= n1+n2-2
= 15+16-2
=29
t0.052,29  2.05
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The 95% confidence interval of
the difference in the means
Y1  Y2  sY Y
1
t

720

582

90.86
2.05





2
,df


2
 138  186
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Testing hypotheses about the
difference in two means
2-sample t-test
The two sa mple t-tes t compare s the mea ns o f a nu me rica l
var iabl e b etween t wo p opula tions .
24
2-sample t-test
Test statistic:
Y1  Y2
t
SE Y Y
1
SEY1 Y2


1 
2 1
 sp   
n1 n 2 
2
2
2
df
s

df
s
2 2
sp2  1 1
df1  df2
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Hypotheses
H0: There is no difference between the number of
buds in the susceptible and resistant plants.
(1 = 2)
HA: The resistant and the susceptible plants differ in
their mean number of buds. (1  2)
26
Null distribution
t2,df
df = df1 + df2 = n1+n2-2

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Calculating t
x1  x 2  720  582

t

1.52
SE x1 x2
90.86

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Drawing conclusions...
Critical value:
t0.05(2),29=2.05
t <2.05, so we cannot reject the null hypothesis.
These data are not sufficient to say that there is a
cost of resistance.
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Assumptions of two-sample t tests
• Both samples are random samples.
• Both populations have normal distributions
• The variance of both populations is equal.
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Two-sample t-test
Null hypothesis
The two populations
have the same mean
Sample
12
Test statistic
Y1  Y2
t
SE Y Y
1
compare
Null distribution
t with n1+n2-2 df
2
How unusual is this test statistic?

P < 0.05
Reject Ho
P > 0.05
Fail to reject Ho
Quick reference summary:
Two-sample t-test
• What is it for? Tests whether two groups have the same
mean
• What does it assume? Both samples are random samples.
The numerical variable is normally distributed within both
populations. The variance of the distribution is the same in
the two populations
• Test statistic: t
• Distribution under Ho: t-distribution with n1+n2-2 degrees of
freedom.

1 
2 1
SEY Y  sp   
Y

Y
1
2
n1 n 2 
• Formulae:
t
1
SE Y Y
1
df1s12  df2 s22
s 
df1  df2
2
p
2

2
Comparing means when
variances are not equal
Welch’s t test
We lch 's ap p rox imate t-tes t com pare s the mea ns o f two
no rma lly distrib uted po pulati ons tha t ha ve un eq ua l
var ian ce s.
33
Burrowing owls and dung traps
QuickTime™ and a
TIFF (LZW) decompressor
are needed to see this picture.
34
Dung beetles
35
Experimental design
• 20 randomly chosen burrowing owl nests
• Randomly divided into two groups of 10
nests
• One group was given extra dung; the other
not
• Measured the number of dung beetles on the
owls’ diets
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Number of beetles caught
• Dung added:
Y  4.8
s  3.26
• No dung added:
Y  0.51

s  0.89
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Hypotheses
H0: Owls catch the same number of dung
beetles with or without extra dung (1 = 2)
HA: Owls do not catch the same number of
dung beetles with or without extra dung
(1  2)
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
Welch’s t
t
Y1  Y2
2
1
df 
2
2
s s

n1 n2
s12 s22 2
  
n1 n2 
 s2 n 2 s2 n 2 
 1 1   2 2  
 n1  1
n2  1 


Round down df to
nearest integer

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Owls and dung beetles
t
Y1  Y2
2
1
2
2
s
s

n1 n2

4.8  0.51
2
3.26 0.89

10
10
2
 4.01

40
Degrees of freedom
s12 s22 
  
n1 n 2 
2
df 
 s2 n

2
s
n
 1 1    2 2  
 n1 1
n 2 1 


2
2

3.26 2 0.89 2 2



10 
 10
 3.26 2 10

2
0.89
10
 
 

 10 1
10 1 


2
2
 10.33
Which we round down to df= 10
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Reaching a conclusion
t0.05(2), 10= 2.23
t=4.01 > 2.23
So we can reject the null hypothesis with
P<0.05.
Extra dung near burrowing owl nests
increases the number of dung beetles eaten.
42
Quick reference summary:
Welch’s approximate t-test
• What is it for? Testing the difference between means of two
groups when the standard deviations are unequal
• What does it assume? Both samples are random samples.
The numerical variable is normally distributed within both
populations
• Test statistic: t
• Distribution under Ho: t-distribution with adjusted degrees
of freedom
Y1  Y2
s s 
• Formulae:
  
t
n n 
2
1
2
2
s s

n1 n2
df 
2
1
2
2
1
2
2
 s2 n 2 s2 n 2 
 1 1   2 2  
 n1  1
n2  1 


The wrong way to make a
comparison of two groups
“Group 1 is significantly different from a
constant, but Group 2 is not. Therefore Group 1
and Group 2 are different from each other.”
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A more extreme case...
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