Transcript Sampling and estimation 2

Sampling and estimation 2
Tron Anders Moger
27.09.2006
Confidence intervals (rep.)
• Assume that X1 ,..,Xn are a random
sample from a normal distribution
• Recall that x has expected value  and
variance 2/n
• The interval x + 1.96/√n is called a 95%
confidence interval for 
• Means that the interval will contain the
population mean 95% of the time
• Often interpreted as if we are 95% certain that
the population mean lies in this interval
Hypothesis testing (rep.)
• Have a data sample
• Would like to test if there is evidence that a
parameter value calculated from the data is
different from the value in a null hypothesis H0
• If so, means that H0 is rejected in favour of some
alternative H1
• Have to construct a test statistic
• It must:
– Have a higher probability for ”extreme” values under
H1 than under H0
– Have a known distribution under H0 (when simple)
Two important quantities
• P-value = probability of the observed value
or something more extreme
assuming null hypothesis
• Significance level α= the value at which
we reject H0
• If the value of the test statistic is ”too extreme”, then H0 is
rejected
• P-value=0.05: We want the probability that the observed
difference is due to chance to be below 5%, or,
equivalently:
• We want to be 95% sure that we do not reject H0 when it
is true in reality
Note:
• There is an asymmetry between H0 and H1: In
fact, if the data is inconclusive, we end up not
rejecting H0.
• If H0 is true the probability to reject H0 is (say)
5%. That DOES NOT MEAN we are 95% certain
that H0 is true!
• How much evidence we have for choosing H1
over H0 depends entirely on how much more
probable rejection is if H1 is true.
Errors of types I and II
• The above can be seen as a decision rule
for H0 or H1.
• For any such rule we can compute (if both
H0 and H1 are simple hypotheses):
1-α
Power 1 - β
H0 true
Accept H0
P(accept | H0)
Reject H0
P(reject | H0)
H1 true
P(accept | H1)
TYPE II error
TYPE I error
Significance level α
P(reject | H1)
β
Sample size computations
• For a sample from a normal population with
known variance, the size of the conficence
interval for the mean depends only on the
sample size.
• So we can compute the necessary sample size
to match a required accuracy
• Note: If the variance is unknown, it must
somehow be estimated on beforehand to do the
computation
• Works also for population proportion estimation,
giving an inequality for the required sample size
Power computations
• If you reject H0, you know very little about the
evidence for H1 versus H0 unless you study the
power of the test.
• The power is 1 minus the probability of rejecting
H0 given that a hypothesis in H1 is true (1-β).
• Thus it is a function of the possible hypotheses
in H1.
• We would like our tests to have as high power
as possible.
Example 1: Normal distribution with
unknown variance
• Assume X 1 , X 2 ,..., X n ~ N (  ,  2 )
• Then
X 
~ tn 1
s/ n
• Thus P( X  tn1, / 2
s
n
   X  tn1, / 2
s
n
) 
• So a confidence interval for  , with significance 
is given by
( X  tn1, / 2 sn , X  tn1, / 2 sn )
Example 1 (Hypothesis testing)
• Hypotheses: H 0 :   0
H1 :   0
X  0
• Test statistic
~ tn 1 under H0
s/ n
• Reject H0 if X  0  tn1, / 2
X  0  tn1, / 2
s
n
s
n
or if
• Alternatively, the p-value for the test can be
computed (if X  0 ) as the  such that
X  0  tn1, / 2
2
n
Example 1 (cont.)
H1 :   0
• Hypotheses: H 0 :   0
X  0
• Test statistic
~ tn 1 assuming   0
s/ n
• Reject H0 if X  0  tn 1,
s
n
• Alternatively, the p-value for the test can be
computed as the  such that X  0  tn 1,
2
n
Energy intake in kJ
•
SUBJECT
1
2
3
4
5
6
7
8
9
10
11
INTAKE
5260
5470
5640
6180
6390
6515
6805
7515
7515
8230
8770
Recommended energy intake: 7725kJ
Want to test if it applies to the 11 women
H0:  (mean energy intake)=7725
H1: 7725
From Explore in SPSS:
i
p
t
E
i
I
M
N
4
6
9
L
5
I
n
U
3
5
0
M
0
V
5
S
2
M
0
M
0
R
0
I
n
0
S
8
1
K
3
9
Test result:
6753.6  7725
t
 2.821
1142.1 / 11
• This quantity is t-distributed with 10 degrees of freedom
(number of subjects -1)
• Choose significance level α=0.05
• From table 8 p.870 in the book, t11-1,0.05/2=2.262
• If the H0 is true, the interval (-2.262, 2.262) covers 95%
of the distribution
• Reject H0 since the test statistic is outside the interval,
or, equivalently, because
X  7725  2.262 *1142.1 / 11  6946
• Can’t find exact p-value from the table
• Could have had α=0.01 or 0.1, but 0.05 is most common
In SPSS: Analyze - Compare means - One-sample t test
Test variable: intake
Test value: 7725
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6
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Differences between means
• Assume X 1 , X 2 ,..., X n ~ N (  x ,  x2 ) and
Y1 , Y2 ,..., Ym ~ N ( y ,  y2 ), all data independent
• We would like to study the difference xy
• Three different cases:
– Matched pairs
– Unknown but equal population variances
– Unknown and possibly different pop.
variances
Matched pairs
• Common situation: Several measurements
on each individual, or on closely related
objects
• These measurements will not be
independent (why?)
• Generally a problem in statistics, but
simple if you only have two measurements
• The key is to use the difference between
the means, instead of each mean
seperately
Example 2: Matched pairs
• In practice, the basis is that x-y=0
1 n
• Set Di  X i  Yi and D  i 1 Di
n
• We get D  0
sD / n
• Where
~ tn 1
n
1
sD 
( Di  D )

i 1
n 1
• Confidence interval for x-y
D  t n 1, / 2 s D / n
Example 2 (Hypothesis testing)
• Hypotheses:
• Test statistic:
H 0 : x   y
H1 :  x   y
D
~ tn 1
sD / n
”Matched pairs T test”
• Reject H0 if
D
 t n 1, / 2
sD / n
or if
D
 t n 1, / 2
sD / n
Example: Energy intake kJ
SUBJECT PREMENST POSTMENS
1
2
3
4
5
6
7
8
9
10
11
5260.0
5470.0
5640.0
6180.0
6390.0
6515.0
6805.0
7515.0
7515.0
8230.0
8770.0
Number of cases read: 11
3910.0
4220.0
3885.0
5160.0
5645.0
4680.0
5265.0
5975.0
6790.0
6900.0
7335.0
Want to test if
energy intake
is different
before and after
menstruation.
H0: premenst=
postmenst
H1: premenst
postmenst
Number of cases listed: 11
Confidence interval and p-values
for paired t-tests in SPSS
• Analyze - Compare Means Paired-Samples T Test.
• Click on the two variabels you want to test, and
move them to Paired variables
Paired Samples Test
Paired Differences
95% Confidence
Interval of the
Difference
Mean
Pair 1
PREMENST POSTMENST
1320,454
55
Std.
Deviation
Std. Error
Mean
366,74551
110,5779
3
Lower
1074,071
56
Upper
1566,837
53
t
11,941
• Conclusion: Reject H0 on 5% sig. level
df
Sig. (2-tailed)
10
,000
Example 3: Unknown but equal
population variances
• We get
X  Y  (x   y )
s
where
s 2p 
2
p

s
2
p
~ t nx  n y  2
nx n y
(nx  1) sx2  (ny  1) s y2
nx  ny  2
• Confidence interval for  x   y
X  Y  tnx  ny 2, / 2
s 2p
nx

s 2p
ny
Example 3 (Hypothesis testing)
• Hypotheses:
H 0 : x   y
X Y
• Test statistic:
• Reject H0 if
2
p
2
p
s
s

nx n y
~ t nx  n y  2
”T test with equal variances”
X Y
2
p
H1 :  x   y
2
p
s
s

nx n y
 tnx  ny  2, / 2
or if
X Y
2
p
2
p
s
s

nx n y
 tnx  ny  2, / 2
Assumptions
1. Independence: All observations are
independent. Achieved by taking random
samples of individuals; for paired t-test
independence is achieved by using the
difference between measurements
2. Normally distributed data (Check:
histograms, tests for normal distribution, QQ plots)
3. Equal variance or standard deviations in the
groups
• Assumptions can be checked in histograms,
box plots etc. (or tests for normality)
• What if the variances are unequal?
Example 4: Unknown and possibly
unequal population variances
• We get
X  Y  (x   y )
2
y
sx2 s

nx n y
where


s x2
nx

s 2y
ny
~ t

2
2
2
( sx2 / nx ) 2 ( s y / n y )

nx  1
ny  1
• Conf. interval for  x   y
X  Y  t , / 2
2
sx2 s y

nx ny
Example 4 (Hypothesis testing)
• Hypotheses:
• Test statistic
H 0 : x   y
X Y
2
sx2 s y

nx n y
H1 :  x   y
~ t
”T test with unequal variances”
• Reject H0 if
X Y
s y2
s

nx n y
2
x
 t , / 2
or if
X Y
2
x
2
y
s
s

nx n y
 t , / 2
Example 5: The variance of a
normal distribution
• Assume X 1 , X 2 ,..., X n ~ N (  ,  2 )
2
(
n

1)
s
• Then
~  n21
•
•
2
 2

(n  1)s 2
2
Thus P  n1,1 / 2   2  n1, / 2   


Confidence interval for  2
 (n  1) s 2 (n  1) s 2 
, 2
 2

  n 1,1 / 2  n 1, / 2 
Example 5: Comparing variances
for normal distributions
• Assume X 1 , X 2 ,..., X n ~ N (  x ,  x2 ) Y1, Y2 ,..., Ym ~ N ( y ,  y2 )
2
2
• We get sx /  x
~ Fnx 1, ny 1
2
2
sy /  y
• Fnx-1,ny-1 is an F distribution with nx-1 and ny-1
degrees of freedom
• We can use this exactly as before to obtain a
confidence interval for  x2 /  y2 and for testing for
example if  x2   y2
• Note: The assumption of normality is crucial!
ID
1
2
12
13
14
15
GROUP
0
0
....
0
0
1
1
ENERGY
6.13
7.05
10.15
10.88
8.79
9.19
....
21
22
1
1
11.85
12.79
Number of cases read:
22
Example:
Energy expenditure
in two groups,
lean and obese.
Want to test if there
is any difference.
H0: lean= obese
H1: lean obese
In SPSS:
• Analyze - Compare Means Independent-Samples T Test
• Move Energy to “Test-variable”
• Move Group to “Grouping variable”
Click “Define Groups” and write 0 and 1 for
the two groups
S
Output:
t
E
e
N
e
e
G
E
l
2
1
4
o
8
9
0
S
s
u
f
a
V
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r
.
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E
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2
o
F
d
p
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w
f
p
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E
E
2
9
6
0
1
6
6
5
8
a
E
6
9
1
6
8
2
1
n
Above 0.05: Read first line (Equal variances assumed)
Otherwise: Read second line (Equal variances not assumed)
Conclusion
• The observed mean for the lean was
8.1, and for the obese 10.3 (mean
difference -2.2, 95% confidence interval
for the difference (-3.4, -1.1))
• The difference between the groups was
significant on a 5%-level (since the CI
does not include the value 0)
• The p-value was 0.001.
• H0 is rejected
Example 6: Population proportions
• Assume X ~ Bin (n, P ) , so that Pˆ 
frequency.
Pˆ  P
• Then
• Thus
X
is a
n
~ N (0,1)
P(1  P ) / n
(approximately, for large n)
Pˆ  P
~ N (0,1)
(approximately, for large n)
Pˆ (1  Pˆ ) / n

ˆ (1  Pˆ )
ˆ (1  Pˆ ) 
P
P
 
P Pˆ  Z  / 2
 P  Pˆ  Z  / 2


n
n


• Thus
• Confidence interval for P

ˆ (1  Pˆ )
ˆ (1  Pˆ ) 
P
P
 Pˆ  Z

ˆ
,
P

Z
 /2
 /2


n
n


Example 6 (Hypothesis testing)
• Hypotheses: H0:P=P0 H1:PP0
Pˆ  P0
• Test statistic P0 (1  P0 )
under H0, for large n
• Reject H0 if P  P0  Z  / 2
P  P0  Z  / 2
~ N (0,1)
P0 (1  P0 )
n
P0 (1  P0 )
n
or if
Example 7: Differences between
population proportions
• Assume X 1 ~ Bin (n1 , P1 ) and X 2 ~ Bin (n2 , P2,)
X2
X1
ˆ
ˆ
so that P1 
and P2 
are
n2
n1
frequencies
Pˆ1  Pˆ2  ( P1  P2 )
P1 (1  P1 ) P2 (1  P2 )

n1
n2
~ N (0,1)
• Then
• Confidence interval for P1-P2
(approximately)

ˆ1 (1  Pˆ1 ) Pˆ2 (1  Pˆ2 ) 
P
 Pˆ  Pˆ  Z


2
 /2
 1

n1
n2


Example 7 (Hypothesis testing)
• Hypotheses: H0:P1=P2 H1:P1P2
• Test statistic
Pˆ1  Pˆ2
~ N (0,1)
Pˆ0 (1  Pˆ0 ) Pˆ0 (1  Pˆ0 )

n1
n2
where ˆ n1 Pˆ1  n 2 Pˆ2
P0 
n1  n 2
• Reject H0 if
Pˆ1  Pˆ2
 Z / 2
Pˆ0 (1  Pˆ0 ) Pˆ0 (1  Pˆ0 )

n1
n2
• Spontanous abortions among nurses helping
with operations and other nurses
• Want to test if there is difference between the
proportions of abortions in the two groups
• H0: Pop.nurses=Pothers H1: Pop.nursesPothers
Op.nurses
Other nurses
No. interviewed
67
92
No. births
36
34
No. abortions
10
3
27.8
8.8
Percent aportions
Calculation:
• P1=0.278 P2=0.088 n1=36 n2=34
Total no. abortions
10  3
p

 0.186
Total no. pregnancie s 36  34
0.278  0.088
z=
1
1
(  )0.186(1  0.186)
36 34
 2.04
• P-value 0.0414=4.1%, reject H0 on 5%sig.level (can’t do this in SPSS)
• 95% confidence interval for P1-P2:
Pˆ1 (1  Pˆ1 ) Pˆ2 (1  Pˆ2 )
ˆ
ˆ
( P1  P2 )  1.96 *

 (0.015,0.190)
n1
n2
Next week:
• Next lecture will be about modelling
relationships between continuous
variables
• Linear regression