Transcript Decision Errors - University of Toronto

The t distribution
• Suppose that a SRS of size n is drawn from a N(μ, σ)
population. Then the one sample t statistic
x
t
s n
has a t distribution with n -1 degrees of freedom.
• The t distribution has mean 0 and it is a symmetric
distribution.
• The is a different t distribution for each sample size.
• A particular t distribution is specified by the degrees of
freedom that comes from the sample standard deviation.
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Tests for the population mean  when  is
unknown
• Suppose that a SRS of size n is drawn from a population
having unknown mean μ and unknown stdev. . To test the
hypothesis H0: μ = μ0 , we first estimate  by s – the sample
stdev., then compute the one-sample t statistic given by
x  0
t
s n
• In terms of a random variable T having the t (n - 1)
distribution, the P-value for the test of H0 against
Ha : μ > μ 0 is
Ha : μ < μ 0 is
P( T ≥ t )
P( T ≤ t )
Ha : μ ≠ μ 0 is 2·P( T ≥ |t|)
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Example
• In a metropolitan area, the concentration of cadmium (Cd) in
leaf lettuce was measured in 6 representative gardens where
sewage sludge was used as fertilizer. The following
measurements (in mg/kg of dry weight) were obtained.
Cd 21 38 12 15 14
8
Is there strong evidence that the mean concentration of Cd is
higher than 12.
Descriptive Statistics
Variable
Cd
N
6
Mean
18.00
Median
14.50
TrMean
18.00
StDev
10.68
• The hypothesis to be tested are: H0: μ = 12 vs
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SE Mean
4.36
Ha: μ > 12.
3
• The test statistics is:
t  x    1812 1.38
s / n 10.68/ 6
The degrees of freedom are df = 6 – 1 = 5
Since t = 1.38 < 2.015, we cannot reject H0 at the 5% level and
so there are no strong evidence.
The P-value is 0.1 < P(T(5) ≥ 1.38) < 0.15 and so is greater
then 0.05 indicating a non significant result.
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CIs for the population mean  when 
unknown
• Suppose that a SRS of size n is drawn from a population
having unknown mean μ. A C-level CI for μ when  is
unknown is an interval of the form
s
s 

*
*
, x t 
x t 

n
n

where t* is the value for the t (n -1) density curve with area C
between –t* and t*.
• Example:
Give a 95% CI for the mean Cd concentration.
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• MINITAB commands: Stat > Basic Statistics > 1-Sample t
• MINITAB outputs for the above problem:
T-Test of the Mean
Test of mu = 12.00 vs mu > 12.00
Variable
N
Mean
StDev
Cd
6
18.00
10.68
SE Mean
4.36
T
1.38
P
0.11
T Confidence Intervals
Variable
Cd
N
6
Mean
18.00
StDev
10.68
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SE Mean
4.36
95.0 % CI
(6.79, 29.21)
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Question 3 Final exam Dec 2000
•
In order to test H0: μ = 60 vs Ha: μ ≠ 60 a random sample of 9
observations (normally distributed) is obtained, yielding x 55
and s = 5. What is the p-value of the test for this sample?
a)
b)
c)
d)
e)
greater than 0.10.
between 0.05 and 0.10.
between 0.025 and 0.05.
between 0.01 and 0.025.
less than 0.01.
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Question
A manufacturing company claims that its new floodlight will
last 1000 hours. After collecting a simple random sample of
size ten, you determine that a 95% confidence interval for the
true mean number of hours that the floodlights will last, , is
(970, 995). Which of the following are true? (Assume all tests
are two-sided.)
I) At any  < .05, we can reject the null hypothesis that the true
mean is 1000.
II) If a 99% confidence interval for the mean were determined
here, the numerical value 972 would certainly lie in this
interval.
III) If we wished to test the null hypothesis H0:  = 988, we
could say that the p-value must be < 0.05.
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Questions
1. Alpha (level of sig. α) is
a) the probability of rejecting H0 when H0 is true.
b) the probability of supporting H0 when H0 is false.
c) supporting H0 when H0 is true.
d) rejecting H0 when H0 is false.
2. Confidence intervals can be used to do hypothesis tests for
a) left tail tests.
b) right tail tests
c) two tailed test
3. The Type II error is supporting a null hypothesis that is false. T/F
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Robustness of the t procedures
• Robust procedures
A statistical inference procedure is called robust if the
probability calculations required are insensitive to violations of
the assumptions made.
• t-procedures are quite robust against nonnormality of the
population except in the case of outliers or strong skewness.
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Simulation study
• Let’s generate 100 samples of size 10 from a moderately
skewed distribution (Chi-square distribution with 5 df ) and
calculate the 95% t-intervals to see how many of them contain
the true mean μ = 5.
• First let’s have a look at the histogram of the 1000 values
generated from this distribution.
400
Frequency
300
200
100
0
0
10
20
30
C1
Variable
C1
N
1000
Mean
4.9758
Median
4.2788
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TrMean
4.7329
StDev
3.1618
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T Confidence Intervals
Variable
C1
N
10
Mean
5.21
StDev
3.89
SE Mean
1.23
95.0 % CI
2.43,
7.99)
(
10
10
10
10
10
10
4.449
5.33
3.267
4.981
3.725
4.487
1.593
4.23
2.312
2.988
1.520
2.332
0.504
1.34
0.731
0.945
0.481
0.738
(
(
(
(
(
(
3.309,
2.31,
1.612,
2.844,
2.638,
2.819,
5.589)
8.36)
4.921)*
7.118)
4.812)*
6.155)
10
10
10
10
10
10
4.650
2.973
4.685
5.594
3.468
5.59
1.854
2.163
2.254
2.984
2.078
3.84
0.586
0.684
0.713
0.944
0.657
1.22
(
(
(
(
(
(
3.324,
1.425,
3.072,
3.459,
1.982,
2.84,
5.977)
4.520)*
6.297)
7.728)
4.955)*
8.34)
10
10
10
5.689
3.724
4.387
3.113
1.741
2.157
0.984
0.551
0.682
(
(
(
3.462,
2.479,
2.843,
7.916)
4.970)*
5.930)
10
10
10
7.01
3.281
4.78
3.44
2.265
3.20
1.09
0.716
1.01
(
(
(
4.55,
1.661,
2.49,
9.47)
4.902)*
7.06)
10
10
6.52
3.614
4.24
2.198
1.34
0.695
(
(
3.49,
2.042,
9.56)
5.186)
. . .
C4
C5
C6
C7
C8
C9
. . .
C14
C15
C16
C26
C27
C28
. . .
C62
C63
C64
. . .
C87
C88
C89
. . .
C99
C100
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The number of intervals not capturing
the true mean (μ = 5) is 6/100.
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Example
• 100 samples of size 15 were drawn from a very skewed
distribution (Chi-square distribution with d. f. 1)
Variable
C1
N
Mean
Median
TrMean
StDev
1500
0.9947
0.4766
0.8059
1.3647
Frequency
1500
1000
500
0
0
5
10
15
C1
• The 95% CIs (t-intervals) for these 100 samples are given below.
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T Confidence Intervals
Variable
N
Mean
C1
15
0.773
C2
15
1.093
C3
15
0.553
C4
15
0.387
C5
15
1.239
...
C23
15
0.491
C24
15
0.582
C25
15
0.550
C26
15
0.634
C27
15
0.508
...
C51
15
1.122
C52
15
0.519
C53
15
1.666
...
C59
15
1.208
C60
15
0.644
C61
15
1.088
StDev
0.939
1.491
0.735
0.732
2.146
SE Mean
0.242
0.385
0.190
0.189
0.554
(
(
(
(
(
95.0 % CI
0.253,
1.293)
0.268,
1.919)
0.146,
0.960)*
-0.019,
0.792)*
0.051,
2.427)
0.619
1.088
0.660
0.769
0.528
0.160
0.281
0.170
0.199
0.136
(
(
(
(
(
0.148,
-0.020,
0.184,
0.208,
0.216,
0.834)*
1.184)
0.915)*
1.060)
0.800)*
1.292
0.664
2.028
0.334
0.171
0.524
(
(
(
0.406,
0.151,
0.543,
1.837)
0.887)*
2.789)
2.297
0.525
1.122
0.593
0.136
0.290
(
(
(
-0.065,
0.353,
0.466,
2.480)
0.935)*
1.709)
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T Confidence Intervals (continuation)
...
C79
C80
C81
C82
C83
C84
...
C99
C100
15
15
15
15
15
15
0.895
0.391
1.038
0.952
0.2763
1.237
0.931
0.767
0.992
1.407
0.2999
1.999
0.240
0.198
0.256
0.363
0.0774
0.516
(
(
(
(
(
(
0.379,
-0.034,
0.488,
0.173,
0.1102,
0.130,
1.411)
0.816)*
1.587)
1.732)
0.4424)*
2.345)
15
15
0.921
0.813
0.865
1.437
0.223
0.371
(
(
0.442,
0.018,
1.400)
1.609)
The number of intervals not capturing the true mean (μ = 1) is 9/100.
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Match Pairs t-test
• In a matched pairs study, subjects are matched in pairs and the
outcomes are compared within each matched pair. The
experimenter can toss a coin to assign two treatment to the two
subjects in each pair. Matched pairs are also common when
randomization is not possible. One situation calling for match
pairs is when observations are taken on the same subjects,
under different conditions.
• A match pairs analysis is needed when there are two
measurements or observations on each individual and we want
to examine the difference.
• For each individual (pair), we find the difference d between
the measurements from that pair. Then we treat the di as one
sample and use the one sample t – statistic to test for no
difference between the treatments effect.
• Example: similar to exercise 7.41 on page 446 in IPS.
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Data Display
Row
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Student
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Pretest
30
28
31
26
20
30
34
15
28
20
30
29
31
29
34
20
26
25
31
29
Posttest
29
30
32
30
16
25
31
18
33
25
32
28
34
32
32
27
28
29
32
32
improvement
-1
2
1
4
-4
-5
-3
3
5
5
2
-1
3
3
-2
7
2
4
1
3
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• One sample t-test for the improvement
T-Test of the Mean
Test of mu = 0.000 vs mu > 0.000
Variable
N
Mean
StDev
SE Mean
improvem 20
1.450
3.203
0.716
T
2.02
P
0.029
• MINITAB commands for the paired t-test
Stat > Basic Statistics > Paired t
Paired T-Test and Confidence Interval
Paired T for Posttest – Pretest
N
Mean
StDev
SE Mean
Posttest
20
28.75
4.74
1.06
Pretest
20
27.30
5.04
1.13
Difference
20
1.450
3.203
0.716
95% CI for mean difference: (-0.049, 2.949)
T-Test of mean difference=0 (vs > 0):
T-Value = 2.02 P-Value = 0.029
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6
Frequency
5
4
3
2
1
0
-4
-2
0
2
4
6
8
improvement
Character Stem-and-Leaf Display
Stem-and-leaf of improvement
Leaf Unit = 1.0
2
-0 54
4
-0 32
6
-0 11
8
0 11
(7)
0 2223333
5
0 4455
1
0 7
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N
= 20
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Two-sample problems
• The goal of inference is to compare the response in two groups.
• Each group is considered to be a sample form a distinct
population.
• The responses in each group are independent of those in the
other group.
• A two-sample problem can arise form a randomized
comparative experiment or comparing random samples
separately selected from two populations.
• Example:
A medical researcher is interested in the effect of added calcium
in our diet on blood pressure. She conducted a randomized
comparative experiment in which one group of subjects receive
a calcium supplement and a control group gets a placebo.
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Comparing two means (with two independent samples)
• Here we will look at the problem of comparing two population means
when the population variances are known or the sample sizes are large.
Suppose that a SRS of size n1 is drawn from an N( μ1, σ1) population
and that an independent SRS of size n2 is drown from an N( μ2, σ2)
population. Then the two-sample z statistics for testing the null
hypothesis H0: μ1 = μ2 is given by

x1  x2   1  2 
z

2
1
 
n1   22 n2

and has the standard normal N(0,1) sampling distribution.
• Using the standard normal tables, the P-value for the test of H0 against
Ha : μ1 > μ2 is P( Z ≥ z )
Ha : μ1 < μ2 is P( Z ≤ z )
Ha : μ1 ≠ μ 2 is 2·P(Z ≥ |z|)
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Example
• A regional IRS auditor runs a test on a sample of returns filed
by March 15 to determine whether the average return this year
is larger than last year. The sample data are shown here for a
random sample of returns from each year.
Last Year
This Year
Mean
380
410
Sample size
100
120
• Assume that the std. deviation of returns is known to be about
100 for both years. Test whether the average return is larger
this year than last year.
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Solution
• The hypothesis to be tested are: H0: μ1 = μ2 vs
• The test statistics is:
Ha: μ1 < μ2.
z  380  410 0 2.22 1.645
1002 1002
100 120
• The P-value = P(Z < -2.22) = 0.0139 < 0.05, therefore we can
reject H0 and conclude that at the 5% significant level, the
average return is larger this year than last year.
• A 95% CI for the difference is given by:




x  x  Z*
1 2




2 2
1
2
n n ,
1
2
301.96 1002 1002  30 26.5
100 120
 (3.5, 56.5)
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Comparing two population means
(unknown std. deviations)
• Suppose that a SRS of size n1 is drawn from a normal
population with unknown mean 1 and that an independent
SRS of size n2 is drawn from another normal population with
unknown mean 2. To test the null hypothesis H0: 1 = 2, we
compute the two sample t-statistic
t
x1  x2  1  2 
s
2
1
 
n1  s22 n2

• This statistic has a t-distribution with df approximately equal
to smaller of n1 – 1 and n2 - 1. We can use this distribution to
compute the P-value.
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Example
• The weight gains for n1 = n2 = 8 rats tested on diets 1 and 2 are
summarized here. Test whether diet 2 has greater mean weight
gain. Use the 5% significant level.
n
Std dev.
mean
Diet 1
8
.033
3.1
Diet 2
8
0.070
3.2
• The hypotheses to be tested are: H0: μ1 = μ2 vs Ha: μ1 < μ2 .
• The test statistic is
3.13.20
t
3.65
0.0332  0.0702
8
8
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• The P-value is P(T(7) ≤- 3.65) = P(T(7) ≥ 3.65) , from table D
we have 0.005 < P-value < 0.01 and so we reject H0 and
conclude that the mean weight gain from diet 2 is significantly
greater than that from diet 1 (at the 5% and 1% significant
level).
• A C% CI for the difference between the two means is given by,
x1  x2  t 
s12 s22

n1 n2
• For this example the 95% CI is
0.0332 0.0702
3.2  3.1  2.365

8
8
= (0.0353, 0.165)
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