Statistical inference I and II
Statistical inference
Stanislav Katina
ÚMS, MU
13.03.2024 22:03
Stanislav Katina (ÚMS, MU) Statistical inference I and II 13.03.2024 22:03 1 / 176
Book Aplikovaná štatistická inferencia I
Figure 1: Book Aplikovaná štatistická inferencia I
Stanislav Katina (ÚMS, MU) Statistical inference I and II 13.03.2024 22:03 2 / 176
Table of contents
1 Overview of testing of statistical hypotheses
2 Three test statistics
3 Confidence intervals
4 Overview of the tests
5 Generalised hypotheses
6 One-sample tests about 𝜇
7 Paired tests about 𝜇
8 One-sample tests about 𝜎2
9 One-sample tests about skewness and kurtosis
10 One-sample tests about correlation coefficient
Stanislav Katina (ÚMS, MU) Statistical inference I and II 13.03.2024 22:03 3 / 176
6 One-sample tests about 𝜇
6.5 One-sample tests about 𝜇 (cont.)
𝐻01 vs 𝐻11. Let 𝑋 ∼ 𝑁(𝜇, 𝜎2
), where 𝜃𝜃𝜃 = (𝜇, 𝜎2
) 𝑇
. Then the likelihood function is
equal to
𝐿(𝜃𝜃𝜃|x) = (2𝜋𝜎2
)−𝑛/2
exp (−
1
2𝜎2
𝑛
∑
𝑖=1
(𝑥𝑖 − 𝜇)2
)
and the log-likelihood function
ℓ(𝜃𝜃𝜃|x) = −
𝑛
2
ln(2𝜋) −
𝑛
2
ln 𝜎2
−
1
2𝜎2
𝑛
∑
𝑖=1
(𝑥𝑖 − 𝜇)2
.
If 𝐻01 is true, ̂𝜃𝜃𝜃0 = (𝜃0, ̂𝜃2|0) 𝑇
, where 𝜃0 = 𝜇0 and
̂𝜃2|0 = ̂𝜎2
0 = 1
𝑛 ∑
𝑛
𝑖=1
(𝑥𝑖 − 𝜇0)2
= 𝑠2
𝑛 + (𝑥 − 𝜇0)2
, where 𝑠2
𝑛 = ̂𝜎2
. The variance ̂𝜎2
0 is the
solution of
𝑆2(𝜃𝜃𝜃0) =
𝜕
𝜕𝜎2
ℓ(𝜃𝜃𝜃|x),
where using 𝜇 = 𝜇0, 𝜎2
= ̂𝜎2
0 and
1
2
(
1
̂𝜎4
0
𝑛
∑
𝑖=1
(𝑥𝑖 − 𝜇0)2
−
𝑛
̂𝜎2
0
) = 0.
we get ̂𝜎2
0.
Stanislav Katina (ÚMS, MU) Statistical inference I and II 13.03.2024 22:03 77 / 176
6 One-sample tests about 𝜇
6.6 One-sample tests about 𝜇 (cont.)
If 𝐻01 is not true, MLE of 𝜃𝜃𝜃 is then equal to
̂𝜃𝜃𝜃 = ( ̂𝜃1, ̂𝜃2) 𝑇
= (𝑥, ̂𝜎2
) 𝑇
= (
1
𝑛
𝑛
∑
𝑖=1
𝑥𝑖,
1
𝑛
𝑛
∑
𝑖=1
(𝑥𝑖 − 𝑥)2
)
𝑇
.
Then
𝑢LR = −2 ln(𝜆(x))
= 2(ℓ( ̂𝜃𝜃𝜃|x) − ℓ (𝜃𝜃𝜃0|x))
= 2 ((−
𝑛
2
(ln(2𝜋) + ln ̂𝜎2
+ 1)) − (−
𝑛
2
(ln(2𝜋) + ln ̂𝜎2
0 + 1)))
= 𝑛 ln
̂𝜎2
0
̂𝜎2
.
Stanislav Katina (ÚMS, MU) Statistical inference I and II 13.03.2024 22:03 78 / 176
6 One-sample tests about 𝜇
6.7 One-sample tests about 𝜇 (cont.)
One-sample likelihood ratio test statistic as a function of one-sample Student
𝑡-statistic.
𝑈LR = −2 ln (𝜆 (X))
𝒟
∼ 𝜒2
1, where 𝐻0 is rejected for high values of the ratio
̂𝜎2
0
̂𝜎2 . Then
̂𝜎2
0 = ̂𝜎2
+ (𝑥 − 𝜇0)2
and then
̂𝜎2
0
̂𝜎2 = 1 + (𝑥−𝜇0)2
̂𝜎2 . We now that
𝑠2
= 1
𝑛−1 ∑
𝑛
𝑖=1
(𝑥𝑖 − 𝑥)2
= 𝑛
𝑛−1 ̂𝜎2
, then
̂𝜎2
0
̂𝜎2 is increasing function of |𝑡 𝑊 | and then
𝑢LR = −2 ln (𝜆 (x)) = 𝑛 ln
̂𝜎2
0
̂𝜎2
= 𝑛 ln (1 +
𝑛
𝑛
∑
𝑛
𝑖=1
(𝑥𝑖 − 𝜇0)2
𝑠2
𝑛
) = 𝑛 ln (1 +
𝑢W
𝑛
)
= 𝑛 ln (1 +
𝑡2
𝑊
𝑛 − 1
) ,
where 𝑢W = 𝑛
∑
𝑛
𝑖=1
(𝑥 𝑖−𝜇0)2
𝑠2
𝑛
= 𝑛
∑
𝑛
𝑖=1
(𝑥 𝑖−𝜇0)2
𝑛−1
𝑛 𝑠2 = 𝑛
𝑛−1 𝑡2
𝑊 .
𝐻02 vs 𝐻12. Then 𝑢LR = 0 if 𝑡 𝑊 ≤ 0 and 𝑢LR = 𝑛
2 ln(1 +
𝑡2
𝑊
𝑛−1 ) for 𝑡 𝑊 > 0.
𝐻03 vs 𝐻13. Then 𝑢LR = 0 if 𝑡 𝑊 ≥ 0 and 𝑢LR = 𝑛
2 ln(1 +
𝑡2
𝑊
𝑛−1 ) for 𝑡 𝑊 < 0.
Stanislav Katina (ÚMS, MU) Statistical inference I and II 13.03.2024 22:03 79 / 176
6 One-sample tests about 𝜇
6.8 One-sample tests about 𝜇 (cont.)
One-sample Score statistic as a function of one-sample Student 𝑡-statistic.
We know that 𝑈S = (a 𝑇
𝑆( ̂𝜃𝜃𝜃0)) 𝑇
(a 𝑇
(ℐ( ̂𝜃𝜃𝜃0))
−1
a) a 𝑇
𝑆( ̂𝜃𝜃𝜃0), where ℐ11( ̂𝜃𝜃𝜃0) = 𝑛/ ̂𝜎2
0 and
𝑆1( ̂𝜃𝜃𝜃0) = 𝑆(𝜇0) = 1
̂𝜎2
0
∑
𝑛
𝑖=1
(𝑥𝑖 − 𝜇0). Then
𝑢S = (
1
̂𝜎2
0
𝑛
∑
𝑖=1
(𝑥𝑖 − 𝜇0))
2
̂𝜎2
0
𝑛
= (
1
̂𝜎2
0
(
𝑛
∑
𝑖=1
𝑥𝑖 − 𝑛𝜇0))
2
̂𝜎2
0
𝑛
=
1
𝑛 ̂𝜎2
0
(𝑛𝑥 − 𝑛𝜇0)
2
= 𝑛
(𝑥 − 𝜇0)
2
̂𝜎2
0
=
𝑛 (𝑥 − 𝜇0)
2
𝑠2
𝑛 + (𝑥 − 𝜇0)2
=
𝑛 (𝑥 − 𝜇0)
2
𝑠2
𝑛 (1 + (𝑥−𝜇0)2
𝑠2
𝑛
)
=
𝑢 𝑊
1 + (𝑥−𝜇0)2
𝑠2
𝑛
=
𝑛𝑢 𝑊
𝑛 + 𝑢W
=
𝑛𝑡2
𝑊
𝑛 − 1 + 𝑡2
𝑊
.
Stanislav Katina (ÚMS, MU) Statistical inference I and II 13.03.2024 22:03 80 / 176
6 One-sample tests about 𝜇
6.9 One-sample tests about 𝜇 (cont.)
One-sample Wald statistic as a function of one-sample Student 𝑡-statistic.
We know that 𝑈W = (a 𝑇 ̂𝜃𝜃𝜃 − a 𝑇 ̂𝜃𝜃𝜃0) 𝑇
(a 𝑇
ℐ( ̂𝜃𝜃𝜃)a) (a 𝑇 ̂𝜃𝜃𝜃 − a 𝑇 ̂𝜃𝜃𝜃0) where
ℐ11( ̂𝜃𝜃𝜃) = ℐ( ̂𝜇) = 𝑛/𝑠2
𝑛. Then
𝑢W = (𝑥 − 𝜇0)
2 𝑛
̂𝜎2
= 𝑛
(𝑥 − 𝜇0)
2
𝑠2
𝑛
=
𝑛
𝑛 − 1
𝑡2
𝑊 .
Stanislav Katina (ÚMS, MU) Statistical inference I and II 13.03.2024 22:03 81 / 176
6 One-sample tests about 𝜇
6.10 One-sample tests about 𝜇 (cont.)
Confidence intervals:
Wald 100 × (1 − 𝛼)% empirical confidence interval for 𝜇 is defined as
𝒞𝒮
(W)
1−𝛼 = {𝜇0 ∶ 𝑈W(𝜇) < 𝜒2
1 (𝛼)} .
Likelihood 100 × (1 − 𝛼)% empirical confidence interval for 𝜇 is defined as
𝒞𝒮
(LR)
1−𝛼 = {𝜇0 ∶ 𝑈LR(𝜇) < 𝜒2
1 (𝛼)} .
Score 100 × (1 − 𝛼)% empirical confidence interval for 𝜇 is defined as
𝒞𝒮
(S)
1−𝛼 = {𝜇0 ∶ 𝑈S(𝜇) < 𝜒2
1 (𝛼)} .
Stanislav Katina (ÚMS, MU) Statistical inference I and II 13.03.2024 22:03 82 / 176
6 One-sample tests about 𝜇
6.11 One-sample tests about 𝜇 (cont.)
Knowing the inequality
𝑥
1 + 𝑥
< ln(1 + 𝑥) < 𝑥, where 𝑥 > −1,
and 𝑥 = 𝑇2
𝑊 /(𝑛 − 1), the three test statistics can be ordered
𝑈S < 𝑈LR < 𝑈W.
Then
𝑈S =
𝑛𝑇2
𝑊
𝑛 − 1 + 𝑇2
𝑊
1
𝑛−1
1
𝑛−1
=
𝑛
𝑇2
𝑊
𝑛−1
1 +
𝑇2
𝑊
𝑛−1
=
𝑛𝑥
1 + 𝑥
,
𝑈LR = 𝑛 ln (1 +
𝑇2
𝑊
𝑛 − 1
) = 𝑛 ln(1 + 𝑥),
𝑈W =
𝑛
𝑛 − 1
𝑇2
𝑊 = 𝑛𝑥.
Note: If 𝐻0 is true, 𝑇W = 𝑋−𝜇0
𝑆
√
𝑛
𝒟
∼ 𝑡 𝑛−1.
Stanislav Katina (ÚMS, MU) Statistical inference I and II 13.03.2024 22:03 83 / 176
6 One-sample tests about 𝜇
6.12 One-sample tests about 𝜇 (cont.)
If 𝐻0 is true, then the probabilities of Type I error are
𝛼W = Pr (𝑢W ≥ 𝜒2
1(𝛼)) = Pr (𝐹1,𝑛−1 ≥
𝑛 − 1
𝑛
𝜒2
1(𝛼)) ,
𝛼LR = Pr (𝑈LR ≥ 𝜒2
1(𝛼)) = Pr (𝐹1,𝑛−1 ≥ (𝑛 − 1) [exp (
𝜒2
1(𝛼)
𝑛
) − 1])
and
𝛼S = Pr (𝑈S ≥ 𝜒2
1(𝛼)) = Pr (𝐹1,𝑛−1 ≥
𝑛 − 1
𝑛 − 𝜒2
1(𝛼)
𝜒2
1(𝛼)) .
Note: 𝑈W is often liberal, 𝑈LR slightly liberal a 𝑈S is usually neither liberal nor
conservative. Limiting (asymptotical) distribution, i.e., for sufficiently large 𝑛, for all test
statistics is 𝜒2
1 distribution.
Stanislav Katina (ÚMS, MU) Statistical inference I and II 13.03.2024 22:03 84 / 176
6 One-sample tests about 𝜇
6.13 One-sample tests about 𝜇 (cont.)
Let 𝑋 ∼ 𝑁(𝜇, 𝜎2
), where 𝜎2
is not known. We would like to test the hypotheses:
1 𝐻01 ∶ 𝜇 = 𝜇0 vs 𝐻11 ∶ 𝜇 ≠ 𝜇0,
2 𝐻02 ∶ 𝜇 ≤ 𝜇0 vs 𝐻12 ∶ 𝜇 > 𝜇0,
3 𝐻03 ∶ 𝜇 ≥ 𝜇0 vs 𝐻13 ∶ 𝜇 < 𝜇0.
Under 𝐻0
𝑇W =
𝑋 − 𝜇0
𝑆
√
𝑛
𝒟
∼ 𝑡 𝑛−1,
where 𝑆2
= 1
𝑛−1 ∑
𝑛
𝑖=1
(𝑋𝑖 − 𝑋)
2
and the distribution 𝑡df, where df = 𝑛 − 1, is called
central 𝑡-distribution with 𝑛 − 1 degrees of freedom. 𝑇W is called one-sample Student
test statistic (or one-sample 𝑡-statistic) and test one-sample Student 𝑡-test about 𝜇.
Stanislav Katina (ÚMS, MU) Statistical inference I and II 13.03.2024 22:03 85 / 176
6 One-sample tests about 𝜇
6.14 One-sample tests about 𝜇 (cont.)
If 𝐻0 is not true (under the alternative 𝐻1), this situations leads to non-central
𝑡-distribution with df degrees of freedom and non-centrality parameter 𝜆, 𝑡df,𝜆, where
𝑇W,𝜆 =
𝑍 𝑊 + 𝜆
√ 𝑉 /df
=
√
𝑛 (𝑋 − 𝜇) /𝜎 +
√
𝑛 (𝜇 − 𝜇0) /𝜎
𝑆/𝜎
𝒟
∼ 𝑡df,𝜆,
df = 𝑛 − 1, 𝑍 𝑊 = 𝑋−𝜇
𝜎
√
𝑛
𝒟
∼ 𝑁 (0, 1), non-centrality parameter 𝜆 = (𝛿/𝜎)
√
𝑛,
𝛿 = 𝜇 − 𝜇0 is minimal detected distance between 𝜇 and 𝜇0, 𝑉 = (𝑛−1)𝑆2
𝜎2
𝒟
∼ 𝜒2
df and is
independent of 𝑍 𝑊 . Let cumulative distribution function of 𝑇W,𝜆 be
𝐺df,𝜆 (𝑡) = Pr (𝑇W,𝜆 ≤ 𝑡). If 𝜆 = 0, then non-central 𝑡-distribution is equivalent to the
central (Student) 𝑡-distribution.
Stanislav Katina (ÚMS, MU) Statistical inference I and II 13.03.2024 22:03 86 / 176
6 One-sample tests about 𝜇
6.15 One-sample tests about 𝜇 (cont.)
If 𝐻02 is not true (under the alternative 𝐻12), then the power function
1 − 𝛽 (𝜇, 𝜎) = Pr
𝜇,𝜎
(
𝑋 − 𝜇0
𝑆
√
𝑛 ≥ 𝑡 𝑛−1 (𝛼)) = 1 − 𝐺 𝑛−1,𝜆 (𝑡 𝑛−1 (𝛼)) ,
where (𝜇, 𝜎) means that the probability is calculated under the condition
𝑋 ∼ 𝑁(𝜇, 𝜎2
), 𝜇 > 𝜇0.
If 𝐻03 is not true (under the alternative 𝐻13), then the power function
1 − 𝛽 (𝜇, 𝜎) = Pr
𝜇,𝜎
(
𝑋 − 𝜇0
𝑆
√
𝑛 ≤ −𝑡 𝑛−1 (𝛼)) = 𝐺 𝑛−1,𝜆 (𝑡 𝑛−1 (𝛼)) ,
where (𝜇, 𝜎) means that the probability is calculated under the condition
𝑋 ∼ 𝑁(𝜇, 𝜎2
), 𝜇 < 𝜇0.
If 𝐻01 is not true (under the alternative 𝐻11), then the power function
1 − 𝛽 (𝜇, 𝜎) = Pr
𝜇,𝜎
(
𝑋 − 𝜇0
𝑆
√
𝑛 ≤ −𝑡 𝑛−1 (𝛼/2) ∨
𝑋 − 𝜇0
𝑆
√
𝑛 ≥ 𝑡 𝑛−1 (𝛼/2))
= 1 − 𝐺 𝑛−1,𝜆 (𝑡 𝑛−1 (𝛼/2)) + 𝐺 𝑛−1,𝜆 (−𝑡 𝑛−1 (𝛼/2)) ,
where (𝜇, 𝜎) means that the probability is calculated under the condition
𝑋 ∼ 𝑁(𝜇, 𝜎2
), 𝜇 ≠ 𝜇0.
Stanislav Katina (ÚMS, MU) Statistical inference I and II 13.03.2024 22:03 87 / 176
6 One-sample tests about 𝜇
6.16 One-sample tests about 𝜇 (cont.)
Critical regions and power functions related to 𝑇W
𝐻0 𝐻1 𝒲 1 − 𝛽 (𝜇)
𝜇 = 𝜇0 𝜇 ≠ 𝜇0 𝒲1 = {𝑇W; |𝑇W| ≥ 𝑡 𝑛−1 (𝛼/2)} Pr (𝐹1,𝑛−1,𝜆2 ≥ 𝐹1,𝑛−1 (𝛼))
𝜇 ≤ 𝜇0 𝜇 > 𝜇0 𝒲2 = {𝑇W; 𝑇W ≥ 𝑡 𝑛−1 (𝛼)} Pr (𝑡 𝑛−1,𝜆 ≥ 𝑡 𝑛−1 (𝛼))
𝜇 ≥ 𝜇0 𝜇 < 𝜇0 𝒲3 = {𝑇W; 𝑇W ≤ −𝑡 𝑛−1 (𝛼)} Pr (𝑡 𝑛−1,𝜆 ≤ −𝑡 𝑛−1 (𝛼))
Note: 𝑡2
𝑛−1(𝛼/2) ≈ 𝐹1,𝑛−1(𝛼) and 𝑡2
𝑛−1,𝜆(𝛼/2) ≈ 𝐹1,𝑛−1,𝜆2 (𝛼).
Note: Noncentrality parameter 𝜆 = (𝛿/𝜎)
√
𝑛, 𝛿 = 𝜇 − 𝜇0.
p-value =
⎧
{
⎨
{
⎩
2Pr(𝑇W ≥ |𝑡 𝑊 ||𝐻01), if 𝐻11 ∶ 𝜇 ≠ 𝜇0
Pr(𝑇W ≥ 𝑡 𝑊 |𝐻02), if 𝐻12 ∶ 𝜇 > 𝜇0
Pr(𝑇W ≤ 𝑡 𝑊 |𝐻03), if 𝐻13 ∶ 𝜇 < 𝜇0
𝐻0 𝐻1 ( ̂𝜇 𝐿, ̂𝜇 𝑈 )
𝜇 = 𝜇0 𝜇 ≠ 𝜇0 𝒞𝒮1−𝛼 = {𝜇0 ∶ 𝜇0 ∈ (𝑥 − 𝑡 𝑛−1 (𝛼/2) 𝑠√
𝑛
, 𝑥 + 𝑡 𝑛−1 (𝛼/2) 𝑠√
𝑛
)}
𝜇 ≤ 𝜇0 𝜇 > 𝜇0 𝒞𝒮1−𝛼 = {𝜇0 ∶ 𝜇0 ∈ (𝑥 − 𝑡 𝑛−1 (𝛼) 𝑠√
𝑛
, ∞)}
𝜇 ≥ 𝜇0 𝜇 < 𝜇0 𝒞𝒮1−𝛼 = {𝜇0 ∶ 𝜇0 ∈ (−∞, 𝑥 + 𝑡 𝑛−1 (𝛼) 𝑠√
𝑛
)}
Stanislav Katina (ÚMS, MU) Statistical inference I and II 13.03.2024 22:03 88 / 176
6 One-sample tests about 𝜇
6.17 One-sample tests about 𝜇 (cont.)
−4 −2 0 2 4
0.00.20.40.60.81.0
n = 10
n = 20
n = 30
n = 40
n = 50
powerfunction
µ − µ0
H11 : µ − µ0 ≠ 0, µ0 = 0, σ2
= 4, α = 0.05
−4 −2 0 2 4
0.00.20.40.60.81.0
n = 10
n = 20
n = 30
n = 40
n = 50
powerfunction
µ − µ0
H12 : µ − µ0 > 0, µ0 = 0, σ2
= 4, α = 0.05
−4 −2 0 2 4
0.00.20.40.60.81.0
n = 10
n = 20
n = 30
n = 40
n = 50
powerfunction
µ − µ0
H13 : µ − µ0 < 0, µ0 = 0, σ2
= 4, α = 0.05
Stanislav Katina (ÚMS, MU) Statistical inference I and II 13.03.2024 22:03 89 / 176
6 One-sample tests about 𝜇
6.18 One-sample Student 𝑡-test about mean 𝜇 in
Example (independence of 𝜇 and 𝜎2
, coverage probability)
Let 𝑋 ∼ 𝑁(𝜇, 𝜎2
), where 𝜇 = 20 and 𝜎2
= 100. Calculate Pearson correlation coefficient
𝑟 𝑋,𝑆 based on the simulation study. Draw the points (𝑥 𝑚, 𝑠 𝑚) as scatter-plot (grey
color), where 𝑚 = 1, 2, … , 𝑀, 𝑀 = 100000. Add the points
𝑡 𝑊,𝑚 = ∣ 𝑥 𝑚−𝜇
𝑠 𝑚
√
𝑛∣ < 𝑡 𝑛−1(𝛼/2) (black color), and a boundary, to define the points
(𝑥 𝑚, 𝑠 𝑚), where 𝑡 𝑊,𝑚 = 𝑡 𝑛−1(𝛼/2). Calculate coverage probability 95% of CI for 𝜇 as a
ration ∑ 𝑚
𝐼 (𝑡 𝑊,𝑚 < 𝑡 𝑛−1(𝛼/2)) /𝑀. Use (a) 𝑛 = 5, (b) 𝑛 = 50 and (c) 𝑛 = 100.
Stanislav Katina (ÚMS, MU) Statistical inference I and II 13.03.2024 22:03 90 / 176
6 One-sample tests about 𝜇
6.19 One-sample Student 𝑡-test about mean 𝜇 in
0 10 20 30 40
0510152025
n = 5, r = 0.001942
arithmetic average
standarddeviation
14 16 18 20 22 24 26
68101214
n = 50, r = −0.001793
arithmetic average
standarddeviation
16 18 20 22 24
8910111213
n = 100, r = −0.001838
arithmetic average
standarddeviation
Scatter-plot of 𝑥 𝑚 and 𝑠 𝑚, 𝑚 = 1, 2, … , 𝑀, 𝑀 = 100000 for 𝑛 = 5 (left), 𝑛 = 50
(middle) and 𝑛 = 100 (right). Lines indicate 𝑡 𝑊,𝑚 = ∣ 𝑥 𝑚−𝜇
𝑠 𝑚
√
𝑛∣ < 𝑡 𝑛−1(𝛼/2), where the
coverage probability is approximately equal to the nominal level 1 − 𝛼 = 0.95.
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6 One-sample tests about 𝜇
6.20 One-sample Student 𝑡-test about mean 𝜇 in
Example (noncentral 𝑡-distribution)
Draw the densities of one central and four non-central 𝑡-distributions 𝑡 𝑛−1,𝜆 ( 𝛿 = 𝜇 − 𝜇0
and 𝜆 = 𝛿/(𝜎/
√
𝑛)) to one figure and distinguish them by different color or line type.
Use 𝜇0 = 0, 𝛿 = 0, 0.5, 0.8, 1 and 1.2, 𝜎 = 1.4 and 𝑛 = 26.
0 5 10
0.00.10.20.30.4
x
density
δ = 0
δ = 0.5
δ = 0.8
δ = 1
δ = 1.2
Densities of non-central 𝑡-distribution with different non-centrality parameter 𝜆
𝛿
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