Download Solution Series 13

Probability and statistics
Lecturer: Prof. Dr.
Coordinator: Yilin
not due
Mete SONER
WANG
Solution Series 13
Q1.
We toss 100 times a coin and we get 60 head. We want to do a test to know whether the
coin is fair.
(a)
Test the hypothesis with a 0.01 level of signicance. Should this test be one or twotailed?.
(b)
What is the biggest amount of head should we have in 100 tossings so we cannot discard
H0 :=
(c)
The coin biased towards tail.
Calculate all
p0
so that the null hypothesis
H0 (p0 ) := Probability
of head is
p0 ,
would not be rejected in a test with 0.05 level of signicance.
Hint:
It will be useful to use the central limit theorem in all of this question.
Solution:
Let
(a)
(Xi )ni=1
an i.i.d sequence of Bernoulli(p). Let
X=
P100
i=1
Xi ,
then
X ∼ Bin(100, p).
We want to know whether the coin is fair or not, so our hypothesis are
H0 : The
H1 : The
coin is fair,
coin is not fair.
That is to say
1
H0 : p = p0 = ,
2
H1 : p 6= p0 .
Then we should use a two-tailed test.
Under
H0
and V ar0 (X) = np0 (1 − p0 ) = 25. By central
X−50
can be approximated by the standard normal
5
be the c.d.f. of the standard normal distribution. Now, take c1 and
we have that
E0 (X) = 50
limit theorem, the distribution of
distribution. Let
c2
φ
such that
0.01 ≥ P0 (X ∈
/ (c1 , c2 ))
c1 − 50
X − 50
c2 − 50
= 1 − P0
≤
≤
5
5
5
c2 − 50
c1 − 50
∼1−φ
+φ
.
5
5
1
Probability and statistics
not due
We would like to make the rejection zone as biggest as we can, given that
φ is symmetric,
we just have to make
c1 − 50
c2 − 50
=−
,
5
5
⇒c1 = 100 − c2 .
Finally
c2 − 50
5
0.01 ≥ 1 − φ
c2 − 50
≥ 0.995
⇒φ
5
c2 − 50
⇒
≥ 2.5758
5
⇒c2 ≥ 62.9,
+1−φ
c2 − 50
5
,
then
K1% = [0; 37] ∪ [63; 100].
Given that
(b)
60 ∈
/ K1%
we cannot reject
H0
with 0.01 level of signicance.
We have to take our hypothesis
H0 : p = p0 =
1
2
H1 : p > p0 ,
now we have to nd
c
such that
0.01 ≥ P0 (X ≥ c) ∼ 1 − φ
Then we have to choose
H0
(c)
c − 50
5
.
c ≥ 61.6, from what we have that K1% = [62; 100].
So we reject
if we have 62 or more heads.
Take
p0 ∈ (0, 1)
and
H0 : p = p0 ,
H1 : p 6= p0 .
Note that
E0 (X) = 100p0
and
V ar0 (X) = 100p0 (1 − p0 ).
Then, take
c1
and
that
0.01 ≥ P0 (X ∈
/ (c1 , c2 ))
= 1 − P0
∼1−φ
c1 − 100p0
X − 100p0
c2 − 100p0
p
≤ p
≤ p
10 p0 (1 − p0 )
10 p0 (1 − p0 )
10 p0 (1 − p0 )
!
!
c1 − 100p0
c2 − 100p0
p
p
+φ
.
10 p0 (1 − p0 )
10 p0 (1 − p0 )
2
!
c2
such
Probability and statistics
not due
p
a
we have that c1 (p0 ) = 100p0 − 19.6
p0 (1 − p0 )
p
c2 (p0 ) = 100p0 + 19.6 p0 (1 − p0 ). Then we are interested in the set
n
o
p
p
p : 100p − 19.6 p(1 − p) ≤ 60 ≤ 100p + 19.6 p(1 − p)
= p : (60 − 100p)2 ≤ 19.62 p(1 − p)
=[0.502; 0.691].
Doing the same that in question
Let δ be a test procedure and S1 the critical region of δ , Θ the parameter space.
π(θ|δ), called the power function of the test δ , is determined by the relation
π(θ|δ) = P(X ∈ S1 |θ),
Suppose that
mean
µ
X1 , · · · , Xn
for
and
The function
θ ∈ Θ.
form a random sample from the normal distribution with unknown
and known variance
1,
and it is desired to test the following hypotheses:
H0 : 0.1 ≤ µ ≤ 0.2
H1 : µ < 0.1 or µ > 0.2.
H0 is rejected if either Xn ≤ c1 or
Xn ≥ c2 , and let π(µ|δ) denote the power function of δ . Suppose that the sample size is
n = 25. Determine the values of the constants c1 and c2 such that
Consider a test procedure
δ
such that the hypothesis
π(0.1|δ) = π(0.2|δ) = 0.07.
Solution:
Given
n = 25,
the mean of
Xi
is
µ
and variance is
1,
Xn ∼ N(µ, σ 2 /n) = N(µ, 1/25).
Thus
5(Xn − µ) ∼ N(0, 1).
Let
Y
be a standard normal distributed random variable and
Φ
its c.d.f.:
π(0.1|δ) = P(Xn ≤ c1 |µ = 0.1) + P(Xn ≥ c2 |µ = 0.1)
= P(5Xn − 0.5 ≤ 5c1 − 0.5|µ = 0.1) + P(5Xn − 0.5 ≥ 5c2 − 0.5|µ = 0.1)
= P(Y ≤ 5c1 − 0.5) + P(Y ≥ 5c2 − 0.5)
= Φ(5c1 − 0.5) + (1 − Φ(5c2 − 0.5)).
Similarly,
π(0.2|δ) = Φ(5c1 − 1) + (1 − Φ(5c2 − 1)).
We have to nd
c1
and
c2
such that
Φ(5c2 − 0.5) − Φ(5c1 − 0.5) = 0.93 = Φ(y + 0.5) − Φ(x + 0.5)
3
(1)
Probability and statistics
not due
and
Φ(5c2 − 1) − Φ(5c1 − 1) = 0.93 = Φ(y) − Φ(x)
where we set
x = 5c1 − 1
It is not hard to see that
and
(2)
y = 5c2 − 1.
x + 0.5 = −y .
In fact, taking the dierence of (1) and (2), one gets:
Φ(y + 0.5) − Φ(y) = Φ(x + 0.5) − Φ(x).
t 7→ Φ(t + 0.5) − Φ(t), any value can have at
most two preimages, and the fact that 0.93 = Φ(y)−Φ(x) implies x 6= y . Hence x+0.5 = −y .
By studying the monotonicity of the function
Substitute it in (2):
0.93 = 1 − Φ(−y) − Φ(x) = 1 − Φ(x + 0.5) − Φ(x)
0.07 = Φ(x + 0.5) + Φ(x).
It has a unique solution
x = x0
(by monotonicity of
−x0 − 0.5.
Hence
c1 = x0 /5 + 1/5
and
c2 = −x0 /5 + 1/10.
4
x 7→ Φ(x + 0.5) + Φ(x)).
And
y =