# Homework Solution Week 8 - Department of Statistics and Probability

```Name:
ID:
Homework for 2/24 Due 3/12
1. Let X1 , . . . , Xn be a random sample from a normal distribution N (&micro;, σ 2 ),
where &micro; is the unknown parameter and σ 2 is assumed to be known. The
hypotheses are
H0 : &micro; = &micro;0
v.s.
HA : &micro; &gt; &micro;0
Consider the test that reject H0 if X &gt; c, where X is the sample mean.
c − &micro;0
√
a. Show that the significance level of this test is 1 − Φ
, where
σ/ n
Φ is the cdf of the standard normal random variable, that is, Φ(z) =
P(Z ≤ z) with Z ∼ N (0, 1).
b. Show that if the test is required to have significance level α, then
σ
c = z1−α √ + &micro;0 . Here z1−α satisfies P(Z ≤ z1−α ) = 1 − α.
n
c. Show that if the
sample gives X = x̄, then the p-value in this case is
x̄ − &micro;0
√
1−Φ
σ/ n
d. Show that
if in fact
&micro; = &micro;1 with &micro;1 &gt; &micro;0 , then the power of the test
c − &micro;1
√
is 1 − Φ
. In particular, if the significance level is α, then
σ/ n &micro;0 − &micro;1
√
.
the power is 1 − Φ z1−α +
σ/ n
a. Since the significance level is the probability of making a type I error,
we have
α = P(reject H0 |H0 ) = P(X &gt; c|H0 )
c − &micro;0 σ2
X − &micro;0
√ &gt;
√ H0
=P
X ∼ N &micro;0 ,
under H0
n
σ/ n
σ/ n
c − &micro;0
c − &micro;0
√
√
=P Z&gt;
=1−P Z ≤
σ/ n
σ/ n
c − &micro;0
√
.
=1−Φ
σ/ n
b. From the above derivation, we have
c − &micro;0
√
α=1−P Z ≤
,
or
σ/ n
c − &micro;0
√
P Z≤
= 1 − α.
σ/ n
By the definition of z1−α , we have
z1−α =
c − &micro;0
√
σ/ n
or equivalently,
σ
c = z1−α √ + &micro;0 .
n
c. Since the p-value is the probability of making type I error when we
reject H0 based on the sample, we have (by following a similar argument as in (a))
p = P(reject H0 based on the sample |H0 ) = P(X &gt; x̄|H0 )
X − &micro;0
x̄ − &micro;0 σ2
√
√
=P
H0
&gt;
X ∼ N &micro;0 ,
under H0
n
σ/ n
σ/ n x̄ − &micro;0
x̄ − &micro;0
√
√
=P Z&gt;
=1−P Z ≤
σ/ n
σ/ n
x̄ − &micro;0
√
=1−Φ
.
σ/ n
d. The probability of making type II error is
β = P(accept H0 |HA ) = P(X ≤ c|HA )
c − &micro;1 X − &micro;1
σ2
√ ≤
√ HA
=P
X ∼ N &micro;1 ,
under HA
n
σ/ n
σ/ n c − &micro;1
c − &micro;1
√
√
=Φ
.
=P Z≤
σ/ n
σ/ n
Therefore the power of the test is
power = 1 − β = 1 − Φ
c − &micro;1
√
σ/ n
.
When the significance level is α, we derived from part (b) that
σ
c = z1−α √ + &micro;0 .
n
Thus, by combining the above two equations, the power of this test
is
&micro;0 − &micro;1
√
1 − Φ z1−α +
.
σ/ n
2
2. The drying time of a certain type of paint under specified test conditions is
known to be normally distributed with mean value 75 min and standard
deviation 9 min. Chemists have proposed a new additive designed to
decrease average drying time. It is believed that drying times with this
additive will remain normally distributed with σ = 9. Because of the
expense associated with the additive, evidence should strongly suggest an
improvement in average drying time before such a conclusion is adopted.
Let &micro; denote the true average drying time when the additive is used. The
appropriate hypotheses are H0 : &micro; = 75 versus HA : &micro; &lt; 75. Only if H0
can be rejected will the additive be declared successful and then be used.
Assume a sample of size n = 25 is obtained.
a.
b.
c.
d.
If x̄ = 72.3, what is the conclusion using α = 0.01?
What is α for the test procedure that rejects H0 when X &lt; 69.816?
For the test procedure of part (b), what is β(70)?
If a level .01 test is used with n = 100, what is the probability of a
type I error when &micro; = 76?
Since the alternative is HA : &micro; &lt; &micro;0 (with &micro;0 = 75), we will use the test
with test statistic X and rejection region X &lt; c.
a. We can make a decision by two methods:
(i) Using rejection region:
According to Appendix [1’] (b), we see that the rejection region
for this pair of hypotheses is R = {X &lt; c}, where c is determined
σ
by the significance level α according to c = zα √ + &micro;0 .
n
Here α = 0.01, thus z0.01 = −2.33. Since &micro;0 = 75, σ = 9 and
n = 25, we have
9
c = (−2.33) &middot; √ + 75 = 70.806.
25
Since x̄ = 72.3 &gt; 70.806 = c and the rejection region is X &lt; c,
we do not reject (that is, accept) H0 .
(ii) Using p-value:
According to Appendix [1’] (c), we see that the p-value when
x̄ = 72.3 is given by
x̄ − &micro;0
72.3 − 75
√
√
p=Φ
=Φ
= Φ(−1.5) = 0.0668.
σ/ n
9/ 25
Since p-value &gt; α = 0.01, we do not reject H0 .
b. According to Appendix [1’] (a), the significance level α in this case is
c − &micro;0
69.816 − 75
√
√
α=Φ
=Φ
= Φ(−2.88) = 0.0020.
σ/ n
9/ 25
3
c. In this case, we assume &micro; = 70. Since we use the test in part (b), we
have c = 69.816. According to Appendix [1’] (d) (with &micro;1 = 70), the
power of the test is given by
c − &micro;1
69.816 − 70
√
√
Φ
=Φ
= Φ(−0.10) = 0.4602.
σ/ n
9/ 25
Therefore,
β(70) = 1 − 0.4602 = 0.5398.
4
3. Let &micro; denote the true average tread life of a certain type of tire. Consider
testingH0 : &micro; = 30, 000 versus HA : &micro; &gt; 30, 000 based on a sample of size
n = 16 from a normal population distribution with σ = 1500.
a. If x̄ = 30, 960 and a level α = 0.01 test is used, what is the decision?
b. If a level .01 test is used, what is β(30, 500)?
c. If x̄ = 30, 960, what is the smallest α at which H0 can be rejected
(based on n = 16)?
Since the alternative is HA : &micro; &gt; &micro;0 (with &micro;0 = 30000), we will use the
test with test statistic X and rejection region X &gt; c.
a. We can make a decision by two methods:
(i) Using rejection region:
According to Problem 1 (b), we see that the rejection region for
this pair of hypotheses is R = {X &gt; c}, where c is determined
σ
by the significance level α according to c = z1−α √ + &micro;0 .
n
Here α = 0.01, thus z0.99 = 2.33. Since &micro;0 = 30000, σ = 1500
and n = 16, we have
1500
c = 2.33 &middot; √ + 30000 = 30873.75.
16
Since x̄ = 30960 &gt; 30873.75 = c and the rejection region is
X &gt; c, we reject H0 .
(ii) Using p-value:
According to Problem 1 (c), we see that the p-value when x̄ =
30960 is given by
x̄ − &micro;0
30960 − 30000
√
√
p=1−Φ
=Φ
σ/ n
1500/ 16
= 1 − Φ(2.56) = 1 − 0.9948
= 0.0052.
Since p-value &lt; α = 0.01, we reject H0 .
b. In this case, we assume &micro; = 30500. According to Problem 1 (d) (with
&micro;1 = 30500 and α = 0.01), the power of the test is given by
&micro;0 − &micro;1
30000 − 30500
√
√
1 − Φ z1−α +
= 1 − Φ z0.99 +
σ/ n
1500/ 16
= 1 − Φ(1) = 1 − 0.8413 = 0.1587.
Therefore,
β(30500) = 1 − 0.1587 = 0.8413.
5
c. In this case, the smallest α at which H0 can be rejected (based on
n = 16) is exactly the p-value. (This is just another way to interpret
p-value.). In part (a), we already find the p-value:
x̄ − &micro;0
30960 − 30000
√
√
p=1−Φ
=Φ
σ/ n
1500/ 16
= 1 − Φ(2.56) = 1 − 0.9948
= 0.0052.
6
4. The melting point of each of 16 samples of a certain brand of hydrogenated
vegetable oil was determined, resulting in x̄ = 94.32. Assume that the
distribution of the melting point is normal with σ = 1.20.
a. Test H0 : &micro; = 95 versus HA : &micro; 6= 95 using a two-tailed level .01 test.
b. If a level .01 test is used, what is β(94), the probability of a type II
error when &micro; = 94?
Since the alternative is HA : &micro; 6= &micro;0 (with &micro;0 = 95), we will use the test
with test statistic X and rejection region |X − &micro;0 | &gt; c.
a. We can make a decision by two methods:
(i) Using rejection region:
According to Appendix [1”] (b), we see that the rejection region
for this pair of hypotheses is R = {|X − &micro;0 | &gt; c}, where c is
σ
determined by the significance level α according to c = z1− α2 √ .
n
Here α = 0.01, thus z0.995 = 2.575. Since &micro;0 = 95, σ = 1.2 and
n = 16, we have
1.2
c = 2.575 &middot; √ = 0.7725.
16
Since |x̄−&micro;0 | = |94.32−95| = 0.68 &lt; 0.7725 = c and the rejection
region is |X − &micro;0 | &gt; c, we do not reject (that is, accept) H0 .
(ii) Using p-value:
According to Appendix [1”] (c), we see that the p-value when
x̄ = 94.32 is given by
|x̄ − &micro;0 |
|94.32 − 95|
√
√
p=2 1−Φ
=2 1−Φ
σ/ n
1.2/ 16
= 2(1 − Φ(2.27)) = 0.0232.
Since p-value &gt; α = 0.01, we do not reject H0 .
b. In this case, we assume &micro; = 94. According to Appendix [1”] (d)
(with &micro;1 = 94 and α = 0.01), the power of the test is given by
&micro;0 − &micro;1
&micro;0 − &micro;1
√
√
1 − Φ z1− α2 +
+ Φ −z1− α2 +
σ/ n
σ/ n
95 − 94
95 − 94
√
√
= 1 − Φ z0.995 +
+ Φ −z0.995 +
1.2/ 16
1.2/ 16
= 1 − Φ(5.91) + Φ(0.76) = 1 − 1 + 0.7764
= 0.7764.
Therefore,
β(94) = 1 − 0.7764 = 0.2236.
7
Name:
ID:
Homework for 2/26 Due 3/12
1. The desired percentage of SiO2 in a certain type of aluminous cement is
5.5. To test whether the true average percentage is 5.5 for a particular
production facility, 16 independently obtained samples are analyzed. The
sample mean is x̄ = 5.25 and the sample standard deviation is s = 0.28
Suppose that the percentage of SiO2 in a sample is normally distributed.
a. Does this indicate conclusively that the true average percentage differs from 5.5, at significance level α = 0.01?
b. If the true average percentage is &micro; = 5.6 and a level α = 0.01 test
based on n = 16 is used, what is the probability of detecting this
departure from H0 , that is, the power of the test? (Assume the
population standard deviation is σ = 0.3.)
a. The hypotheses are
H0 : &micro; = 5.5
and HA : &micro; 6= 5.5
We can make a decision by two methods:
(i) Using rejection region:
The rejection region for this pair of hypotheses is R = {|X−&micro;0 | &gt;
c}, where c is determined by the significance level α according
s
to c = t1− α2 ,n−1 √ .
n
Here α = 0.01, thus t0.995,15 = 2.947. Since s = 0.28 and n = 16,
we have
0.28
c = 2.947 &middot; √ = 0.20629.
16
Since |x̄ − &micro;0 | = |5.25 − 5.6| = 0.35 &gt; 0.20629 = c and the
rejection region is |X − &micro;0 | &gt; c, we reject H0 .
(ii) Using p-value:
The p-value when x̄ = 94.32 is given by
|x̄ − &micro;0 |
|5.25 − 5.5|
√
√
p = 2 1 − Fn−1
= 2 1 − F15
s/ n
0.28/ 16
= 2(1 − F15 (3.571)) = 0.0028,
where Fn−1 is the cdf of the t-distribution with degrees of freedom n − 1. Since p-value &lt; α = 0.01, we reject H0 .
If we use TABLE 4 to find the p-value, we will have
F15 (3.571) ≥ F15 (2.947) = 0.995
Thus p &lt; 2(1 − 0.995) = 0.01. This will also lead to rejection of
H0 .
b. In this case, we assume σ = 0.3 is known. We also assume &micro; = 5.6.
According to Appendix [1”] (d) (with &micro;1 = 5.6 and α = 0.01), the
power of the test is given by
&micro;0 − &micro;1
&micro;0 − &micro;1
√
√
+ Φ −z1− α2 +
1 − Φ z1− α2 +
σ/ n
σ/ n
5.5 − 5.6
5.5 − 5.6
√
√
= 1 − Φ z0.995 +
+ Φ −z0.995 +
0.3/ 16
0.3/ 16
= 1 − Φ(1.24) + Φ(−3.91) = 1 − 0.8925 + 0.0000
= 0.1075.
9
2. The article “Uncertainty Estimation in Railway Track Life-Cycle Cost”
(J. of Rail and Rapid Transit, 2009) presented the following data on time
to repair (min) a rail break in the high rail on a curved track of a certain
railway line.
159
120
480
149
270
547
340
43
228
202
240
218
A normal probability plot of the data shows a reasonably linear pattern,
so it is plausible that the population distribution of repair time is at least
approximately normal. The sample mean and standard deviation are 249.7
and 145.1, respectively.
a. Is there compelling evidence for concluding that true average repair
time exceeds 200 min? Carry out a test of hypotheses using a significance level of .05.
b. Using σ = 150, what is the type II error probability of the test used
in (a) when true average repair time is actually 300 min? That is,
what is β(300)?
a. The hypotheses are
H0 : &micro; = 200
and HA : &micro; &gt; 200
We can make a decision by two methods:
(i) Using rejection region:
The rejection region for this pair of hypotheses is R = {X &gt; c},
where c is determined by the significance level α according to
s
c = t1−α,n−1 √ + &micro;0 .
n
Here α = 0.05, thus t0.95,11 = 1.796. Since &micro;0 = 200, s = 145.1
and n = 12, we have
145.1
c = 1.796 &middot; √
+ 200 = 275.23.
12
Since x̄ = 249.7 &lt; 275.23 = c and the rejection region is X &gt; c,
we do not reject H0 .
(ii) Using p-value:
The p-value when x̄ = 249.7 is given by
x̄ − &micro;0
249.7 − 200
√
√
p = 1 − Fn−1
= 1 − F11
s/ n
145.1/ 12
= 1 − F11 (1.187) = 1 − 0.8699 = 0.1301,
where Fn−1 is the cdf of the t-distribution with degrees of freedom n − 1. Since p-value &gt; α = 0.05, we do not reject H0 .
10
If we use TABLE 4 to find the p-value, we will have F11 (0.876) &lt;
F11 (1.187) &lt; F11 (1.363), or equivalently, 0.80 &lt; F11 (1.187) &lt;
0.90. Thus 0.10 &lt; p &lt; 0.20. This will also lead to non-rejection
of H0 .
b. In this case, we assume σ = 150 is known. We also assume &micro; = 300.
Since &micro;1 = 300 and α = 0.05, the power of the test is given by
200 − 300
&micro;0 − &micro;1
√
√
= 1 − Φ z0.95 +
1 − Φ z1−α +
σ/ n
150/ 12
= 1 − Φ(−0.66) = 1 − 0.2546
= 0.7454.
Therefore,
β(300) = 1 − 0.7454 = 0.2546.
11
```