STAT 431: HOMEWORK 3
Problem 1. (5 points) Suppose X1 , . . . , Xn are i.i.d. draws from a population with mean µ
and known variance σ 2 . Consider the following two situations:
Situation
1: You perform the one-sided hypothesis test
H0 : µ ≥ µ0
and
H1 : µ < µ0
at signicance level α.
Situation
test
2: Using the same data set, your friend performs the one-sided hypothesis
H0 : µ < µ0
and
H1 : µ ≥ µ0 .
at signicance level α.
True or false: You reject your null hypothesis if and only if your friend does not reject his/her
null hypothesis. Explain your answer.
Without loss of generality, let's assume σ is known. Suppose we are testing the
hypotheses at signicance level α. In situation 1, we reject H0 whenever X < µ0 − zα √σn
(i.e., we reject H0 if X is in the red shaded region in the left-panel of the gure below.) In
situation 2, we fail to reject H0 whenever X < µ0 + zα √σn (i.e., we fail to reject H0 if X is in
the blue shaded region in the right panel of the gure below).
Solution:
1
Note that if we reject H0 in situation 1, we will fail to reject H0 in situation 2 (in terms of
the gure, if X is in the red region, it is also in the blue region). However if we fail to reject
H0 in situation 2, we do not necessarily reject H0 (in terms of the gure, if X is in the blue
shaded region, it is not necessarily in the red shaded region).
Hence, the statement is false.
Problem 2. (5 points) According to historical records, a large department store found that
1 out of every 14 people entering the store engaged in shoplifting. To help alleviate the
problem, the store hired additional security guards. Suppose the store now chooses 300
shoppers at random and follows their movements by camera. If 18 of the 300 shoppers are
found to engage in shoplifting, is there signicant evidence to conclude that the strategy is
working?
Solution: We want to test H0 : p ≥ 1/14 against H1 : p < 1/14. Our test statistic is
p̂ − 1/4
q
1
300
1
4
× × 1−
1
4
= −0.769
and our p-value is computed below.
> z <− (18 / 300 − 1/ 14) / s q r t (1 / 300 * 1/14 * 13/ 14)
> pnorm ( z , mean = 0 , sd = 1)
[ 1 ] 0.2210609
2
Since the p-value is larger than 0.05, we cannot reject the null hypothesis. So while the sample proportion of shoplifters is smaller than the historical record (6% compared to 7.14%), it
is not signicantly smaller. Thus there is not enough evidence (at any reasonable signicance
level) to conclude that the strategy is working.
Problem 3.
(a) (5 points) Let X be a random variable with a continuous cumulative distribution
function F (so that the quantile function F −1 is well-dened). Dene a new random
variable Y = F (X). Show that Y has a uniform distribution on the interval [0, 1].
Solution:
show that
To show that Y has a uniform distribution on the unit interval we need to
P r(X ≤ x) = x for x ∈ [0, 1] .
Since F −1 is well-dened we have that F is a bijection between the values that X
may take and the interval [0, 1]. What does this imply? First, it means that the
event {X ≤ x} for x in the range of X is equivalent to the event {F (X) ≤ F (x)}.
Second, it means that for all y ∈ [0, 1] there is an x ∈ Range(X) such that y = F (x).
The rst fact implies that
F (x) = P r(X ≤ x) = P r(F (X) ≤ F (x)) = P r(Y ≤ F (x)) .
So we have that P r(Y ≤ F (x)) = F (x) for all x ∈ Range(X). The second fact
implies that P r(Y ≤ y) = y for all y ∈ [0, 1], completing our proof.
(b) (5 points) Let X1 , X2 , . . . , Xn be i.i.d. random samples from a N (µ, σ 2 ) distribution,
with known σ 2 . Consider testing
H0 : µ = 0
versus
H1 : µ > 0
√ 0 . For observed data x1 , x2 , . . . , xn the p-value is
based on the test statistic Z = X̄−µ
σ/ n
x̄−µ
√ 0 . Thus, over repeated random samples with the same
p = 1 − Φ(z), with z = σ/
n
sample size n, the p-value
P = 1 − Φ(Z)
is a random variable. Use part (a) to show that P has a uniform distribution on the
interval [0, 1], and use this result to further show that a test which rejects H0 when
the p-value is less than α has the probability of making Type I error equal to α.
Solution: Using part (a) we know that Φ(Z) ∼ U[0, 1] so we can write P = 1 − U
where U ∼ U[0, 1]. We begin by showing that P is uniform on [0, 1] when we assume
that H0 is true. One way to see this is that U 0 = U − E[U ] = U − 1/2 is symmetric
3
about 0 and thus U − 1/2 and 1/2 − U have the same distribution. Consequently,
1/2+(U −1/2) = U and 1/2+(1/2−U ) = 1−U have the same distribution. Another
way would be to show that P and U have the same distribution function.
Problem 4. Let X1 , X2 , . . . , Xn be i.i.d. random samples from N (µ, σ2 ). Consider the
problem of testing
H0 : σ 2 = σ02
and H1 : σ 2 > σ02 ,
at signicance level α, using the test that rejects H0 if
(n − 1)S 2
> χ2n−1,α ,
σ02
where S 2 is the sample variance.
a (3 points) Find an expression for the power of this test in terms of the χ2n−1 distribution if the true σ 2 = cσ02 , where c > 1.
2
2
Solution: The expression for power at σ = cσ0 is
P
(n − 1)S 2
> χ2n−1,α σ 2 = cσ02
σ02
.
has a chi-squared distribution under H0 but not under H1 . HowRecall that (n−1)S
σ02
2
ever, we do know that (n−1)S
is chi-squared distributed with n−1 degrees of freedom
cσ02
when this particular alternative, σ 2 = cσ02 is true. Using this fact we can express the
power in terms of the distribution function of the χ2n−1 , Fχ2n−1 (·):
2
Power
(cσ02 )
=P
(n − 1)S 2
> χ2n−1,α σ 2 = cσ02
2
σ0
χ2n−1,α
(n − 1)S 2
=P
>
cσ02
c
2
χn−1,α
= 1 − Fχ2n−1
.
c
2
σ =
cσ02
Notice that as c → ∞ the power increases to 1 and that as c → 1 then the power
converges to the level, α.
(b) (2 points) Find the power of this test if α = 0.05, n = 16, and c = 4, that is, the true
σ 2 is four times the value being tested under H0 .
Solution:
To compute this we run the following R code,
4
> alpha <− 0 . 0 5
> n <− 16
> c <− 4
> pow <− 1 − p c h i s q ( q c h i s q (1 − alpha , df = n −1)/c , df = n −1)
> pow
[ 1 ] 0.9752557
Problem 5. On Groundhog Day, February 2, a famous groundhog in Punxsutawney, PA is
used to predict whether a winter will be long or not based on whether or not he sees his
shadow. We collected data on whether he saw his shadow or not, and kept them in the le
groundhog.txt.
Now, suppose we want to know whether this information says anything about whether or
not we will experience a rainy winter in California. For this, we found rainfall data, and
saved it in the le rainfall.csv.
(a) (4 points) Make a boxplot of the average rainfall in Northen California comparing
the years the groundhog sees his shadow versus the years he does not.
3
4
5
6
7
Solution:
N
Y
ghog
<− read . csv ( " groundhog . t x t " )
r a i n f a l l <− read . csv ( " r a i n f a l l . csv " )
a l l . data <− merge ( ghog , r a i n f a l l ,
by . x = " year " , by . y = "WY" )
a l l . data $Avg <− a l l . data $ Total /12
boxplot (Avg ~ shadow , data = a l l . data )
5
(b) (3 points) Construct a 90% condence interval for the dierence between the mean
rainfall during the month of February in years the groundhog sees his shadow and
years he does not.
Solution: Now we just consider February rainfall. Let Ȳ be the average February
rainfall in years when the groundhog saw his shadow and N̄ the average when he
did not. Since there are very few observations, we'll probably want to assume that
February rainfall is normal so that we can use t critical values to construct our 90%
CI. Our CI will take the form

s
Ȳ − N̄ − tν,0.05
Where ν =
s2Y
nY
s2Y
+
s2N
s
, Ȳ + N̄ − tν,0.05
s2Y
+
s2N


nN
2 2 2
2 2
s2N
sY
sN
+ nN /
/(nY − 1) + nN /(nN − 1) ≈ 5.77.
nY
nY
nN
We can use R to compute the interval
6
nY
>
>
>
>
>
>
>
>
>
>
>
i .Y <− a l l . data $shadow=="Y"
i .N <− a l l . data $shadow=="N"
s .Y <− sd ( a l l . data $Feb [ i .Y] )
s .N <− sd ( a l l . data $Feb [ i .N] )
n .Y <− sum( i .Y)
n .N <− sum( i .N)
Y. bar <− mean ( a l l . data $Feb [ i .Y] )
N. bar <− mean ( a l l . data $Feb [ i .N] )
s
<− s q r t ( s .Y^2/n .Y + s .N^2/n .N)
nu
<− s ^4/ ( ( s .Y^2/n .Y) ^2/ ( n .Y−1)+(s .N^2/n .N) ^2/ ( n .N−1) )
c (Y. bar −N. bar − qt ( 0 . 9 5 , df=nu ) * s ,Y. bar −N. bar + qt ( 0 . 9 5 , df=
nu ) * s )
[ 1 ] − 3.878695 8.933445
The 90% CI is [−3.88, 8.93].
(GRADING NOTE: If the student uses a pooled standard error estimator
correctly that answer should also receive full credit.)
(c) (3 points) At level, α = 0.05, would you reject the null hypothesis that the average
rainfall in Northern California during the month of February was the same in years
the groundhog sees his shadow versus years he does not? What assumptions are you
making in forming your condence interval and in your hypothesis test?
The interval formed in part (b) includes zero which means we cannot
reject the null hypothesis that the average rainfall in Northern California during the
month of February was the same in years the groundhog sees his shadow versus years
he does not at level α = 0.1. Consequently, we cannot reject this null at α = 0.05
either. We have assumed that the February rainfall data for years the g'hog sees
his shadow and for years that he doesn't are independent samples and are normally
distributed. However, we do not assume they have equal variance (it might actually
be a good idea to do this since there are so few data, this would give us a better
estimate of s... full credit will be given if you use the equal variance assumption and
compute s appropriately).
Solution:
Problem 6. Suppose we are given two independent random samples of sizes n1 and n2 from
Bernoulli populations with parameters p1 and p2 . Let p̂1 and p̂2 be the corresponding sample
7
proportions. Consider the problem of testing the hypotheses:
H0 : p1 − p2 = 0 versus H1 : p1 − p2 6= 0.
(a) (1 point) Read Section 9.2.1 from the book, and write down the test statistic for the
above hypothesis.
Solution:
There are two possible test statistics we can use. The rst is
p̂1 − p̂2
p̂1 − p̂2 − (p1 − p2 )
=q
.
t= q
p̂2 (1−p̂2 )
p̂2 (1−p̂2 )
p̂1 (1−p̂1 )
p̂1 (1−p̂1 )
+
+
n1
n2
n1
n2
Since the null hypothesis states that p1 = p2 we can use the pooled proportion
p̂ =
n1 p̂1 + n2 p̂2
n1 + n2
to estimate the variance which gives us that
p̂1 − p̂2 − (p1 − p2 )
t= r
.
1
1
p̂(1 − p̂) n1 + n2
Either is OK to use for a hypothesis test while the second is not appropriate when
constructing a condence interval. (GRADING NOTE: part (b) is solved using
the rst version but full credit should be given if you use the second. Part
(c) using the second version should only get partial credit)
(b) Now, suppose a random sample of 220 female and 210 male coee drinkers was
selected and interviewed. The result was that 71 women and 58 men indicated a
preference for decaeinated coee.
(i) (2 points) Should we conclude, at a 5% signicance level, that the proportion of
female coee drinkers who prefer decaeinated coee diers from the proportion
of male coee drinkers who prefer decaeinated coee?
Let pM and pF be the proportion of male and female coee drinkers
who prefer decaeinated coee. We want to test H0 : pM = pF vs H1 : pM 6= pF .
Observe that
Solution:
p̂M
pM (1 − pM ) pF (1 − pF )
− p̂F ∼ N pM − pF ,
+
nM
nF
8
We compute
p̂M − p̂F
q
p̂M (1−p̂M )
nM
+
= −1.05
p̂F (1−p̂F )
nF
If Z ∼ N (0, 1) then the p-value is
P (|Z| ≥ 1.395) = 0.291,
and we fail to reject the hypothesis that pM = pF .
(ii) (2 points) Construct a 95% condence interval for the dierence in proportions
between male and female coee drinkers who prefer decaeinated coee.
Solution:

s
p̂M − p̂F − z α
2
The condence interval has the form
p̂M (1 − p̂M ) p̂F (1 − p̂F )
+
, p̂M − p̂F + z α2
nM
nF
s

p̂M (1 − p̂M ) p̂F (1 − p̂F ) 
+
nM
nF
We compute the 95% CI as [−0.148, 0.025].
> s <− s q r t ( ( phat .m * (1 − phat .m) ) /n .m + ( phat . f * (1 −
phat . f ) ) /n . f )
> c ( phat .m − phat . f − qnorm ( 0 . 9 7 5 , 0 , 1) * s , phat .m −
phat . f + qnorm ( 0 . 9 7 5 , 0 , 1) * s )
[ 1 ] − 0.13298585 0.03991226
Problem 7. Consider the problem of testing
H0 : µ1 = µ2
versus
H1 : µ1 > µ2 ,
when the samples are independently drawn from two normal populations N (µ1 , σ12 ) and
N (µ2 , σ22 ), where σ12 and σ22 are assumed to be known. Let n1 and n2 be the sample sizes
and x̄ and ȳ be the sample means, respectively. The α-level test for H0 rejects if
x̄ − ȳ
z=p 2
> zα .
σ1 /n1 + σ22 /n2
(a) (4 points) Show that the power of the α-level test as a function of µ1 − µ2 is given by
µ1 − µ2
π(µ1 − µ2 ) = Φ −zα + p 2
σ1 /n1 + σ22 /n2
9
!
.
We rst observe that under the alternative µ1 − µ2 > 0, our test statisp
tic (x̄ − ȳ)/ σ12 /n1 + σ22 /n2 is normal with standard deviation 1 and mean (µ1 −
p
µ2 )/ σ12 /n1 + σ22 /n2 . We now start with the general denition of power and use this
fact to write it in terms of the standard normal CDF:
Solution:
π(µ1 − µ2 ) = P (reject H0 | H1 )
=P
!
x̄ − ȳ
=p 2
> zα µ1 − µ2 > 0
σ1 /n1 + σ22 /n2
x̄ − ȳ − (µ1 − µ2 )
µ1 − µ2
=P = p 2
> zα − p 2
2
σ1 /n1 + σ2 /n2
σ1 /n1 + σ22 /n2
!
µ1 − µ2
= 1 − Φ zα − p 2
σ1 /n1 + σ22 /n2
!
µ1 − µ2
= Φ −zα + p 2
.
σ1 /n1 + σ22 /n2
!
µ1 − µ2 > 0
(b) (3 points) For detecting a specied dierence µ1 − µ2 = δ > 0 show that for a xed
total sample size n1 + n2 = N , the power is maximized when
n1 =
σ1
N
σ1 + σ2
n2 =
σ2
N.
σ1 + σ2
(Here we are ignoring integer restrictions on n1 and n2 ).
In order to maximize π(µ1 − µ2 ) in terms of n1 , n2 given that we have
a total sample size of N we need to make the term inside of Φ(·) as big as possible.
Since µ1 − µ2 = δ > 0, this is equivalent to making the term σ12 /n1 + σ22 /n2 as small
as possible such that n1 + n2 = N . Now notice that we can write n2 as N − n1 and
write this term to minimize as the function
Solution:
f (n1 ) = σ12 /n1 + σ22 /(N − n1 ) .
One can easily check that this function is strictly convex for n1 > 0 and thus setting
the derivative to zero (rst order condition) and solving for n1 will yield the unique
10
minimizer:
f 0 (n1 ) = σ22 /(N − n1 )2 − σ12 /(n21 ) = 0
⇒ n1 = (σ1 /σ2 )(N − n1 )
⇒ n1 (σ2 + σ1 )/σ2 = (σ1 /σ2 )N
⇒ n1 = N σ1 /(σ1 + σ2 ) .
From the relation n2 = N − n1 we see that n2 = N σ2 /(σ1 + σ2 ), which is our desired
result.
(c) (3 points) Show that the smallest integer total sample size required to guarantee
power at least 1 − β , when µ1 − µ2 = δ > 0, is given by
(zα + zβ )(σ1 + σ2 )
N=
δ
2
.
If we are interested in the smallest N to guarantee a certain power, then
we should use the result from part (b) to choose the allocation of the total sample
across the two populations. Since Φ−1 is well dened we have that
Solution:
!
δ
Φ −zα + p 2
σ1 /n1 + σ22 /n2
=1−β
δ
=⇒ −zα + p 2
= zβ
σ1 /n1 + σ22 /n2
δ
=⇒ p
= zα + zβ
(σ1 + σ2 )σ1 /N + (σ1 + σ2 )σ2 /N )
√
N
zα + zβ
=
=⇒ p
δ
(σ1 + σ2 )2
2
(zα + zβ )(σ1 + σ2 )
=⇒ N =
.
δ
Now since this might not return an integer, we must take the ceiling of this term to
get the smallest integer N to guarantee us power of at least 1 − β .
11
Problem 8. (5 points) To study the eectiveness of a certain commercial liquid protein diet,
the Food and Drug Administration sampled nine individuals who were entering a two-week
weight loss program. Their weights immediately before and 6 months after completing the
program are recorded below:
Person Weight before Weight after
1
197
185
2
212
220
3
188
180
4
226
217
5
170
185
6
194
197
7
233
219
8
166
170
9
205
202
Based on the data, would you conclude that the diet is eective for weight loss? Use a 5%
signicance level. What assumptions have you made?
This is a matched pairs design (since each person contributes two measurements,
one before the weight loss program and one after the program). For person i, let di be
the dierence between their weight after the program and their weight before the program
(weight after minus weight before). Let µd be the (population) average weight loss following
the program. We want to test H0 : µd ≥ 0 vs H1 : µd < 0. 1 Let D and sD be the sample
average and standard deviations of the observed dierences d1 , . . . , d9 . Then under the null
we have
Solution:
D
√ ∼ t9−1
sd / 9
and can proceed as we have previously.
We compute our t-statistic to be -0.504 and the p-value to be 0.301 So we cannot reject
the null that the average weight after the program is at least the average weight before the
program (i.e. there's weak evidence that the plan doesn't work).
The crucial assumption is that the di 's are i.i.d normal random variables. Note, that we do
not need the weights themselves to be normally distribution, just the dierence.
1From
the standpoint of the regulatory agency, approving a weight-loss regimen that doesn't work (or
increases weight) is much worse than rejecting a regimen that does reduce weight. In light of this, we set H0
so that the rst situation corresponds to a Type I error.
12
Problem 9. (5 points) A study was instigated to see if southern California earthquakes of
at least moderate size (having values of at least 4.4 on the Richter scale) are more likely to
occur on certain days of the week than on others. The following data were obtained for 1100
earthquakes:
Day
Sun Mon Tues Weds Thurs Fri Sat
Number of Earthquakes 156 144 170 158
172 148 152
(a) Test the hypothesis that an earthquake is equally likely to occur on any of the seven
days of the week. Use a 5% signicance level.
If earthquakes are equally likely to occur on any of the seven days, we
would expect to see counts of 157.1429 in each cell of the table. Under the null we
the approximate distribution of our test statistic
Solution:
7
X
(observedi − expected )2
i=1
expectedi
i
∼ χ26 .
We compute χ20.05,6 = 12.591 and our test-statistic is 4.269. Since 4.269 < 12.591, we
fail to reject the hypothesis that earthquakes are equally likely to occur on any of the
seven days of the week.
> n . c <− 7
> obs <− c ( 1 5 6 , 144 , 170 , 158 , 172 , 148 , 152)
> expected <− rep (sum( obs ) /n . c , times = n . c )
> c r i t i c a l . value <− q c h i s q ( 0 . 9 5 , df = n . c − 1)
> c r i t i c a l . value
[ 1 ] 12.59159
> t e s t . s t a t <− sum( ( obs − expected ) ^2/ expected )
> test . stat
[ 1 ] 4.269091
(b) What is the p-value of the data?
The p-value is the probability that χ26 random variable exceeds our observed test-statistic.
Solution:
> 1 − p c h i s q ( t e s t . s t a t , df = n . c − 1)
[ 1 ] 0.640312
13
(c) What kind of condence interval might be used here?
We may try to construct a condence interval for any parameter that summarizes
the "dierences" between the parameters. For example, the variance of of the probability of earthquakes per day, or the dierence between the max and min probability.
However, nding a condence interval may be dicult using the methods we have
talked about. None of the standard intervals t this directly.
14