Solution of AMS312_2008 Practice Mid-term 1
AMS312.01
Practice Midterm Exam #1
Spring, 2009
Instructions: This is a close book exam. Anyone who cheats in the exam shall receive a
grade of F. Please provide complete solutions for full credit. Good luck!
1. A biologist wishes to estimate the effect of an antibiotic on the growth of a particular
bacterium by examining the mean amount of bacteria present per plate of culture when a
fixed amount of the antibiotic is applied. Previous experimentation with the antibiotic on
this type bacterium indicates that the standard deviation of the amount of bacteria present
per plate is approximately 12 cm 2 .
(1) Please derive the general formula for sample size calculation in this type of scenario
based on a maximum error of E at a confidence level of 100(1-α) %.
(2) Plug the information given in this particular problem to determine the number of
plates necessary to estimate the mean amount of bacteria present within 6 cm 2 with a
probability of 99%.
Solution: This is a sample size determination problem for the inference of  based on
the maximum error E.
x 
Z
N (0,1)
(1) P.Q.

n
P(| x   | 5)  0.95
E
x 
E
P(


)  1
 n  n  n
Z
E
 Z  n  ( 2 ) 2
2
 n
E
(2)We just need to plug in the numbers:
2.575*12 2
)  27
E=6,   1  0.99  0.01 ,   12 , Z  Z0.05  2.575 , n  (
2
6
i .i .d .
, X n ~ N (  ,  2 ) , be a random sample from the normal population. Please
2. Let X 1 ,
prove that
(1) X ~ N (  ,
(2) Z 
2
n
).
X 
~ N (0,1) .
/ n
Proof：
(1) M X (t )  E (e )  E (e
tX
X
Hence
t
1
n
 Xi
)  E (e
t
n
 Xi
n
)   E (e
i 1
1
N ( ,  2 )
n
1
t
Xi
n
)  ( E (e
t
X
n
))  e
n
t 
1 2 2
t
2 n
Solution of AMS312_2008 Practice Mid-term 1
(2) M Z (t )  E (e
Hence Z
X 

t
)  E (e
 t


t

X
)e
 t
t


E (e
X
)e
 t

t
MX ( )  e

 t

1
e
t 
 
t   2  
2

2
e
1 2
t
2
N (0,1)
3. University officials are planning to audit 1586 new appointments to estimate the
proportion p who have been incorrectly processed by the Payroll Department.
(1) How large does the sample size need to be in order for X/n, the sample proportion,
to have an 85% chance of lying within 0.03 of p? Please first derive the general
formula for sample size calculation based on a maximum error of E and a
confidence level of 100(1-α)%.
(2) Past audits suggest that p will not be larger than 0.10. Using this information,
recalculate the sample size asked for in Part (1).
Solution:
(1)
General Formula:
x
p
E
E
n
P(


)  1
pˆ (1  pˆ ) / n
pˆ (1  pˆ ) / n
pˆ (1  pˆ ) / n
nE 2
E
 ( Z  ) 2 , that is:
 Z  Hence,
p(1  p)
pˆ (1  pˆ ) / n
2
2
(Z  )2
n
2
2
E
p(1  p) 
( Z ) 2
2
4E 2
We plug in the numbers in this problem: E=0.03,   1  0.85  0.15
Hence n should be at least 576.
(Z  )2
(2) Now we know a boundary of p. We can use the formula: n  22 p(1  p ) . Since
E
it is an increasing function of p when p is no larger than 0.10, therefore:
(Z  )2
( Z ) 2
2
2
n
p(1  p) 
0.1*(1  0.1)
E2
4E 2
That is n should be at least 208.
i .i .d .
4. Let X 1 ,
, X n ~ N (  ,  2 ) , be a random sample from the normal population where
both  and  2 are unknown. Please derive
(1) The maximum likelihood estimators for  and  2 .
(2) The method of moment estimators for  and  2 .
(3) The best estimator for  assuming that  2 is known.
Solution:
(1) MLEs for  and  2 :
2
Solution of AMS312_2008 Practice Mid-term 1
Likelihood function:
n
n
n
i 1

1
L   f ( X i ; , 2 )  
2 2
i 1
n
ln L  (n) ln( 2 2 ) 
(X
i 1
i
( X i   )2
e
2 2
n
(
1
2 2
)e
 ( X i   )2
 i1
2 2
  )2
2 2
n
 ln L
 Xi
 2 ( X i   )  0  ˆ 

n
i 1
n
 ln L
1
 ( n ) 2 
2
2 

EX  X 
(2)
MOME:
X
i
n
 ( X i   )2
i 1
4
2
EX 2 
 ˆ 2 
(X
i 1
n
(3) Use the Cramer-Rao lower bound:
i
n

(X
i 1
i
 X )2
n
n
From problem 2(a), we already know that

1
f X ( , ) 
2
d ln f

d
i 1
n
 X )2
X
ln f  ln(
 0  ˆ 2 
 ( X i  ˆ )2
 ˆ  X
n
 X i2
n
2 2
1
)
2 2
( Xi  )
1
N ( ,  2 )
n
( X i   )2
e
2 2
( X i   )2
2 2
2
d 2 ln f
1
 2
2
d

Hence, the lower bound of variance is
2
n
.
5. John Pauzke, president of Cereals Unlimited Inc., wants to be very certain that the
mean weight μ of packages satisfies the package label weight of 16 ounces. The
packages are filled by a machine that is set to fill each package to a specified weight.
However, the machine has random variability measured by σ2. John would like to have
strong evidence that the mean package weight is above 16 ounces. George Williams,
quality control manager, advises him to examine a random sample of 25 packages of
cereal. From his past experience, George knows that the weight of the cereal packages
3
Solution of AMS312_2008 Practice Mid-term 1
follows a normal distribution with standard deviation 0.4 ounces. At the significance
level α = 0.05:
(a) What is the decision rule (rejection region) in terms of the sample mean?
(b) What is the decision of your test if a sample of 25 packages of cereal yields a mean of
16.3 ounces? What is the p-value of your test?
(c) What is the power of the test when μ= 16.23 ounces? (Please derive the formula for
the power of the test first, and then calculate the desired power.)
Solution:
(a) x   or Z  x/n is a pivotal quantity for this test problem.
If we select Z as the pivotal quantity, Z 0 
x  0
/ n
 x  0 C *  0

P

|   0 
P (Reject H0|H0)=P (Z0>|μ=μ0) = 
 / n  / n

Therefore we should reject H0 in favor of Ha if Z0> where  =1.645.

0.4
From the above, we have C *  0 
1.645  16 
1.645  16.13 . Therefore we
5
n
should reject H0 in favor of Ha if x >C* where C* =16.13.
(b) Since 16.3>16.13, we should reject H0 in favor of Ha. P-Value is the tail area.
16.3  16 1.5

 3.75 . Therefore P-Value = 0.0001.
0.4 / 5
0.4
(c) Power= 1-=P(reject H0|Ha)=P(Z0>|μ=μa)=P(
x  0
>|μ=μa)
/ n
x  0  a  0
  0
x  a
  0

 Z  a
|   a ) = P(
 Z  a
|   a )
/ n / n
/ n
/ n
/ n
  0
16.23  16
= P( Z  Z  a
) = P( Z  1.645 
) = P ( Z  1.23) =1-0.1093=0.8907
/ n
0.4 / 25
= P(
6. Jerry is planning to purchase a sporting goods store. He calculated that in order to
cover basic expenses average daily sales must be at least $525. He checked the daily sales
of 36 randomly selected business days. And he found that the average daily sale for these
days is $565 with a standard deviation of $150.
(a) At the significance level =0.05, can Jerry conclude that the average daily sale is
higher than $525? What is the p-value?
4
Solution of AMS312_2008 Practice Mid-term 1
(b) In order to estimate the average daily sale of the store to within $20 with 95%
reliability, how many days should Jerry sample? Please derive the general formula first.
(c) If Jerry could only check the daily sales of 9 randomly selected business days (instead
of 36 randomly selected days). Suppose the daily sale for these 9 days are 510, 537, 548,
592, 503, 490, 601, 499 and 640 respectively. At the significance level =0.05, can Jerry
conclude that the average daily sale is higher than $525? What assumptions are necessary
for your test? What is the p-value of your test?
Solution. Inference on one population mean. n=36. Population variance  2 is unknown.
If you know the data, then you do normality test (e.g., Sharpio-Wilk test) to see if the
sample is from normal distribution. If the population is normal, then we only use tdistribution.
If it’s not normal but the sample size is large (>=30), the pivotal quantity
Z
X 
~ N (0,1) (by Central Limit Theorem and Slusky Theorem).
S/ n
(a) X  565 , S=150.
 H 0 :   0  525

 H a :   0
Note: If X  505 (<525), then you should notice that H a is not suitable.
Test statistic: Z 0 
X  0 565  525

 1.6
150 / 6
S/ n
At the significance level , we reject H 0 if Z 0  z . Here z  z0.05  1.645 . Since
z0  1.6  1.645  z , we cannot reject H 0 .
P-value= P(Z  z0 | H 0 )  P(Z  1.6)  0.0548   . We can not reject H 0 .
Note: P-value= P( X  565 | H 0 )  P( Z  1.6 | H 0 ) .
(b). First we derive the general formula.
5
Solution of AMS312_2008 Practice Mid-term 1
P( X    E )  1  
P( E  X    E )  1  
E
X 
E


)  1
/ n / n / n
z 
E

 z / 2  n  (  / 2 ) 2
E
/ n
P(
Next we plug in the values to obtain the answer for the given problem.
n(
z / 2 2 1.96*150 2
) (
)  216.09  217 .
E
20
(c) Suppose from the Sharpio-Wilk test, we know the data/sample is from a normal
population.
 H 0 :   0  525

 H a :   0
X  546.67 , S=53.09.
Test statistic: T0 
X  0 546.67  525

 1.22 .
53.09 / 3
S/ n
At the significance level , we reject H 0 if T0  tn 1,  t8,0.05  1.860 . Since
T0  1.22  t8,0.05  1.860 , we cannot reject H 0 .
P-value= P(t  t0 | H 0 )  P(t  1.22)  0.1 . Since P-value > α = 0.05, we cannot reject H 0 .
6