STA 4322/5328 – Spr/2010 Major Exam1
PRINT NAME _____________
For all problems, the sample mean, sample variance, and sample proportion are:
Sample Mean : Y 
1 n
 Yi
n i 1
Sample variance :

1 n
 Yi  Y
n-1 i 1

2
^
Sample Proportion : p 
Y
where Y ~ Bin (n, p)
n
Q.1: Suppose Y1,…,Yn denote a random sample from a population with an exponential distribution whose
density is given by (where Y~Exp()):
1  e  y /  y  0,   0
f ( y)  
elsewhere
0
Note : If Y ~ Exp( ) then E (Y )   V (Y )   2
If Y(1) = min(Y1,…,Yn) then density function for Y(1) is:
n  e  ny1 /   1 /  *e  y1 /  *
g (1)  y1   
elsewhere
0
y1  0,   0
P.1.a. Give the mean and variance of Y(1)
ANSWER: E Y(1)    * 

n
 
2
V Y(1)    *   
n
2
P.1.b. Give a function of Y(1) that is an unbiased estimate of :
 1  hY(1) 
^
^
ANSWER:  1  nY(1)
P.1.c. Give the variance of your estimator in Part P.1.b.
2
^ 
2
2
ANSWER: V  1   V nY(1)   n V Y(1)   n 2   2
n
 
^
P.1.d. Give the variance of the unbiased estimator  2  Y
2
 ^  V (Y ) 
ANSWER: V  2  

n
n
 
Q.2: An engineer is interested in estimating the mean breaking strength of a new material to be used in
satellites. Because the material is so expensive, she can only obtain n = 9 test measurements. She obtains a
sample mean of 3000 psi and a sample standard deviation of 300 psi.
P.2.a: Obtain a 95% confidence interval for the true mean breaking strength for this material, assuming
measurements are approximately normally distributed.
 s 
 300 
ANSWER: Y  t.025,9 1 
  3000  2.306
  3000  230.6
 n
 9
P.2.b: In the future, based on larger study, she would like to estimate true mean breaking strength within
50 psi with 95% confidence. How many test measurements will she need? (Use z in calculations)
  
 300 
ANSWER 50  1.96
  1.96
 
 n
 n
 300 
2
n  1.96
  n  (1.96 * 6)  139
50


Q.3: A political analyst is interested in comparing the proportions of adult males and females who favor
universal health care. He obtains (independent) random samples of 1000 males and 1000 females,
and determines that 450 of the males and 550 of the females favor universal health care. Obtain a 95%
Confidence Interval for the difference in the proportions of males and females in the population who favor
universal health care by completing the following parts.
^
^
P.3.a: Point Estimate: p M  p F 
450 550

 0.45  0.55  0.10
1000 1000
P.3.b: Standard Error of Point Estimate:  ^
^
pM  pF

0.45(0.55) 0.55(0.45)

 0.02225
1000
1000
P.3.c: 95% Confidence Interval for pM-pF: -0.10 ± 1.96(0.02225)  -0.10 ± 0.0436  (-.1436,-.0564)
^
^
Q.4: Suppose that we have 2 estimators  1 , 2 of a parameter  with the following properties:
^ 
^ 
E  1   E  2   
 
 
^ 
^ 
V  1   4 V  2   8
 
 
^
^
^
^ ^ 
Cov 1 , 2   0 Let  3  a  1  (1  a)  2


^
P.4.a. Show that  3 is an unbiased estimator of 
^
^ 
 ^

^ 
^ 
E  3   E  a  1  (1  a)  2   aE  1   (1  a) E  2   a  (1  a)  
 


 
 
^ 
P.4.b. What is V   3  if we set a=1/2?
 
ANSWER:
^
^ 
 ^

^ 
^ 
V  3   V  a  1  (1  a) 2   a 2V  1   (1  a) 2 V  2   a 2 (4)  (1  a) 2 (8)
 


 
 
^ 
a  0.5  V  3   0.25(4)  0.25(8)  3
 
P.4.c. What value (a*) of a will give the minimum variance? What is the variance?
^
^ 
 ^

^ 
^ 
V  3   V  a  1  (1  a) 2   a 2V  1   (1  a) 2 V  2   a 2 (4)  (1  a) 2 (8)
 


 
 
^
 
V  3 
   2a(4)  2(1  a)(8)( 1) set
 0  8a*  16  16a *  a*  2 / 3
a
24
1
 ^  4
 V  3    (4)   (8) 
 2.667
9
  9
9
Q.5: A news organization reports that 45% of Americans are aware that the Winter Olympics are aware that the
Winter Olympics are in British Columbia this month, with a margin of error of ±4.5% (with 95% confidence).
Approximately what was the sample size for this sample (Hint: First convert your percentages to proportions
and use p=0.5 in standard error of sample proportion).
0.045  1.96
0.5(0.5)
n

n
1.96(0.5)
 21.778  n  475
0.045
Q.6. A lot acceptance sampling plan for large lots specifies that 144 items be randomly selected and that the lot
be accepted if less than or equal to 20 of the items selected do not conform to specifications. What is the
approximate probability that a lot will be accepted if the true proportion of non-conforming items in the lot is
0.20? Use the continuity correction in your calculation.
Y ~ Bin (n  144, p  0.20)  E (Y )  144(0.20)  28.80, V (Y )  144(0.20)(1  0.20)  23.04
20.5  28.80


PY  20 Y ~ Bin (n  144, p  0.20)   PY  20.5 Y ~ N (28.80,23.04)   P Z 
 1.73   .0418
23.04


Q.7: If Y1,…,Yn ~N(2) and are independent, then:
P.7.a: Give the sampling distribution of:
n  1S 2

2
~  n21
P.7.b: Derive the form of a 95% Confidence Interval for  (be very explicit in showing all steps in the
derivation).
 1


(n  1) S 2
2
1 
2


0.95  P  n21,.975 



P

 2

n 1,. 025 
2
2
2



(
n

1
)
S



n 1,. 025 
 n 1,.975
 (n  1) S 2
 (n  1) S 2
(n  1) S 2 
(n  1) S 2
 P 2
2  2
 P



 
  n21,.025
 n 1,.975 
 n21,.975
 n 1,.025





P.7.c: We observe a sample with n=8 and S2=9. Obtain a 95% Confidence Interval for 
 (8  1)(9) (8  1)(9) 
  (1.98,6.11)
 72,.975  16.013  72,.025  1.690  95% CI for  : 
,


1.690
16.013 