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1
Homework 5 STAT305B Spring 2017 Due 4/6(R) SOLUTION
Table 1. Some Handy-Dandy Test Statistics
For X ~ N ( x ;  X , X2 ) and associated iid data collection variables {X k }nk 1 :


(A): Z  (  X   X ) /( X / n ) ~ N (0,1)
;
(B): T   X   X ~ tn1

 X2 / n
2


(C): n  X2 /  X2 ~  n2 when  X is used ;
(D): (n  12) X ~  n21 when  X is used.
2
 /  X2
~ f n ,n when  X & Y are used ;
 X /  X2
(E): F   2X1
1
2
2
1
2
X

 X2 1 /  X2 1


(F): F   2
~ f n11,n2 1 when  X & Y used.
2
 X2 /  X2
PROBLEM 1(30pts) Suppose that you are interested in X=The act of measuring the power supply voltage for a laptop
design rated at 19.4 volts. This interest results from numerous complaints from customers that their supply voltage was

significantly different than the stated 19.4 volts. Using data related to 100 laptops, you found that  X  x  19.38 and

 X  s  .07 .

(a)(10pts) Because n=100 is large, you can assume that  X ~ N (  X , X / n ) . Since you do not know  X , use the tstatistic (B) in Table 1 to compute the 95% 2-sided CI estimator for  X . [NOTE: The only non-numerical entities in the

CI should be the random variables  X and  X .]
Solution:
0.95  Pr[t1  T99  t2 ] . Hence: t1=tinv(.025,99) = -1.9842 and t2=tinv(.9755,99) = 1.9842.





Write [t1  T99  t2 ]  t1   X   X  t2    X  t2  X2 / n   X   X  t1  X2 / n . Hence, the CI estimator is



 X2 / n

X

 

 0.19842 X ,  X  0.19842 X .
(b)(5pts) Compute the CI estimate associated with (a).
Solution: 19.38  0.19842(.07) , 19.38  0.19842(.07)  [19.366 , 19.394]
(c)(10pts) Compute the 95% 2-sided CI estimator for  X . [NOTE: The only non-numerical entity in the CI should be the
random variable  X .]


2
Solution: The appropriate statistic is (D). Hence, 0.95  Pr x1   99
 x2 gives x1= chi2inv(.025,99) = 73.36 and x2=
chi2inv(.975,99) =128.42. Hence,




  (n  1) X2
(n  1) X2
(n  1) X2
2
x


x




2
X
 1

 X2
x2
x1

 
 
99
99 



X X
  0.878 X   X  1.1617 X
   X
128.42
73.36 
 


(d)(5pts) Compute the CI estimate associated with (c).


Solution: 0.878 X   X  1.1617 X  0.878(.07)   X  1.1617(.07  0.0615   X  0.0813 . Hence, the CI estimate is

 0.0615 , 0.0813
 
 

2
PROBLEM 2(20pts) The length of the skulls of 15 fossil skeleton of an extinct species of bird has a sample mean of 6.58
cm and a sample standard deviation of 0.29 cm. Assume that the data collection random variables { X k }15
.
k 1 ~ iid N (  X ,  X )
(a)(7pts) Develop the 95% 2-sided confidence interval estimator for the unknown true mean,  X .
Solution: Pr[T  t1 ]  Pr[T  t2 ]  .025 . So: [tinv(.025,14),tinv(.975,14)] = [ -2.1448 2.1448]



  X







[t1  T  t 2 ]  t1   X
 t 2   t1 ( X / 15 )   X   X  t 2 ( X / 15 )   X  t 2 ( X / 15 )   X   X  t1 ( X / 15 ) .
 X / 15



 

(b(3pts) Use your result in (a) to compute the CI estimate.
Solution: CI   X  0.554 X   X   X  0.554 X  =[6.42 , 6.74]
(c)(7pts) Develop the 90% 2-sided confidence interval estimator for the unknown true standard deviation,  X .
Solution: Pr[  214  x1 ]  Pr[  214  x2 ]  .05 : [chi2inv(.05,14) , chi2inv(.95,14)] = [ 6.57 23.68]






[ x1  n  X2 /  X2  x2 ]  [ x1 / n  X2  1 /  X2  x2 / n  X2 ] , which results in:  X n / x2   X   X n / x1
(d)(3pts) Use your result in (c) to compute the CI estimate.


Solution:  X n / x2   X   X n / x1 =[0.223 , 0.423]

3
PROBLEM 3(20pts) The F35-A (shown at right) has a design (dry) thrust of
28,000 lbf. with a claimed  2 uncertainty of  400 lb f . The USAF ran dynamometer tests on n=5 engines, resulting in a sample mean of 27,780 lbf and a
sample standard deviation of 250 lbf. Assume that thrust is normally distributed,
and that the tested engines were randomly selected.
http://en.wikipedia.org/wiki/Lockheed_Martin_F-35_Lightning_II#Specifications_.28F-35A.29 ]
(a)(10pts) Your colleague feels that the sample mean is close enough to the design mean that, in this part,
you will assume that, indeed,  X  28000 lb f Arrive at an estimate of the 95%2-sided CI for  X .
Solution: x1=chi2inv(.025,5)=0.8312 & x2=chi2inv(.975,5) =12.8325
2
2
2 
2 
 2
 2
Hence: 0.95  Pr[ x1  5 2 /  2  x2 ]  Pr  5   2  5   Pr  5    5  so CI   5 , 5  = [156.05 613.15].
 x2
x1 

x2
x1 

x2
x1 
(b)(5pts) Compute the CI estimate , but without assuming that  X  28000 lb f .


x2
x1 
2
2 

Solution: x1=chi2inv(.025,4)=0.4844 & x2=chi2inv(.975,4) =11.1433. So CI   4 , 4  = [149.78 718.39]

(c)(5pts) Compute the percent increase in the CI estimate width associated with using the sample mean.
Solution: %Increase   (718.39  149.78)  613.15  156.05  100%  24.4%

613.15  156.05

4
PROBLEM 4(30pts) A sample poll of 1000 eligible voters resulted in the headline: “By a margin of 52% to 48%
Americans will elect Hillary Clinton over Donald Trump.” To analyze this result, let X=”the act of recording whether or
not a randomly polled voter would vote for Clinton. Let the event that the voter votes for Clinton be denoted [ X  1] .
n

n  2

(a)(10pts) Recall that:  X2  1  ( xk  x ) 2 . Show that  X2 
( p  p ) where p  x . [Hence,  X  0.4998 .]
n 1
n  1 k 1
n
[HINT: Notice that, since these are Bernoulli random variables, the quantity
n
x x
k 1
2
k
k 1
k
 n x .]
n
n
n
n
n
Solution:  X2  1  ( xk  x ) 2  1  ( xk2  2 x xk  x 2 )  1   xk2  2 x  xk   x 2  .
n  1 k 1
n  1 k 1
n  1  k 1
k 1
k 1

Note that because xk {0,1} we have
n
n
x x
k 1
2
k
k 1
k
 n x . Hence,
1

n x  2n x 2  n x 2   n 1 1 n x  n x 2   n n 1 x  x 2   1000
.52 .522   0.2498 , or  X  0.4998 .
 X2 
n 1
999

(b)(5pts) Recall that  p   X . In polling, the standard error refers to  2 p 100% . Compute the numerical value of this
n
standard error.


.4998
Solution:  p  X 
 0.0158 , and so we obtain  2(.0158) 100% =  3.16% .
n
1000
(c)(10pts) Carry out a statistical analysis to arrive at the 95% 2-sided CI estimate for the parameter Pr[ X  1]  p   X .



To this end, note that p  X is the average of a large number of iid random variables. Hence, T999  p  p  Z ~ N (0,1) .]
 p



p p
Solution: 0.95  Pr[  z  Z  z ]  Pr  1.96    1.96  Pr p   p 1.96  p  p   p 1.96 .
 p


 
 

Hence, CI  p   p 1.96  p  p   p 1.96 . So: CI  .52  (.0158)1.96  p  p  (.0158)1.96 = [0.489 , 0.551].




(d)(5pts) Based on your above analysis, comment as to the accuracy of the headline.
Comment:
I would argue that the claim is speculative, at best. The standard error puts Clinton somewhere between 48.84% and
55.16%. Furthermore, the CI estimate for p% is 48.9% to 55.1%. So it is very conceivable that the election could be much
closer than a 4% difference.