Download Homework 3 SOLUTION

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HOMEWORK 3 STAT 305A Spring 2017 Due 2/23(R) SOLUTION
PROBLEM 1 (35pts) [See book Problem 5-55 on p.183] In an acid-base titration, a base or acid is gradually added to
the other until they have completely neutralized each other. Let X and Y denote (the act of measuring) the milliliters of
 120  5 2
,
acid and base, respectively, needed for equivalence. Assume that ( X , Y ) ~ N  
 100 
 XY

 XY  
 .
2 2  
(a)(5pts) For  XY  0.6 find  XY .
Solution:  XY 
 XY
 XY
  XY   XY X  Y  0.6(5)( 2)  6
(b)(5pts) Find Pr[ X  116] .
Solution: Pr[ X  116] = normcdf(116,120,5) = 0.2119 ml.
(c)(10pts) Find Pr[ X  116 | Y  102] . [Hint: See book p.182.]
Solution:
 X 
5
( y  Y )  120  0.6 (102  100)  123 and
2
 Y 
2
  X2 (1   XY
)  52 (1  0.62 )  16 . So X | Y  2 ~ N (123,4 2 ) .
From p.182 we have  X |Y  y   X   XY 
 X2 |Y  y
Hence, Pr[ X  116 | Y  102] = normcdf(116,123,4) = 0.0401
(d)(5pts) Find Pr[ X  118  Y  96] . To this end, write a Matlab code that uses the Matlab command ‘mvncdf’.
Solution:
M=[120,100];
C=[25,6;6,4];
X=[118,96];
Prd=mvncdf(X,M,C) Ans: Prd = 0.0203
y
(e)(10pts) Find Pr[118  X  122  96  Y  104] . To this end, (i) show
the event as a shaded area in an x-y plot. Then (ii) write a Matlab code that uses 104
the Matlab command ‘mvncdf’. Include your code HERE.
96
Solution:
p1=mvncdf([122,104],M,C);
p2=mvncdf([122,96],M,C);
x
122
118
p3=mvncdf([118,104],M,C);
p4=mvncdf([118,96],M,C); %This area was counted twice
Figure 1(e) Plot showing the event.
Pre=p1-p2-p3+p4
Ans: Pre = 0.3066
NOTE: You can also use mvncdf([118,96],[122,104],M,C) WOW! All in just ONE command. But- the simplicity of this
command can prevent one from gaining a better understanding of the cdf concept. Beware!
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PROBLEM 2 (40pts) In relation to PROBLEM 1, assume that, in truth, ( X , Y ) has the bivariate normal distribution
given in PROBLEM 1.
(a)(5pts) Write a Matlab code that will simulate n  100 measurements
of ( X , Y ) , and then produce a scatter plot of the resulting data.
Solution: [See code under 2(a) in Appendix]
xy=mvnrnd(M,C,25);
x=xy(:,1); y=xy(:,2);
figure(21)
plot(x,y,'*')
grid
Figure 2(a) Scatter plot for n  100
[Grader: I am including code here for your benefit.]
(b)(5pts) From your data in (a) obtain an estimate of  XY  0.6 . Then compute the percent error of your estimate.
Solution: R=corrcoef(xy); rhoXYhat=R(1,2); rhoXYhat = 0.6314 . e  100(.6314  .6) / .6  5.23%
R=corrcoef(xy); rhoXYhat=R(1,2)
(c)(10pts) Use your data to obtain a linear prediction model

Y  mX  b . The overlay this model on a scatter plot of the data.


Aslo, give the numerical values of your m and b .
Solution: [See code under 2(c) in Appendix]
[mhat bhat] = [ 0.2347 71.8097 ]
Mhat=mean(xy); Chat=cov(xy);
mhat=Chat(1,2)/Chat(1,1); bhat=Mhat(2)-mhat*Mhat(1);
yhat=mhat*x + bhat; hold on
plot(x,yhat,'r','LineWidth',2)
Figure 2(c) Scatter plot for n  100 and prediction line.
(d)(10pts) Use 104 simulations to arrive at the pdf for the

estimator m . Also, give the numerical values of  m and  m .
Solution: [See code under 2(d) in Appendix] m  0.2402 ,  m  0.0326
mhat=zeros(1,nsim); bhat=zeros(1,nsim);
for k=1:nsim
Code in (c) with m & b replaced by m(k) & b(k)
end
[hm,bm]=hist(mhat,50); dbm=(bm(2)-bm(1));fmhat=hm/(nsim*dbm);
figure(23); bar(bm,fmhat); grid; xlabel('m-value');
title('PDF for mhat'); mu_mhat=mean(mhat);std_mhat=std(mhat);
[mu_mhat , std_mhat]
[NOTE: n=25 was used in Spring 2016! I changed to n=100.]

Figure 2(d) PDF for m using sample size n  100 .
(e)(5pts) Compute the true value of m. Then comment on how it compares to your value for  m in (d).
Solution: m   XY /  X2  6 / 52  0.2400 . This is almost exactly  m  0.2407 .


(f)(5pts) Assume that m ~ N (0.24 , 0.07 2 ) . Find Pr[0.23  m  0.25]
Solution: Pre=normcdf(.25,.24,.07) - normcdf(.23,.24,.07)
Pre = 0.1136
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PROBLEM 3 (25pts) A press fit refers to fitting two parts together, so as to ensure that they will not slip relative to one
another. From https://en.wikipedia.org/wiki/Interference_fit we have the following example:
A 10 mm (0.394 in) shaft made of 303 stainless steel will form a tight fit with allowance of 3–10 µm (0.0001–0.0003 in).
A slip fit can be formed when the bore diameter is 12–20 µm (0.0005–0.0008 in) wider than the rod.
Suppose that shafts are machined to conform to a shaft diameter distribution S ~ N ( S  10mm ,  S ) and the bores are
machined to conform to a bore diameter distribution B ~ N ( B  10.0065mm ,  B ) . Denote the clearance between the
shaft and bore as C  B  S .
(a)(3pts) Use the linearity of E(*) to arrive at  C in units of  m .
Solution: C  E (C )  E ( B  S )  E ( B)  E ( S )  10.0065  10.0000  0.0065 mm  6.5 m .
(b)(2pts) Suppose we require that  C  1 m . From this constraint, obtain an expression for  B2 as a function of  S2 .
Solution: Because we are given no covariance information, we will assume that B and S are independent. Then
2
2
 C2  1   S2   B2 . Hence,  B  1   S .
(c)(5pts) Suppose that the shaft machining cost as a function of  S is T ( S )  1 /  S ($) and the bore machining cost as a
function of  B is T ( B )  3 /  B ($). Then the total cost of a shaft/bore unit is Ttot  T ( S )  T ( B )  (1 /  S )  (3 /  B ) . From
this cost function, along with the constraint you found in (b), one can use the method of Lagrange multipliers
[ https://en.wikipedia.org/wiki/Lagrange_multiplier ] to show that the total cost will be minimized for  S  0.5700 m
and  B  0.8218 m . Rather than using this elegant mathematical method, proceed to arrive at these numbers as
follows: (i) define the Matlab array  S  0.1 : 0.001 : 0.9 (ii) create your  B array per your constraint in (b) (iii) compute
your Ttot array per the above equation (iv) use the command [Tmin , k0 ]  min( Ttot ) to find the minimum cost and the
associated array index, and (v) recover your values of  S (k 0 ) and  B (k 0 ) . Include your code and answers HERE.
Solution:
min( Ttot ) = $5.4056 ;  S =0.5700 um
Code: sigS=0.1:0.001:0.9;
[Tmin,k0]=min(T);
;
 B = 0.8216 um
sigB=sqrt(1-sigS.^2);
T=sigS.^-1 + 3*sigB.^-1;
[Tmin , sigS(k0) , sigB(k0)]
(d)(10pts) Suppose that one of your colleagues claims that the company has been using  S   B  1 m for ages, and it is
still in business. Hence, to change to the tighter specifications you are suggesting in (c) is simply a waste of money. In this
part compute the probability that the clearance will not be in the range 3  10 m for both his specs. and yours.
Solution: From (b): C ~ N (6.5m ,1m) . So My failure prob. = p1=normcdf(3,6.5,1) + (1-normcdf(10,6.5,1))= 4.6526e-04
Using  C  12  12  2 m his is p2=normcdf(3,6.5,sqrt(2)) + (1-normcdf(10,6.5,sqrt(2))) = 0.0133
(e)(5pts) Compute the total cost of a shaft/bore pair using his numbers. Then use your answers in (c) and (d) to compute
the anticipated cost of lost units in a run of 10,000 bore/shaft pairs of using your specs. and his.
Solution: His cost is Ttot  (1 /  S )  (3 /  B )  (1 / 1)  (3 / 1)  $4.00 . For a run of 10,000 units:
Using my specs. we could anticipate ~5 failed units for a total cost of 5  $5.42  $27
Using his specs we could anticipate 133 failed units for a total cost of 133  $4.00  $532 (20x my cost!)
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Matlab Code
%PROGRAM NAME: hw4.m
clear all
%PROBLEM 1:
%1(d)
M=[120,100];
C=[25,6;6,4];
X=[118,96];
Prd=mvncdf(X,M,C)
%1(e):
p1=mvncdf([122,104],M,C);
p2=mvncdf([122,96],M,C);
p3=mvncdf([118,104],M,C);
p4=mvncdf([118,96],M,C);
Pre=p1-p2-p3+p4
%PROBLEM 2:
%2(a):
n=100;
xy=mvnrnd(M,C,n);
x=xy(:,1); y=xy(:,2);
figure(21)
plot(x,y,'*')
grid
xlabel('Acid Amount (x)')
ylabel('Base Amount (y)')
%2(b):
R=corrcoef(xy);
rhoXYhat=R(1,2)
%2(c):
Mhat=mean(xy);
Chat=cov(xy);
mhat=Chat(1,2)/Chat(1,1);
bhat=Mhat(2)-mhat*Mhat(1);
yhat=mhat*x + bhat;
hold on
plot(x,yhat,'r','LineWidth',2)
%2(d):
nsim=10^4;
mhat=zeros(1,nsim); bhat=zeros(1,nsim);
for k=1:nsim
xy=mvnrnd(M,C,100);
x=xy(:,1); y=xy(:,2);
Mhat=mean(xy);
Chat=cov(xy);
mhat(k)=Chat(1,2)/Chat(1,1);
bhat(k)=Mhat(2)-mhat(k)*Mhat(1);
end
[hm,bm]=hist(mhat,50);
dbm=(bm(2)-bm(1));
fmhat=hm/(nsim*dbm);
figure(23)
bar(bm,fmhat)
grid
xlabel('m-value')
title('PDF for mhat')
mu_mhat=mean(mhat);
std_mhat=std(mhat);
[mu_mhat , std_mhat]