1. (15%) Sampling for defectives £rom large lots of manufactured ...
1. (15%) Sampling for defectives £rom large lots of manufactured ...
1. (15%) Sampling for defectives £rom large lots of manufactured ...
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<strong>1.</strong> (<strong>15%</strong>) <strong>Sampling</strong> <strong>for</strong> <strong>defectives</strong> £rom <strong>large</strong> <strong>lots</strong> <strong>of</strong> <strong>manufactured</strong> product yields anumber <strong>of</strong><br />
-<br />
<strong>defectives</strong>, Y, that follows a binomial probability distribution. A sampling plan consists <strong>of</strong><br />
specifying the number <strong>of</strong> items n to be included in a sample and an acceptance number "a".<br />
The <strong>lots</strong> is accepted if Y I a and rejected if Y > a. Letp denote the prop<strong>of</strong>iion <strong>of</strong><br />
<strong>defectives</strong> in the lot.<br />
(a) For n = 5 and a = 0, calculate the probability <strong>of</strong> lot acceptance if (a)p= O,.(b)p = .l,(c)<br />
p= .3,(d) p = .5,(e)p = <strong>1.</strong>0.<br />
I<br />
(b) A graph showing the probability <strong>of</strong> lot acceptace as a function <strong>of</strong> lot £radon defective is<br />
called the operating characteristic curve <strong>for</strong> the sample plan. Construct the operating<br />
characteristic curve <strong>for</strong> the plan n = 5, a = 0.<br />
2. (10%) Let m(t)=(1/6)ef+(~6)ez'+(3/6)df. Find the following:<br />
(a) E(Y)<br />
(b) V(Yr<br />
(c) The distribution <strong>of</strong> Y<br />
3. (10%) A diagnostic test <strong>for</strong> a disease is said to be 90% accurate in that if a person has the<br />
disease, the test will detect it with probability 0.9. Also, if a person does not have the disease,<br />
the test will report that he or she does not have it with probability 0.9. Only 1% <strong>of</strong> the<br />
population has the disease in question. If a person is chosen at random £rom the population<br />
and the diagnostic test indicates'that she has the disease, what is the probability that she does<br />
have the disease? Are you surprised by the answer? Would you call this diagnostic test<br />
reliable?<br />
4. (<strong>15%</strong>) Let Y, be the amount <strong>of</strong> pollutant per sample collected above the stack wifhout the<br />
cleaning device and Y, be the amount collected above the stack with the cleaner. The joint<br />
density <strong>of</strong> Y, and Y, is<br />
osy,s2, O ~ Y , ~ ~ , ~ Y , ~ Y I<br />
elsewhere.<br />
The random variable (YI -Yz) represents the amount by which the weight <strong>of</strong> pollutant can be<br />
reduce by using the cleaning device.<br />
(a) Find E(Y, -Y,).<br />
(b) FindV(Y, -Y,).
5. (20%) Let Xbe a random variable whose p.d.f.@robability density function) f is either the<br />
U(0,1), Uni<strong>for</strong>m, to be denoted byfo, or the triangular over the interval [0,1], to be denoted by<br />
fi, that is f;(x) = 4x, <strong>for</strong>0 2 x ~112; f;(x) = 4-4x, <strong>for</strong>112 2 x 2 1; and 0, otherwise.<br />
(a) Test the hypothesis Ho: f =fo v.s. HI: f =fi at level <strong>of</strong> significance a = 0.05.<br />
(b) Compute the power <strong>of</strong> the test.<br />
6. (<strong>15%</strong>) Answer the following questions:<br />
(a) A random variable Xis said to be memoryless if P{X > s+t I X > t) = P{Y > s) <strong>for</strong> all<br />
s, t <strong>1.</strong>0. Show that Xis memoryless when Xhas an exponential distribution with parameter h.<br />
(b) Let X be the interarrival time <strong>of</strong> buses at a station. IfX is exponentially distributed,<br />
describe how the memoryless property affects the waiting time <strong>of</strong> a passenger.<br />
7. (<strong>15%</strong>) After running a multiple regression analysis with five independent variables, the<br />
following ANOVA table is olitained<br />
Source<br />
Regression<br />
Residual<br />
Total<br />
d f<br />
45<br />
SS<br />
224<br />
270<br />
MS<br />
F<br />
(a) Complete the above ANOVA table.<br />
(b) Calculate R2 and adjusted R2.<br />
(c) Test the hypothesis Ho:PL=P2=P3=P4=P5=0 with a =0.05.
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,0000'1<br />
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6666'0 6666'0 6666'0 6666'0 6666'0 6666'0 6666'0 6666'0 8666'0 86660<br />
8666'0 8666'0 8666'0 8666'0 8666'0 8666'0 8666'0 8666'0 8666'0 8666'0<br />
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TABLEIV -<br />
Values <strong>of</strong> <strong>1.</strong> df<br />
-
TABLE VII (cont.)<br />
Valuer <strong>of</strong> x:
TABLEVIII (cont)<br />
Values <strong>of</strong> Fa<br />
dfd<br />
u<br />
dfn<br />
0.10<br />
0.05<br />
25 0.025<br />
0.01<br />
0.005<br />
0.10<br />
0.05<br />
26 0.025<br />
0.01<br />
0.005<br />
0.10<br />
0.05<br />
27 0.025<br />
0.01<br />
0.005<br />
0.10<br />
0.05<br />
28 0.025<br />
0.01<br />
0.005<br />
0.10<br />
0.05<br />
29 0.025<br />
0.01<br />
0.005<br />
0.10<br />
0.05<br />
30 0.025<br />
0.01<br />
0.005<br />
0.10<br />
0.05<br />
60 0.025<br />
0.01<br />
0.005<br />
0.10<br />
0.05<br />
120 0.025<br />
0.01<br />
0.005