¦„SPECIAL MS WRITTEN EXAMINATION IN BIOSTATISTICSÈ…
¦„PART I…È

¦May 12, 1995:  9:00 a.m. Û,2 1:00 p.m.È





‚INSTRUCTIONS:ƒ

þì™a)™
This is an ‚closed bookƒ examination.
™b)™
Answer any ‚threeƒ questions.
™c)™
Put the answers to different questions on separate sets of paper.
™d)™
Put your „CODE LETTER…, ‚not your nameƒ, on each page.
™e)™
Return the examination with a signed statement of the honor pledge on a page separate ™
™™
from your answers.
™f)™
You are required to answer „only what is asked… in the questions,„ not …‚all you know about the ™
™™
topicsƒ. 


þè
Question 1.

þì™Consider stratified simple random (without replacement) sampling from a population of size N that has been divided into H strata, each containing N¬hÈ population members, such that N = Û 0µHȶh=1È N¬hÈ.  The sample is sued to estimate the proportion, P = Û 0µHȶh=1È W¬hÈP¬hÈ , of population members that possess þìsome attribute, where W¬hÈ = N¬hÈ/N and P¬hÈ is the proportion of the members of the h-th stratum who possess the same attribute.  A point estimate of P is obtained using,

™P¬woÈ = Û 0µHȶh=1È W¬hÈp¬hÈ ,

where p¬hÈ is the proportion of the sample in the h-th stratum that possess the same attribute.

a)™Show that the variance of the sampling distribution of p can be written as,

™Var(p¬woÈ) = Û 0µHȶh=1È W«Û27ɬhÈ ºN¬hÈÛ,2n¬hÈÂN¬hÈÛ,2Û18Ë ºP¬hÈÛ(8Û18Û,2 P¬hÈÛ)8Ân¬hÈË ,

™where n¬hÈ is the sample size in the h-th stratum, such that n = Û 0µHȶh=1È n¬hÈ .

b)™If proportionate allocation is used for this sample, show that this variance becomes

™Var(p¬woÈ) = ºÛ18Û,2fÂn«Û27ÈË Û 0µHȶh=1È ºn¬hÈN¬hÈÂN¬hÈÛ,2Û18Ë P¬hÈ(1Û,2P¬hÈ)

c)™Show that the ratio (Ût;), defined as the variance of p¬woÈ assuming the proportionately stratified þî
sample above divided by the variance of the sample proportion, p, under an unstratified simple random (without replacement) sample, can be written as,


Ût; = ºÛ 0µHȶh=1È W¬hÈ ºN¬hÈÂN¬hÈÛ,2Û18Ë P¬hÈ(1Û,2P¬hÈ)ºNÂNÛ,2Û18Ë P(1Û,2P )Ë




þî
Points:™™
™a) 9,™b) 8,™c) 8.

Question 2.


™For a certain manufacturing process the random variable Y¬xÈ (the amount in kilograms manufactured per day) has mean E(Y¬xÈ) = Ûa; x + Ûb; x«Û27È and variance V(Y¬xÈ) = Ûs;«Û27È, where x is the „‚known…ƒ amount of raw material in kilograms used per day in the manufacturing process.  The n ‚mutually independentƒ data pairs (x, Y¬xÈ), x = 1, 2, ..., n, are available to estimate the unknown parameters of interest.  Thus, Y¬Û17È, Y¬Û27È, ..., Y¬nÈ are a set of n ‚mutually independentƒ random variables.  Further, let    S¬kÈ =  Û 0µnȶx=1Èx«kÈ, k = 1, 2, 3, ...; again, note that the S¬kÈ's are non-stochastic quantities with ‚„knownƒ… values.


a)™Derive ‚explicit expressionsƒ for the ‚unweighted least squaresƒ estimators Ûa;µÛ^8È and Ûb;µÛ^8È of the unknown þî
parameters Ûa; and Ûb;.

þî
b)™Derive ‚explicit expressionsƒ for E(Ûb;µÛ^8È) and V(Ûb;µÛ^8È), the mean and variance of Ûb;µÛ^8È.

c)™If Y¬xÈ Û[2 N(Ûa; x + Ûb; x«Û27È, Ûs;«Û27È), x = 1, 2, ..., n, with the Y¬xÈ's being mutually independent random þî
variables, compute an „‚exactƒ… 95% confidence interval for Ûb; if n = 4, Ûb;µÛ^8È = 2, and Ûs;«Û27È = 1.

þî














Points:  ™™
a) 8,™b) 8,™c) 9.™™
þè
Question 3.


™Let X¬Û17È, X¬Û27È, ..., X¬nÈ be a random sample from a N(Ûm;, 1) population.  Suppose that it is desired to test H¬Û07È:  Ûm; = 0 versus H¬Û17È:  Ûm; = Ûm;¬Û17È > 0, where Ûm;¬Û17È is a known positive number.


a)™Show that the most powerful test of size Ûa; for testing H¬Û07È versus H¬Û17È involves a rejection region R þî
of the form
™™
™™™™™#R = { XµÛ,2È : XµÛ,2È Û;2 k¬Û17Û,2Ûa;È| H¬Û07È} = Ûa;.

þî
b)™For Ûa; = 0.025, what is the value of k¬Û.8Û97Û77Û57È when n = 100?

c)™Is the test developed in part (a) a uniformly most powerful (UMP) test of size Ûa;?

d)™Suppose that we decide to reject H¬Û07È when XµÛ,2È Û;2 0.80.  What is the ‚„ƒpower… of this test when       þî
Ûm;¬Û17È = 1 and n = 25?

þî
e)™As in part (d), again suppose that we decide to reject H¬Û07È when XµÛ,2È Û;2 0.80.  What is the þî
minimum sample size required to insure that we reject H¬Û07È:  Ûm; = 0 with a probability of at least 0.975 when, in fact, Ûm;¬Û17È = 1?
þî













Points:™™
a) 5,™b) 5,™c) 5, ™d) 5,™e) 5.
þè
Question 4.


Let Y¬Û17È, Y¬Û27È, ..., Y¬nÈ constitute a random sample of size n (Û;2 2) from a N(Ûm;, Ûs;«Û27È) population.

a)™If Ûs;«Û27È is ‚knownƒ to be equal to 1, what is the ‚minimumƒ sample size n«Û*8È required so that

™™
™pr { | YµÛ,2ÈÛ,2Ûm; |  < 0.20} Û;2 0.95,

™where YµÛ,2È= n«Û,2Û17È  Û!0µnȶi=1È Y¬iÈ ?


b)™Consider using YµÛ,2È to test H¬OÈ:  Ûm; = 0 versus H¬AÈ:  Ûm; > 0 when Ûa; = 0.025.  If Ûs;«Û27È is ‚knownƒ to be þî
equal to 1, what is the ‚minimumƒ sample size n«Û*8È required to that the probability of a Type II error is no more than 0.16 when Ûm; = 0.50?

þî
c)™Let S«Û27È = (nÛ,21)«Û,2Û17È  Û!0µnȶi=1È (Y¬iÈÛ,2YµÛ,2È)«Û27È .  If


™™
™k = ¤ºnÛ,2Û18Û)8ÂÛ28ËÕ    ºÛG;Û(8ºnÛ,2Û18ÂÛ28Ë)ÂÛG;Û(8ºnÂÛ28Ë)Ë  ,

™prove that the estimator Ûs;µÛ^8È = kS is an unbiased estimator of Ûs;.












Points:™™
a) 9,™b) 9,™c) 7.