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Mathematical Statistics

by: Jacinto Carter Sr.

Mathematical Statistics STA 3523

Jacinto Carter Sr.
GPA 3.79


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This 22 page Class Notes was uploaded by Jacinto Carter Sr. on Thursday October 29, 2015. The Class Notes belongs to STA 3523 at University of Texas at San Antonio taught by Staff in Fall. Since its upload, it has received 14 views. For similar materials see /class/231437/sta-3523-university-of-texas-at-san-antonio in Statistics at University of Texas at San Antonio.

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Date Created: 10/29/15
STAT 3523 MATRIX APPROACH TO REGRESSION ANALYSIS FALL 2007 Random Vectors and Matrices A random vector or a random matrix contains elements that are random variables Let Y denote an n x 1 column vector Y1 Yn The expected value of Y is the vector E Y given by ME The Variance Covariance or dispersion matrix of Y is given by VarY1 Cowl17 0011Y17Yn Con7Y VarY Con7Yn COUYE 21 2 2 CowlMK CovYnYg VarOn OI CovY E ozj Where 02 CovYZ Let A be a constant matrix and W AY be a random vector Using the de nitions of the mean vector and covariance matrix we have 1 E A A 2 AEY 3 CovW AEA Simple Linear Regression Model in Matrix Form Consider the simple linea regression model Yi o lXiEz7il27n Or Y1 5051X181 Y2 5051X282 KL 081Xn5n The above may be written in matrix form as YX66 Where 1 X1 Y1 E1 1 X Y X 2 g BO 6 51 Yn en 1 Xn The assumptions on the error random variables may be written as E6 0 00116 02I Least Squares Estimation To obtain the LSE s we minimize the sum of the squared deviations Q 21106 50 lXZDg The criterion may be written as Q Y XmY X Y Y 2p X Y MX XW To nd the estimates we take the derivative of Q wrt 6 set the resulting equation to 0 and solve We have 3Q 2X Y 2X Xm which yields the normal equations X Xb X Y The solution gives the least squares estimate as b a XX 1XY provided the inverse XX 1 exists The matrix will be nonsingular if the regressors are linearly independent Horn the expression above we immediately have Y X3 XXX 1XY Veri cation of the Results of Simple Linear Regression in Matrix Form We noticed that in matrix form the simple linear regression model was written as YX66 with E6 0 and Var6 agI Also 1 X1 Y1 1 X Y E X 2 6 50 I 39 51 Yn 1 Xn TheLSEof is bXX 1XY Notethat Y1 1 1 1 Y2 XY X1 X2 Xn Yn 212136 Z211X2 Y2 Nle gt 5X1 zu wit me kWhm 1 an 0 WXCKUH AZX39ZX 3 4 a 4 WXZWXZFQKZWZXZ 2 1 2 1 AX u U X uZ g K53 X 1713 gX 1 I LXIJXX 4 I 3 SI EISI em eoueH u 53 G 4 2X 12 gX I I 14 AAON I2 Z I2 39ZUXZ ngu I IIqu 39XX JO queuluneaep em eq I m X I gX ii ZX5 s s X ZX IX 2 1 Z XX X u u X I I I I pue Multiple Regression Set up in Matrix Notation Let there be k predictors with Yz 5051X1i 2X2i kaiei7 i12 Or Y1 5051X1152X21 39 ka1 1 Y2 5051X1252X2239 5ka2 2 Yn 0 1X1n 2X2nquot398kan n YX66 with Y1 Yi Y2 Yn 1 X11 X21 in 1 X12 X22 Xk2 X 1 X13 X23 ng 1 Xln X2n an SLIDES SET 5 TESTS OF HYPOTHESES Victor De Oliveira Oct 20 2008 A hypothesis H is a claim or conjecture about a feature of a population often expressed in terms of the values of a parameter For instance H M 15 H M 15 H 02 2 H 02 lt 2 Suppose we have two contrasting hypotheses H0 and H1 H0 would be called the null hypothesis and represents the current belief of status quo H1 would be called the alternative hypothesis and represents a new belief or a research hypothesis And suppose we also have data directly related to the hypotheses H0 and H1 The test of hypothesis problem consists of deciding whether the data offer support in favor of H0 or in favor of H1 This problem is analogous to the problem faced by a jury member in a trial involving a person accused of a crime In this case we have H0 the accused is innocent H1 the accused is guilty data the evidence presented by the defense and prosecutor 1 A test of hypothesis is a rule to decide based on the data when we should reject H0 it also follow from the test when not to reject H0 Any test has two ingredients 0 The test statistic which is the function of the data that the test uses to decide between H0 and H1 0 The rejection region which is the set of values of the test statistic that lead to the rejection of H0 For any test of hypothesis there are two possible types of errors 0 type I error reject H0 when H0 is true 0 type I error do not reject H0 or accept H1 when H1 is true By Viewing the testing of hypothesis problem as a decision problem with two possible actions we have depending on the state of nature and our decision four possible end results H0 true H1 true reject H0 type I error correct decision do not reject H0 correct decision type ll error As for any statistical procedure there are good and bad tests and some tests are better than others We would assess the goodness of a test by its probabilities of type I and type ll errors or P type I error and 6 P type ll error Example Let X1 X25 be a random sample from the NM4 distribution where M is unknown We want to test the hypotheses H02M1 versus H02M3 for this initial example we assume these two are the only possible values M can take Consider the following test Test 1 reject H0 if X gt 2 The test statistic of this test is X and the rejection region is 2 To compute oz and 6 it is key to know what is the distribution of X X N N1425 when H0 is true X N N3 425 when H1 is true Then on Ptype I error of test 1 Preject H0 when H0 is true PX gt 2 when M 1 PZ gt 25 00062 if we do this test a large number of times we will make a type I error on average 62 out of every 10000 times Also 61 Ptype ll error of test 1 Pnot to reject H0 when H1 is true PX lt 2 when 0 3 PZ lt 25 00062 3 the two probabilities are equal in this particular example but they are usually different Note that except for extreme cases in general it does not hold that or 6 1 why 7 Consider Test 2 reject H0 if X gt 22 Then by a similar calculation as above we have a2 Ptype I error of test 2 00013 62 Ptype ll error of test 2 00228 Which test is better 7 In terms of type I error test 2 is better than test 1 but the opposite holds in terms of type ll error This example illustrate the following fact it is not possible to modify a test to reduces the probabilities of both types of error If one changes the test to reduce the probability of one type of error the new test will inevitable have a larger probability for the other type of error The classical approach to deal with this dilemma and select a good test is the following The type I error is the one considered more serious so we would like to use a test that has small 04 Then we x a value of or close to zero either subjectively or by tradition say 005 or 001 and then nd the test that has this or as its probability of type I error This or will also be called the signi cance level of the test 4 Remarks 0 For any testing problem there are several possible tests but some of them make little or no sense For the previous example tests of the form reject H0 when X gt c with c gt 1 are worth considering but tests of the form reject H0 when X lt 6 make no sense for this example why 7 0 We will consider only problems where the hypotheses refer to the values of a parameter of interest generically called 6 In addition the null hypothesis will always be of the form H0 9 60 where 60 is a known value And the alternative hypothesis will always be one of the three possible forms H1 9 lt 60 H1 9 gt 60 or H1 9 3e 90 The rst two are called one sided alternatives and the last one is called two sided alternative 0 The classical approach to test hypothesis does not treat H0 and H1 symmetrically and does so in several ways Because of that the issue of what hypothesis to consider as the null and what as the alternative has an important bearing on the conclusion and interpretation of a test of hypotheses Example A mixture of pulverized fuel ash and cement to be used for grouting should have a compressive strength of more than 1300 KNmz The mixture cannot be used unless experimental evidence suggests that this requirement is met Suppose that compressive strengths for specimens of this mixture are normally distributed with mean M and standard deviation 0 60 a What are the appropriate null and alternative hypotheses 7 H0 M 1300 versus H1 M gt 1300 b Let E be the sample mean of 20 specimens and consider the test that rejects H0 if X gt 133126 What is the probability of type I error Since X N N1300 60220 vvhen H0 is true we have a 133gt 133126xvhen M 1300 133126 1300 60 0 c What is the probability of type I error when M 1350 7 HZgt 1 ng 3um Since X N N1350 60220 when M 1350 we have 61350 13jr 1331 26xyhen M 1350 133126 1350 1323 60x20 g yszg rhu d How the test in part b needs to be Changed to obtain a test with a probability of type I error 7 What would be 61350 for this new test 7 Need to Change the cut off value 133121 with a new one say 0 for which it holds 7 1300 005 PX gt 0 when 0 1300 PZ gt 6 60 x 20 This implies that 071300 2005 1645 and solving for c we have 60m c 132207 The above show an important point there is a one to one correspon denoe between the cut off value of a test and the probability of type I error The new test has a larger or than the test in b so it must have a smaller 6 Indeed 132207 1350 61350 PX 3 132207 when M 1350 Z 6mm 3 PZ g 208 00188 Test About M in Normal Populations Case When 02 is Known Let X1 Xn be a random sample from a N01 02 distribution 2 is assume known We want to test the null hypothesis where 0 H0 M HQ with 0 a known constant against one of the follow ing alternatives hypothesis H lt 0 H1 3 M gt 0 M i 0 The right test to use would depend on the alternatives hypothesis that is used To start consider the case when we want to test the hypotheses H0MM0 versus H12MltM0 The right kind of test in this case is to reject H0 when X lt c for some constant 6 why 7 The question then is how to choose the cut off value c 7 We use the classical approach described before we x a small probability oz and choose the value of c so the resulting test has this oz as its probability of type I error signi cance level 0 M0 0 From this follows that zlm zO and c M0 ZaUW ozPXltcwhenMM0PZlt 8 Then for this case the test of hypotheses having signi cance level or is X MO Ux reject H0 if X lt 0 zap or equivalently if Z lt sz Following a similar reasoning and computation we have that for the case when we want to test the hypotheses H0MM0 versus H12MgtM0 the test with signi cance level or is reject H0 if X gt 0 Zap or equivalently if Z gt za Note that in the previous two cases the rejection region is one sided and of the same form as the alternative hypothesis Finally to test the hypotheses H0MM0 versus leuy uo with signi cance level or we use the test X M0 0 Note that in this case the rejection region is two sided as is the al reject H0 if Z gt 2042 ternative hypothesis Values of Z that are signi cantly different from zero either positive or negative provide evidence in favor of H1 so they call for the rejection of H0 Note also that in this case we use the Z value corresponding to or 2 instead of oz Sample Size Calculation Suppose we want to test the hypotheses H0MM0 versus H12MltM0 and for that we can take observations X1 Xn assumed to be a random sample from a N01 02 distribution with 0 known but we have not decided yet how many we would take After we collect the data we would like to use the test with signi cance level or as usual and in addition we want the test to have probability of type ll error equal to 6 when M 1 oz 6 and 1 are given The question is what the sample size n needs to be for these to hold 7 As we saw before the test with signi cance level or is reject H0 if X lt 0 zap or equivalently if Z 3375 lt zO Also if we want that this test has probability of type ll error when M 1 equal to 6 it must hold that 6 6011 PX gt Mo ZaUx When M M1 Mo Mi 0 132 gt 20 where Z N N0 1 which is equivalent to Mo Mi 0 The latter equation implies that za zg and solving for n in this equation we get The required sample size is the above value rounded up The same sample size as above would be used for the case we are testing the hypotheses H0MM0 versus H12MgtM0 But for the case we are testing the hypotheses H0MM0 versus leuy uo the required same sample size is 0Za2 Zg2 n M0 M1 Example The drying time of a certain type of paint is known to be normally distributed with mean 75 min and standard deviation 9 min A chemical lab has proposed a new additive that claim to decrease the mean drying time It is believed that drying times with the additive will remain normally distributed with the same standard deviation To check the labs claim we want to test the hypothesis H0 0 75 versus H1 M lt 75 a Test this hypothesis based on a random sample of 25 observations for which i 723 using or 001 The test with signi cance level or 001 is reject H0 if Z lt z001 233 For the observed data we have 2 ii M0i723 75i ObsiaA i 95 7 Since zobs does not fall in the rejection region 15 7i 233 the 15 conclusion is not to reject H0 b What is or for the test that rejects H0 if Z lt 288 7 oz PZ lt 288 0002 c For the test in b compute the probability of type ll error when M 70 First note that Z lt 288 is equivalent to X lt 75 288 6982 why 7 Then 670 PX gt 6982 when M 70 PZ gt 01 05398 d If the test in part b is used what is the sample size required to ensure that 670 001 7 M W 2 8795 0 M1 75 70 so we need to collect n 88 observations Test About M When n is Large Suppose X1 X7 is a random sample from any distribution not necessarily normal with mean M and variance 02 both unknown And suppose we want to test the null hypothesis H0 M M0 The test statistic in this case would be X M0 5M5 The key point to note is that if n is moderate or large we have by Z the CLT that Z apgox N0 1 when H0 is true Depending on the alternative hypothesis that is being tested the test with signi cance level or is H1 reject H0 if M lt M0 Z lt za M gt 0 Z gt 2a M 7g 0 lZl gt 2042 Test About M in Normal Populations Case When 02 is Unknown Let X1 Xn be a random sample from a N01 02 distribution 2 where now 0 is unknown We want to test the null hypothesis H0 M 0 against one of the usual alternatives hypothesis H1 The test statistic in this case would be X 0 T SA and the key point to note is that T N an when H0 is true Depending on the alternative hypothesis that is being tested the test with signi cance level oz is H1 reject H0 if M lt 0 T lt to 7n71 M gt 0 T gt town M 7g 0 lTl gt tgmq


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