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# 126 Class Note for STAT 401 at PSU

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Date Created: 02/06/15

STAT 401 Ch 8 Tests of Hypotheses Sometimes we want answers to specialized questions such as is y gt a speci ed value 0 Ex1 a A trucking rm suspects the claim made by a tire man ufacture that certain tires last on average at least 28000 miles This can be addressed in terms of a hypothesis testing prob lem where we are asked to decide between H0 y 2 28 000 and Ha y lt 28000 Note that CI though they provide a region containing 0 are not designed for choosing between two hy potheses We will see that such decision involve considerations that do not arise in the construction of CI The hypothesis denoted by H0 is called the null hypothesis while HCl is the alternative There is a rule for which hypothesis gets designated the null hypothesis Consider the Ex1 b The tire manufacturer wants to claim that certain tires last on average at least 28000 miles The decision is between H0 y 3 28000 andHa y gt 28000 The rule for designating the null and alternative hypotheses is Alternative is the hypothesis the investigator wants to claim as true In other words Alternative is the hypothesis the investi gator wants evidence for Convention The null hypothesis will be stated as equality e g H0 y 28000 vs H y lt 28000 Speci cation of 21 Testing Procedure A test procedure is speci ed in terms of i a test statistic and ii a rejection rule region Ex1 a Continued In this case the test statistic will be 2 and the rejection rule speci es that H0 is rejected wheneverX gt C some constant Ex2 To investigate whether certain detonators used with explo sives in coal mining meet the requirement that at least 90 will ignite 20 are tested and it is found that 17 function correctly Hypotheses Hozp9Haplt9 A X Test statistic p g or Just X Rejection rule Reject H0 if X g C same constant Question How is C and thus the testing procedure speci ed The answer to this question comes by consideration of the different types of error and assigning priorities as will be eX plained Type I and type II errors It is a fact that mistakes cannot be avoided Thus H0 can be rejected when it is true or not be rejected when it is false 2 Ex2 Cont Suppose the rejection rule is Reject H0 ifX g 16 Also suppose that p 9 so H is true Then from binomial tables we get PX 16p 9314 20 133 That is H will be rejected 133 of the time even though it is true Now suppose that p 8 so HCl is true Then PX 16p 8311 20 589 That is H0 will be rejected 589 of the time meaning we commit an error 411 of the time Remark 1 If H0 were false in a more pronounced way ie p 5 then the probability of committing an error ie not rejecting H0 would be smaller The two types of error that can be committed are called type I and type II and can be shown in the table Truth H0 Ha Outcome H0 Correct decision Type II of test Ha Type I Correct decision Ex2 Cont Suppose we change the R to Reject H if X g 17 Then Ptype I error PX g 17p 2911 20 323 Ptype II error atp 8 1 PX g 17p 9n 20 1 794 206 Remark Changing the RR fromX g 16 toX g 17 the prob ability of type I error T from 133 to 323 while the probability of type II error I from 411 to 206 Fact We cannot control both types of errors with the same sample size Philosophy Type I error is more serious than type 11 Thus we always want to contrail Prob Type I error The level at which we decide to control the P type I error is denoted by X usually on 105 or 01 and is called level of signi cance Fact Deciding on X determines the RR and thus also the test ing procedure Ex2 Cont Deciding on or 133 determines that the R is X g 16 Remark The discreteness of the binomial distributions does not allow the usual choices of QC Tests About the Population Mean Case 1 Sample from Nu6262 known Let X1X be a rsample from N y 62 and suppose we want to test H0 y no no is some speci ed value vs some alternative hypothesis The test statistic is J but more conveniently we use f o Gx Test statistic Z Remark It is important to realize that Z N N 0 1 only if H is true ie only if the true value of y is the speci ed 0 Suppose we decide on some signi cance level 06 The R depends on the alternative hypothesis Ha RRatlevelOL gto 2220c lto 23 20 u 75 0 Z 2 Zoe2 Z 2 Zoe2 or Z S Zoc2gt Ex3 A tire company wants to change the tire design Econom ically the modi cation can be justi ed of average life time with new design exceeds 20000 miles A rsample of n 16 new tires is tested Assume life times are N y 15002 The 16 tires yield 2 203 758 Should the new design be adopted Test at 06 01 Sol Here H0 y 2030003 Ha y gt 203000 why The test statistic is tt0 20758 20000 c 1500x16 R at on 01 Z gt 201 233 Since 202 4 233H0 is not rejected implying the new design is not adopted Remark The fact that 203 758 gt 203 000 but the testing procedure decides against Ha y gt 20 000 serves to Z 202 a highlight the bias of the testing procedure in favor of H and b raise questions regarding the performance characteristics of the procedure For example it would be of interest to know what is the P type 11 error when y 21 a 000 Ex3 Cont Find the probability of not rejecting H when y 21 a 000 Sol Ptype II errory 21 a 000 P X 0 lt 21 000 7 Z 2 cm 0 C Plt ltmza 213000 X 213000lt 0 21000 21000 Z cw cw 0 21 000 20 000 21 000 qgt Ha qgt 233 GW 150016 CI 34 3669 Remarks a The probability of type 11 error at a value y is denoted by mu Thus in above example we found 321a 000 3669 b In HW problems and in quizzes you can use directly the formulas in p319 Thus 203000 213000 1500x16 See formulas for But for Ha y lt yo and HCl y 75 no If the probability of type 11 error is large for our purposes we can increase the sample size in order to reduce it 3213000c1gt233 gtCIgt 343669 7 Ex3 Cont Find the sample size needed to have 321a 000 1 The equation we must solve is zaOl 1 or z1 Za0ILI 2 orn Here 0 1500233 128 203000 213000 As usual we round up and use n 30 2 l 5422 2932 Remark See p319 for sample size calculation for Ha y 75 0 Case 11 Large Samples n gt 30 Here the rsample X1X is allowed to come from any population distribution with mean u and variance 62 both un known Interest in testing H0 y no vs some alternative Test statistic f m Sm IfHo is true then Z 53 N9a 1 The R at level or are Z Ha RRatlevelOL gto 222a lto ZS Zoc o ZIZZdZ 8 Ex4 A trucking rm suspects the claim that certain tires last at least 28000 miles From a rsample of n 40 tires J 274633 S 1348 Test atOL 01 H0 2283000 vsHa zylt 283 000 X yo Sol Test stat1st1c Z 2 WW 2 252 R Z lt z01 233 In this caseZ 252 lt 233 so H0 is rejected Case 111 Sample from N y 952 unknown When 62 is un known we use as test statistic for testing H0 y no vs some alternative If H0 is true T N ln1 and the RR s are Ha R at level 06 gt 0 T gt tam lt 0 T lt locn1 75 0 TI gt loc2n1 Ex5 The maximum acceptable level of eXposure to microwave radiation in US is an average of 10 microwatts per cm It is feared that a large television transmitter may be pushing the average level of radiation above the safe limit A rsample of n 25 givesfz 103s 20 TestHO y 10 vsHa y gt10 at 06 1 f uo SLTtttit T 75 0 es sa1s1c S RR T gt 11724 2 1318 Here 75 4 1318 so H0 is not re jected Note No sample size determination for ltest Tests for the Binomial p We consider testing H0 p p0 vs some alternative only for the case of large samples In this case large samples means up 2 5 and n1 pg 2 5 The test statistic is A P Po POOPo 7 HBO is true then Z 53 N0a 1 The TR s at level or are Z Ha RRat level or p gt 0 Z 2 Zoe p lt 0 Z S ZOL P75P0 ZI ZZaz Ex6 It is thought that more than 70 of all faults in transmis sion lines are caused by lightning To gain evidence in support of this contention a rsample of 200 faults from a large data base yields that 151 of 200 are due to lightning Test H0 p 7 vs Hazpgt7at0L01 10 Sol Here I3 151200 2 755 Test statistic A 7 z p 697 73200 Since 2007 2 5 and 2003 2 5 the RR is Z gt 201 233 Here 1697 4 233 so H0 is not rejected To evaluate the performance characteristics of the testing procedure we want to look at the probability of type 11 error Formulas for the probability of type 11 error are given in p329 Here we give the formula with a brief derivation for Hapgtp0 Bp Ptype 11 error when p p Here p is some value gt p0 A P Po PZltZ pp39PWlt II Iquotgt A 1 PPltP0Zoc WIPPIgt 1 P0 P39 ZocPO1 Po gt p 1 p n 11 Ex6 Cont Find cD7 8233 73200gt 82200 CI 866 1936 Pvalues We have seen that the rejection rule depends on the chosen or value For example in Example 6 the test statistic for H0 p 7 vsHa p gt 7 is Z 1697 andHo was not rejected at on 01 since 201 233 However had we chosen or 05 H0 would have been rejected since 205 1645 When we report only the outcome of the test for one particular signi cance level we are not being as informative as we can be Thus it is good practice to always report the socalled value pvalue Pvalue is the smallest level of signi cance at which H0 would be rejected for a given data set Reporting the pvalue is very informative and in fact the p value can be used instead of the test statistic to decide the out come of the test If pvalueg or gt reject H0 at level or If pvaluegt or gt don t reject H0 at level OL PValue for a Ztest Let Z denote the test statistic of any one of the Ztests ie Case 1 Case 11 or the Binomial case Then the pvalue is given by 12 1 Z for uppertailed test P value Z for lowertailed test 21 Z for twotailed test Ex6 Cont The test statistic for H0 p 7 vs Ha p gt 7 right tailed test is Z 1697 Thus the pValue is 1 CIgt1697 2 1 CI17 0446 13 PValue for a T test This is determined in exactly the same way except that you use test statistic T instead of Z and the ltable instead of the Ztable Ex5 Cont The test statistic for H0 p 10 vs Ha p gt 10 right tailed test is T 75 In this case the ltable is not detailed enough 14

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