Lecture 23. Chapter 11: Testing Hypotheses About Proportions. Nancy Pfenning Stats Recall: last time we presented the following examples:

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1 Lecture 23 Nacy Pfeig Stats 1000 Chapter 11: Testig Hypotheses About Proportios Recall: last time we preseted the followig examples: 1. I a group of Pitt studets, 42 were left-haded. Is this sigificatly lower tha the proportio of all Americas who are left-haded, which is.12? 2. I a group of studets, 45 chose the umber seve whe pickig a umber betwee oe ad twety at radom. Does this provide covicig statistical evidece of bias i favor of the umber seve, i that the proportio of studets pickig seve is sigificatly higher tha 1/20 =.05? 3. A uiversity has foud over the years that out of all the studets who are offered admissio, the proportio who accept is.70. After a ew director of admissios is hired, the uiversity wats to check if the proportio of studets acceptig has chaged sigificatly. Suppose they offer admissio to studets ad 888 accept. Is this evidece of a chage from the status quo? Each example metios a possible value for p, which would idicate o differece/o chage/status quo. The ull hypothesis H 0 states that p equals this traditioal value. I cotrast to the ull hypothesis, each example suggests that a alterative may be true: a sigificace test problem always pits a alterative hypothesis H a agaist H 0. H a proposes that the proportio differs from the traditioal value p 0 H a rocks the boat/upsets the apple cart/ marches to a differet drummer. A key differece amog our three examples is the directio i which H a refutes H 0. I the first, it is suggested that the proportio of all Pitt studets who are left-haded is less tha the proportio for adults i the U.S., which is.12. I the secod, we woder if the proportio of studets pickig the umber seve is sigificatly more tha.05. I the third, we iquire about a differece i either directio from the stated proportio of.70. We ca list our ull ad alterative hypotheses as follows: 1. H 0 : p =.12 H a : p < H 0 : p =.05 H a : p > H 0 : p =.70 H a : p.70 I geeral, we have H 0 : p = p 0 vs. H a : p < > p 0 Note that your textbook may have expressed the first two ull hypotheses as H 0 : p.12 ad H 0 : p.05. These expressios serve well as logical opposites to the alterative hypotheses, but our strategy to carry out a test will be to assume H 0 is true, which meas we must commit to a sigle value p 0 at which to ceter the hypothesized distributio of ˆp. Thus, we will write H 0 : p = p 0 i these otes. Alteratives with < or > sigs are called oe-sided alteratives; with they are two-sided. Whe i doubt, a two-sided alterative should be used, because it is more geeral. Note: I statistical iferece, we draw coclusios about ukow parameters. Thus, H 0 ad H a are statemets about a parameter (p), ot a statistic (ˆp). We ca t argue about ˆp; its value has bee measured ad take as fact. Note: Just as success ad failure i biomial settigs lost their cootatios of favorable ad ufavorable, H a may or may ot be a desired outcome. It ca be somethig we hope or fear or simply suspect is true. However, because p 0 is a traditioally accepted value, we ll stick with H 0 uless there is covicig evidece to the cotrary: H 0 is iocet util prove guilty. How ca we produce evidece to refute H 0? By usig what probabilty theory tells us about the behavior of the R.V. sample proportio ˆp: it is cetered at p, has spread is ormal. p(1 p), ad for large eough its shape 88

2 Our strategy will be to determie if the observed value ˆp is just too ulikely to have occurred if H 0 : p = p 0 were true. If the probability of such a outcome (called the P-value) is too small, the we ll reject H 0 i favor of H a. The P-value of a test about a proportio p is the probability, computed assumig that H 0 : p = p 0 is true, that the test statistic (ˆp) would take a value at least as extreme that is, as low or as high or as differet as the oe observed. The smaller the P-value, the stroger the evidece agaist H 0. Sice 1 has H a : p <.12, the P-value is the probability of a sample proportio of left-haders as low as 42 =.113 or lower, comig from a populatio where the proportio of left-haders is.12. Based o what we leared aboutthe samplig distributio of ˆp, we kow that ˆp here, assumig H 0 is true, has mea.12(.88) p =.12, stadard error, ad a approximately ormal shape sice (.12) 45 ad (.88) 326 are both greater tha 10. [Also, we have i mid a much larger populatio of Pitt studets, certaily more tha 10() = 0.] P-value = P (ˆp.113) P (Z ) = P (Z.41) = (.88) p(1 p) Note: For cofidece itervals, sice ˆp has stadard deviatio with p ukow, we estimated it with s.e.(ˆp) = of ˆp is ˆp(1 ˆp). Now, we carry out our test assumig H 0 : p = p 0 is true, so the stadard deviatio p 0(1 p 0) ad the test statistic is z = ˆp p 0 p0 (1 p 0 ) Sice it s ot at all ulikely (probability about 34%) for a radom sample of from a populatio with proportio.12 of left-haders to have a sample proportio of oly.113 left-haders, we have o cause to reject H 0 : p =.12. The proportio of left-haders at Pitt may well be the same as for the whole coutry,.12. Because we rely o stadard ormal tables to determie the P-value, we trasform from a observed value ˆp to a stadardized value z = ˆp p 0 p0 (1 p 0 ) H a, as illustrated below, first i terms of ˆp, the i terms of z.. The way to compute the P-value depeds o the form of 89

3 P-value for Tests of Sigificace about p Observed Stadardized H a : p < p 0 H a : p < p 0 P-value P-value = P (ˆp R.V. ˆp stat ) = P (Z z) x p 0 ˆp z = ˆp p 0 p0 (1 p 0 ) 0 Z H a : p > p 0 H a : p > p 0 P-value = P (ˆp RV ˆp stat ) P-value = P (Z z) p 0 ˆp ˆp RV 0 z = ˆp p 0 p0 (1 p 0 ) Z H a : p p 0 H a : p p 0 P-value P-value = combied area = 2P (Z z ) ˆp RV Z p 0 ˆp oe of these 90 0 ˆp p z = p0 0 oe of these (1 p 0 )

4 Summary of Test of Sigificace about p Say a simple radom sample of size is draw from a large populatio with ukow proportio p of successes. We measure ˆp = X ad carry out the test as follows: < 1. Set up H 0 : p = p 0 vs. H a : p > p 0 2. Verify that the populatio is at least 10 times the sample size, ad that p 0 10 ad (1 p 0 ) 10. ˆp p The calculate stadardized test statistic z = 0 p0 (1 p 0 ) 3. Fid P-value = P (Z R.V. z statistic ) for H a : p < p 0 = P (Z R.V. z statistic ) for H a : p > p 0 = 2P (Z R.V. z statistic ) for H a : p p 0 4. Determie if the results are statistically sigificat: if the P-value is small, reject H 0 i favor of H a, ad say the data are statistically sigificat ; otherwise, we have failed to produce covicig evidece agaist H 0. [For specified α, reject H 0 if P-value < α.] 5. State coclusio i cotext of the particular problem. Let s follow these steps to solve the secod problem. I a group of studets, 45 chose the umber seve whe pickig a umber betwee oe ad twety at radom. Does this provide covicig statistical evidece of bias i favor of the umber seve, i that the proportio of studets pickig seve is sigificatly higher tha 1/20 =.05? First calculate ˆp = 45 = H 0 : p =.05 H a : p > We have i mid a very large populatio of all studets. We check that (.05) = 19 ad (.95) = 352 are both greater tha 10. Next, z = = P-value = P (Z 6.19) = P (Z 6.19) 0.05(.95) 4. Sice the P-value is very small, we reject H 0 ad say the results are statistically sigificat. 5. There is very strog evidece of bias i favor of the umber seve. I also suspected bias i favor of the umber sevetee. I a group of studets, 25 chose the umber sevetee whe pickig a umber betwee oe ad twety at radom. Does this provide covicig statistical evidece of bias i favor of the umber sevetee, i that the proportio of studets pickig sevetee is sigificatly higher tha 1/20 =.05? First calculate ˆp = 25 = H 0 : p =.05 H a : p > z = (.95) = P-value = P (Z 1.50) = P (Z 1.50) = We could call this a borderlie P-value. Next lecture, we ll discuss guidelies for how small the P-value should be i order to reject H 0, ad we ll solve the third example. Ofte, a cut-off probability α is set i advace, i which case we reject H 0 if the P-value is less tha α. 91

5 Lecture 24 Testig Hypotheses About Proportios Last time, we leared the steps to carry out a test of sigificace: < 1. Set up H 0 : p = p 0 vs. H a : p > p 0 2. I order to verify that the uderlyig distributio is approximately biomial, check that the populatio is at least 10 times the sample size. I order to justify use of a ormal approximatio to biomial ˆp p proportio, check that p 0 10 ad (1 p 0 ) 10. Calculate stadardized test statistic z = 0 3. Fid P-value = P (Z R.V. z statistic ) for H a : p < p 0 = P (Z R.V. z statistic ) for H a : p > p 0 = 2P (Z R.V. z statistic ) for H a : p p 0 p0 (1 p 0 ) 4. Assess sigificace: if the P-value is small, reject H 0 i favor of H a, ad say the data are statistically sigificat ; otherwise, we have failed to produce covicig evidece agaist H 0. [For specified α, reject H 0 if P-value < α.] 5. State coclusios i cotext. Last time we bega to solve the followig example: Whe studets are asked to pick a umber at radom from oe to twety, I suspect their selectios will show bias i favor of the umber sevetee. I a group of studets, 25 chose the umber sevetee. Does this provide covicig statistical evidece of bias i favor of the umber sevetee, i that the proportio of studets pickig sevetee is sigificatly higher tha 1/20 =.05? The ull ad alterative hypotheses were H 0 : p =.05 H a : p >.05 ad so the z-statistic was z = = 1.50 ad the P-value was = P (Z 1.50) = P (Z 1.50) = (.95) Step 4 says to reject H 0 if the P-value is small. How small is small? Sometimes, it is decided i advace exactly how small the P-value would have to be to lead us to reject the ull hypothesis: a cut-off probability α is prescribed i advace. The, if the P-value is less tha α, we reject H 0, ad say the results are statistically sigificat at level α. Otherwise, we do ot have sufficiet evidece to reject H 0. Is there evidece of bias i favor of the umber sevetee at the α =.05 level? The P value is.0668, which is ot less tha.05, so by this criterio it is ot small eough to reject H 0. It could be that studets did t have ay systematic preferece for the umber sevetee, ad the proportio of sevetees selected was a bit high oly by chace. A uiversity has foud over the years that out of all the studets who are offered admissio, the proportio who accept is.70. After a ew director of admissios is hired, the uiversity wats to check if the proportio of studets acceptig has chaged sigificatly. Suppose they offer admissio to studets ad 888 accept. Is this evidece at the α =.05 level that there 92

6 has bee a real chage from the status quo? How about at the.02 level? First we fid that ˆp = 888 =.73 is the sample proportio of studets who accepted admissio. 1. Set up H 0 : p =.70 vs. H a : p Both coditios are satisfied. z = (.3) = Because of the two-sided alterative, our P-value is = 2P (Z 2.27 ) = 2P (Z 2.27) = 2(.0116) = Sice.0232 <.05, we have evidece to reject at the 5% level. But.0232 is ot less tha.02, so we do t have evidece to reject at the 2% level. 5. If we set out to gather evidece of a chage i either directio for overall proportio of studets acceptig admissio, we would say yes with a cutoff of.05, o with a cutoff of.02. Thus, this test is rather icoclusive. A uiversity has foud over the years that out of all the studets who are offered admissio, the proportio who accept is.70. After a ew director of admissios is hired, the uiversity wats to check if the proportio of studets acceptig has icreased sigificatly. Suppose they offer admissio to studets ad 888 accept. Is this evidece at the α =.05 level that there has bee a sigificat icrease i proportio of studets acceptig admissio? How about at the.02 level? Agai we fid that ˆp = 888 =.73 is the sample proportio of studets who accepted admissio. 1. The subtle re-phrasig of the questio ( icreased istead of chaged ) results i a differet alterative hypothesis. H 0 : p =.7 vs. H a : p >.7 2. The z statistic is uchaged: z = (.3) = Because of the oe-sided alterative, our P-value is = P (Z 2.27) = P (Z 2.27) = (.0116) 4. Sice.0116 <.05, we agai have evidece to reject at the 5% level. This time,.0116 is also less tha.02, so we also have evidece to reject at the 2% level. 5. If we set out to gather evidece of icreased overall proportio of studets acceptig admissio, we would say yes, we have produced evidece of a icrease, whether the α =.05 or α =.02 level is used. The previous examples demostrate that 1. It is more difficult to reject H 0 for a two-sided alterative tha for a oe-sided alterative. I geeral, the two-sided P-value is twice the oe-sided P-value. The oe-sided P-value is half the two-sided P-value. 2. It is more difficult to reject H 0 for lower levels of α. Calculatig the P-value i Step 3 gives us the maximum amout of iformatio to carry out our test we kow exactly how ulikely the observed ˆp is. If a cut-off level α is prescribed i advace, the it is possible to bypass the calculatio of the P-value i Step 3. Istead, the z-statistic is compared to the critical value z associated with α. For example, if we have a two-sided alterative ad α is set at.05, the the rejectio regio would be where the test-statistic z exceeds 1.96 i absolute value. The disadvatage to this method is that it provides oly the bare miimum of iformatio eeded to decide whether to reject H 0 or ot. We will ot employ the rejectio regio method i this course, but studets should be aware of it i case they ecouter it i other cotexts. A method that falls somewhere i betwee those which provide maximum ad miimum iformatio is the followig: close i o the P-value by surroudig the z statistic with eighborig values z from the 93

7 ifiite row of Table A.2. The advatage to this method is that it familiarizes us with the use of Table A.2, which will be eeded whe we carry out hypothesis tests about ukow populatio mea of a quatitative variable. Note that z = correspods to a area of.90 symmetric about zero, so each tail probability, that z takes a value less tha or greater tha , is.05. z = correspods to a area of.95 symmetric about zero, so each tail probability, that z takes a value less tha or greater tha , is.025. z = correspods to a area of.98 symmetric about zero, so each tail probability, that z takes a value less tha or greater tha , is.01. z = correspods to a area of.99 symmetric about zero, so each tail probability, that z takes a value less tha or greater tha , is.005. These tail probabilities may be peciled i at the top or bottom eds of the colums i Table A.2 for easy referece. We will ow re-solve some of our earlier examples, usig Table A.2 istead of Table A.1. A uiversity has foud over the years that out of all the studets who are offered admissio, the proportio who accept is.70. After a ew director of admissios is hired, the uiversity wats to check if the proportio of studets acceptig has icreased sigificatly. Suppose they offer admissio to studets ad 888 accept. Is this evidece at the α =.05 level that there has bee a sigificat icrease i the proportio of studets acceptig? First we foud that ˆp = 888 =.73 is the sample proportio of studets who accepted admissio ad set up H 0 : p =.70 vs. H a : p.70. Next we calculated z = = (.3) Our P-value is = P (Z 2.27). Accordig to Table A.2, z = 2.27 is betwee z = ad z = Therefore, our p-value, P (Z 2.27), is betwee.025 ad.01, which meas it must be less tha.05. We ca reject H 0 at the 5% level. [Recall: Table A.1 showed the precise P-value to be.0116, which is i fact betwee.025 ad.01.] A uiversity has foud over the years that out of all the studets who are offered admissio, the proportio who accept is.70. After a ew director of admissios is hired, the uiversity wats to check if the proportio of studets acceptig has chaged sigificatly. Suppose they offer admissio to studets ad 888 accept. Is this evidece at the α =.05 level that there has bee a real chage (i either directio) from the status quo? First we foud that ˆp = 888 =.73 is the sample proportio of studets who accepted admissio, ad we set up H 0 : p =.70 vs. H a : p.70. Next we calculated z = = (.3) Because of the two-sided alterative, our P-value is = 2P (Z 2.27 ) = 2P (Z +2.27). Accordig to Table A.2, z = 2.27 is betwee z = ad z = Therefore, P (Z 2.27) is betwee.025 ad.01, ad the P-value, 2P (Z 2.27), is betwee 2(.025) ad 2(.01), that is, betwee.05 ad.02. We still ca reject H 0 at the 5% level, but ot at the 2% level. I a group of Pitt studets, 42 were left-haders, which makes the sample proportio.113. Is this sigificatly lower tha the proportio of Americas who are left-haders, which is.12? Earlier we foud the z-statistic to be =.41 ad the P-value to be P (Z.41)..12(.88) Cosultig Table A.2, we see that -.41 is less extreme tha 1.645, so the P-value is larger tha.05. Agai, we have failed to produce ay evidece agaist H 0. 94

8 Whe studets are asked to pick a umber at radom from oe to twety, I suspect their selectios will show bias i favor of the umber sevetee. I a group of studets, 25 chose the umber sevetee. Does this provide covicig statistical evidece of bias i favor of the umber sevetee, i that the proportio of studets pickig sevetee is sigificatly higher tha 1/20 =.05? The ull ad alterative hypotheses were H 0 : p =.05 z = (.95) H a : p >.05 ad so the z-statistic was = 1.50 ad the P-value was = P (Z 1.50). Istead of usig Table A.1 to fid the precise P-value, we ote from Table A.2 that 1.50 is less tha 1.645, so the tail probability must be greater tha =.05. Thus, our P-value= P (Z 1.50) is greater tha.05 ad we do ot have covicig evidece of bias. Note: Earlier we foud the exact P-value to be P (Z 1.50) =.0668, which is ideed greater tha.05. Note: I a previous, we bega by assumig that the proportio of freshme takig itro Stats classes is.25. Accordig to survey data, we foud the sample proportio of freshme to be.08. By had we calculated the probability of a sample proportio this low, comig from a populatio with proportio.25: it was approximately zero. I characterized this as virtually impossible ad decided ot to believe that the overall proportio of freshme is.25. Alteratively, I could use MINITAB to test the hypothesis that populatio proportio is.25, vs. the less tha alterative. Sice year allows for more tha two possibilities, it is ecessary to use Stat, the Tables, the Tally to cout the umber of freshme (35). The use the Summarized Data optio i the 1 Proportio procedure, specifyig 445 as the Number of Trials ad 35 as the Number of Successes. I opted to use test ad iterval based o ormal distributio, sice that s how I origially solved the problem by had. The p-value is zero, ad I reject the ull hypothesis i favor of the alterative. I agai coclude that the proportio of freshme i itro Stats classes (at least i the Fall) is less tha.25. Tally for Discrete Variables: Year Year Cout other 14 N= 445 *= 1 Test ad CI for Oe Proportio Test of p = 0.25 vs p < 0.25 Sample X N Sample p 95.0% Upper Boud Z-Value P-Value Exercise: I a previous Exercise, we explored the samplig distributio of sample proportio of females, whe radom samples are take from a populatio where the proportio of females is.5. We oted the sample proportio of females amog surveyed Stats studets, ad calculated by had the probability of observig such a high sample proportio, if populatio proportio were really oly.5. We used this probability to decide 95

9 whether we were willig to believe that populatio proportio is i fact.5. For this Exercise, address the same questio by carryig out a formal hypothesis test usig MINITAB. Be sure to specify the appropriate alterative hypothesis. State your coclusios clearly i cotext. Lecture 25 Type I ad Type II Error Whe we set a cutoff level α i advace for a hypothesis test, we are actually specifyig the log-ru probability we are willig to take of rejectig a true ull hypothesis, which is oe of the two possible mistake decisios that ca be made i a hypothesis test settig. Recall our testig-for-disease example i Chapter 7, i which the probability of a false positive was.015, probability of false egative was.003. All the possibilities for Decisio ad Actuality are show i the table below. If we decide to use.015 as our cut-off probability (p-value <.015 meas reject H 0 ; otherwise do t reject), the.015 is the probability of makig a Type I Error the probability of rejectig the ull hypothesis, eve though it is true. That meas the probability of correctly acceptig a true ull hypothesis is =.985. I medical situtatios, this is the specificity of the test. Actuality Decisio Healthy(H 0 true) Diseased(H a true) Healthy correct icorrect(false eg.) (do t reject H 0 ) prob.=specificity=.985 Type II error(prob.=.003) Diseased icorrect(false pos.) correct (accept alt.hyp.) Type I error(prob.=.015) prob.=sesitivity =power=.997 I our example, we were told the probability of a false egative, or Type II Error. Thus, the probability of a correct positive for a ill perso (called the sesitivity of the test) = 1 mius the probabilty of Type II error. Statisticias refer to this probability as the power of the test. I a z test about populatio proportio p, the probability of a Type II error [icorrectly failig to reject the ull hypothesis whe the alterative is true] ca oly be calculated if we are told specifically the actual value of the populatio proportio. Thus, we eed to kow the alterative proportio which cotradicts the ull hypothesized proportio. What we do ot eed i order to calculate the probability of Type II error is the value of a observed proportio ˆp. Our probability is about the test itself, ot about the results. Rather tha focusig o makig such calculatios, we will istead thik carefully about the implicatios of makig Type I or Type II errors. For our medical example above, the probability of icorrectly tellig a healthy perso that he or she does have AIDS is higher tha the probability of icorrectly tellig a ifected perso that he or she does ot have AIDS. If a healthy perso iitially tests positive (Type I error), the the cosequece (besides cosiderable axiety) is a subsequet, more discerig test, which has a better chace of makig the correct diagosis secod time aroud. If a ifected perso tests egative (Type II error), the the cosequeces are more dire, because treatmet will be withheld, or at best delayed, ad there is the risk of further ifectig other idividuals. Thus, i this case it makes sese to live with a higher probability of Type I error i order to dimiish the probability of Type II error. 96

10 Cosider the followig legal example: the ull hypothesis is that the defedat is iocet ad the alterative is that the defedat is guilty. The trial weighs evidece as i a hypothesis test i order to decide whether or ot to reject the ull hypothesis of iocece. What would Type I ad II errors sigify i this cotext? A Type I error meas rejectig a ull hypothesis that is true, i other words fidig a iocet perso guilty. Most people would agree that this is much worse tha committig a Type II error i this cotext, which would be failig to covict a guilty perso. Dr. Stephe Fieberg of CMU did extesive work for the govermet is assessig the effectiveess of lie-detector tests. He cocluded that probabilities of committig both types of error were so high that he ad a pael of ivestigators recommeded discotiuig the use of such tests. A peek at a brai ca umask a liar tells about the most recet techology for ew sorts of lie detectors to replace the old-fashioed polygraph. Role of Sample Size Suppose oe demographer claims that there are equal proportios of male ad female births i a certai state, whereas aother claims there are more males. They use hospital records from all over the state to sample 10,000 recet births, ad fid 5120 to be males, or ˆp =.512. They.5(.5) 10,000 test H 0 : p =.5 vs. H a : p >.5 ad calculate z = = 2.4, so the P-value is.0084, quite small. Does this mea (a) they have evidece that the populatio proportio of male births is much higher tha.5; or (b) they have very strog evidece that the populatio proportio of male births is higher tha.5? The iterpretatio i (b) is the correct oe; (a) is ot. Especially whe the sample size is large, we may produce very strog evidece of a relatively mior differece from the claimed p 0. Coversely, if is too small, we may fail to gather evidece about a differece that is quite substatial. A Statistics recitatio istructor suspects there to be a higher proportio of females overall i Stats classes. She observes 12 females i a group of 20 studets, so ˆp =.6. Does this cofirm her suspicios? She would test H 0 : p =.5 vs. H a : p >.5. First she verifies that 20(.5) ad 20(1.5) are both 10, just barely satisfyig our coditio for a ormal approximatio. Also, she has i mid a populatio i the hudreds or eve thousads, so the biomial model applies. She calculates z =.6.5 =.89. The P-value is.1867, providig her with o statistical evidece.5(.5) 20 to support her claim. I fact, the populatio proportio of females really is greater tha.5, but this sample size was just too small to prove it. I cotrast, a lecture class of 80 studets with ˆp =.6 would produce a z statistic of 1.79 ad a P-value of Remember that z = ˆp p 0 p 0(1 p 0) = (ˆp p 0) p0 (1 p 0 ) We reject H 0 for a small P-value, which i tur has arise from a z that is large i absolute value, o the friges of the ormal curve. There are three compoets that may result i a z that is large i absolute value, which i tur cause us to reject H 0 : 1. What people ted to focus o as the cause of rejectig H 0 is a large differece ˆp p 0 betwee the observed proportio ad the proportio proposed i the ull hypothesis. This aturally makes z large ad the P-value small. 97

11 2. A large sample size, because is actually multiplied i the umerator of the test statistic z, brigs about a large z ad a small P-value. Coversely, a small sample size may lead to a smaller z ad failure to reject H 0, eve if it is false (a Type II error). 3. If p 0 is close to.5, the p 0 (1 p 0 ) is cosiderably larger tha it is for p 0 close to 0 or 1. (For example, p0 (1 p 0 ) is.5 for p 0 =.5, but it is.1 for p 0 =.01 or.09.) Whe Hypothsis Tests are Not Appropriate Remember that we carry out a hypothesis test, based o sample data, i order to draw coclusios about the larger populatio from which the sample was obtaied. Hypothesis tests are ot appropriate if there is o larger group beig represeted by the sample. I 2002, the govermet requested ad wo approval for 1228 special warrats for secret wiretaps ad searches of suspected terrorists ad spies. Is this sigificatly higher tha 934, which was the umber of special warrats approved i 2001? Statistical iferece is ot appropriate here because 1228 ad 934 represet etire populatios for 2002 ad 2001; they are ot sample data. 928,000 I 2000, 1238,000 =.75 of all bachelor s degrees were eared by whites. Is this sigificatly lower tha.86, the proportio of all bachelor s degrees eared by whites i 1981? We would ot carry out a sigificace test, because the give proportios already describe the populatio. A iteret review of home pregacy tests reports: Home pregacy testig kits usually claim accuracy of over 95% (whatever that may mea). The reality is that the literature cotais iformatio o oly four kits evaluated as they are iteded to be used by wome testig their ow urie. The results we have suggest that for every four wome who use such a test ad are pregat, oe will get a egative test result. It also suggests that for every four wome who are ot pregat, oe will have a positive test result. From this iformatio we ca idetify the probabilities of both Type I ad II errors, accordig to the review, as beig 1 i 4, or 25%. Goorrhea is a very commo ifectious disease. I 1999, the rate of reported goorrhea ifectios was per 100,000 persos. A polymerase chai reactio (PCR) test for goorrhea is kow to have sesitivity 97% ad specificity 98%. What are the probabilities of Type I ad Type II Errors? Give the high degree of accuracy of the test, if a radomly chose perso i the U.S. is routiely screeed for goorrhea, ad the test comes up positive, what is the probability of actually havig the disease? The ull hypothesis would be that someoe does ot have the disease. A Type I Error would be rejectig the ull hypothesis, eve though it is true: testig positive whe a perso does ot have the disease. A Type II Error would be failig to reject the ull hypothesis, eve though it is false: testig egative whe a perso does have the disease. A sesitivity of 97% meas that if someoe has the disease, the probability of correctly testig positive is 97%, ad so the probability of testig egative (whe someoe has the disease) is 3%: this is the probability of a Type II error. A specificity of 98% meas that if someoe does ot have the disease, the probability of correctly testig egative is 98%, ad so the probability of 98

12 testig positive (whe someoe does ot have the disease) is 2%: this is the probability of a Type I error. A two-way table makes it easier to idetify the probability we are seekig (of havig the disease, give that the test is positive). We begi with a total of 100,000 people, of whom 132 have the disease (the remaiig 999,868 do ot). Sesitivity 97% meas 127 of the 132 with goorrhea test positive. Specificity 98% meas 979,871 of the 999,868 people without goorrhea test egative. The remaiig couts ca be filled i by subtractio: Positive Negative Total Goorrhea No Goorrhea 1,997 97,871 99,868 Total 2,124 97, ,000 Of the 2,124 people who test positive, 127 actually have the disease: if someoe tests positive, 127 the probability of havig the disease is 2,124 =.06. Remember, however, that this probability applies to a radomly chose perso beig screeed. If someoe is screeed because of exhibitig symptoms, the probability is of course higher. Exercise : Refer to the article How ot to catch a spy: Use a lie detector, which reports at the bottom of the first colum, Eve if the test were desiged to catch eight of every 10 spies, it would produce false results for large umbers of people. For every 10,000 employees screeed, Fieberg said, eight real spies would be sigled out, but 1,598 iocet people would be sigled out with them, with o hit of who s a spy ad who is t. Based o this iformatio, set up a two-way table, classifyig 10,000 employees as actually beig spies or ot, ad beig sigled out as a spy by the lie detector or ot. Report the probability of a Type I Error ad of a Type II Error. If someoe is idetified by the lie detector as beig a spy, what is the probability that he or she is actually a spy? 99