Comments on Discussion Sheet 18 and Worksheet 18 ( ) An Introduction to Hypothesis Testing

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1 Commet o Dicuio Sheet 18 ad Workheet 18 ( ) A Itroductio to Hypothei Tetig Dicuio Sheet 18 A Itroductio to Hypothei Tetig We have tudied cofidece iterval for a while ow. Thee are method that allow you to ue a iterval to etimate the value of a populatio parameter, like µ or σ, with a kow probability of beig correct. Sometime, we eed to awer a "ye or o" quetio about thee parameter whe we make a certai hypothei about their value. Thi i imilar to uig cofidece iterval but lightly differet. We itroduce how to do a hypothei tet i thi hadout ad compare it to cofidece iterval. We iitially focu o hypothei tet for the mea. A Workig Hypothei ad it Alterative Every hypothei tet begi by makig a aumptio, a hypothei about what i true of a populatio parameter. Thi iitial or workig hypothei i ometime called the ull hypothei, that i, it i the oe that we aume before we have examied ay relevat data. It i deoted H 0. It i aumed to be true util we have reao to believe otherwie. Every ull or workig hypothei ha a logical complemet, that i, a differet hypothei that mut be true if the ull or workig hypothei i ot. Thi i called the alterative hypothei ad i deoted H a. 1. Suppoe our workig hypothei i H 0 : µ 5, that i that the populatio mea of a radom variable y equal 5. If we had a radom ample of data about y of ize 36 that had a ample mea of y 8 ad a ample tadard deviatio of, what would be the value of z y µ give our workig hypothei? Commet: Sice our workig hypothei i H 0 : µ 5, we aume that µ 5. We are alo give that y 8,, ad 36. So we have z y µ So the value of z i z.5, aumig our workig hypothei H 0 : µ 5 i correct o that we ca ubtitute 5 for µ.. If our ull hypothei i H 0 : µ 5, what would be our alterative hypothei, H a. Remember: H a i what logically ha to be true if H 0 i fale. Commet: Whe comparig ay two real umber, a ad b, there are three poibilitie: a b, a < b ad a > b. So whe we compare µ to 5 our ull hypothei, H 0 : µ 5, cotai oe of thoe three logical poibilitie. The other two are µ < 5 ad µ > 5. Thee are the two poibilitie that have to be icluded i our alterative hypothei, H a. Whe µ < 5 or µ > 5 ha to be true, the mathematical horthad for thoe two cae (everythig but µ 5 ) i µ 5. So our alterative hypothei i H a : µ 5. That i, if H 0 : µ 5 i fale, the logically H a : µ 5 mut be true. 3. Give our workig hypothei, H 0 : µ 5, ad the ample data we have i Quetio 1, what i the probability of gettig a value of Z at leat a big or bigger tha the value of Z we foud i Quetio 1? Commet: 1

2 I Quetio 1 we foud that z.5 with the workig (ull) hypothei H 0 : µ 5 ad the ample data give. So we wat to kow what P( z >.5) i. We kow that the amplig ditributio for y i a ormal ditributio (due to the Cetral Limit Theorem), that it mea i E y ( ) µ, ad that it tadard deviatio i σ which for reaoably large ( 30) ample i approximately (which i true here ice ). So z y µ would be ormally ditributed with a mea of 0 ad a tadard deviatio of 1. That i, it would be a tadard ormal ditributio ad probabilitie for it could be looked up i Table o page 109. Sice our workig (ull) hypothei i that H 0 : µ 5 the we are aumig that z y 5 data ad the hypothei H 0 : µ 5 the z y 5 i a tadard ormal radom variable. We have foud that, give the ample i z.5. What i P( z >.5) for a tadard ormal radom variable? We ee that the bigget value i Table i 3.09 ad that P( 0 < z < 3.09).990. Thi mea that P( z < 3.09) ad o P( z > 3.09) 1 P( z < 3.09) However, ice.5 > 3.09 thi mea that P( z >.5) < P ( z > 3.09) ice.5 i further to the right tha 3.09 ad thu the piece of the tail uder the curve above it i le tha the piece of the tail uder the curve above Thi mea that P( z >.5) < P ( z > 3.09) So with the workig hypothei H 0 : µ 5 ad the ample data give, we have P( z >.5) <.0001, that i, it i practically 0.. Give the probability that you foud i Quetio 3, would you coider that the ull hypothei, H 0 : µ 5, i likely to be true or ot. Commet: If the ull (workig) hypothei H 0 : µ 5 i correct ad the ample data give i Quetio 1 i correct the oe of two thig ha to be true: The evet z.5 ha happeed or the hypothei (aumptio) i ot correct. The probability of the evet z.5 i P( z >.5) <.0001 if the workig hypothei i correct. So we have to chooe: either omethig that had le tha oe chace i 10,000 ha happeed or the workig aumptio (ull hypothei) i icorrect. Uder thee circumtace, it eem more reaoable to aume that the ull hypothei i ot likely to be true. I fact, if we aume that it i ot true, the our chace of beig wrog i that aumptio i equal to the probability of the ulikely evet happeig with the ull hypothei true, that i, le tha So it eem that it i very ulikely that H 0 : µ 5 i true. 5. Give what you awered to Quetio, what cocluio would you draw about your workig hypothei baed o the data give i Quetio 1? Commet: If we aume that the workig hypothei, H 0 : µ 5, i ot true, the the alterative i captured i the alterative hypothei we foud i Quetio, H a : µ 5, mut logically be what we aume to be true. So baed o our workig hypothei ad the data we are give, give what we aw i Quetio, we mut aume that H a : µ 5 i true rather tha H 0 : µ 5. ( ) P z > a 6. Fid a umber a uch that P z < a or z > a Quetio 1. ( ).05 where z i the value you foud i Commet: We are coiderig a tadard ormal radom variable, z, that i, a ormally ditributed radom variable with mea 0 ad tadard deviatio 1. We ca look up probabilitie related to it i Table o page 109. Whe we look at P z < a or z > a ( ) ( z > a) ( ), we are lookig at P z < a [ ], the uio of

3 two evet that are dijoit ice we ca't have z < a ad z > a true at the ame time, whatever a i. Thi mea that P[ (z < a ) (z > a)] P( z < a) + P( z > a). Sice the tadard ormal ditributio i ymmetric, we kow P( z < a ) P(z > a), that i the probability of the part of the tail above a uder the curve for the deity fuctio i the ame a the probability of the part of the tail below a uder the curve for the deity fuctio. So P[ (z < a ) (z > a)] P( z < a) + P( z > a) P( z > a) + P( z > a ) P( z > a) Ad we wat P( z < a or z > a).05 o we wat P( z > a ).05 which mea P( z > a) Ufortuately, we ca oly fid P(0 < z < a) i Table. If P( z > a).05, what i P(0 < z < a)? Well, if P( z > a).05 the P( z < a) 1 P( z > a) That mea that P( z < a).5 + P( 0 < z < a).9750 o P(0 < z < a) Now remember that iide Table are probabilitie ad the row ad colum of Table how you which value of z have thoe probabilitie. So we look iide Table to fid a umber a cloe to.750 a we ca get. We ca actually fid a etry i Table that i exactly.750. It occur i the row for z 1.9 ad i the colum for.06. So it correpod to z That i, P(0 < z < 1.96).750 which mea that P( z < 1.96 ).9750 which mea that P( z > 1.96 ).05. Sice the ormal ditributio i ymmetrical (the left ide [below 0] i a mirror image of the right ide [above 0]), thi alo mea that P( z < 1.96) P (z > 1.96 ).05. So P[ (z < a ) (z > a)] P( z < a) + P( z > a) Thi i what we wat. P[ (z < a ) (z > a)] i jut a mathematically more precie way of writig P( z < a or z > a), o we have P( z < 1.96 or z > 1.96).05. What doe it mea to ay " z < 1.96 or z > 1.96 "? It mea that 1.96 < z < 1.96 i ot true. But 1.96 < z < 1.96 i the ame a ayig z < 1.96, o " z < 1.96 or z > 1.96 " mea that z < 1.96 i ot true, o that mea z > 1.96 i true. So we have P( z < 1.96 or z > 1.96) P( z > 1.96 ).05 ad the a we wat i a What would you coclude from the fact that the z you foud i Quetio 1 i greater tha the a you foud Quetio 6. Commet: I Quetio 1 we aw that with the workig (ull) hypothei H0 : µ 5 ad the ample data give the z.5. Notice that.5 > 1.96, o that mea that P( z >.5 ) < P ( z >.5) < P( z > 1.96 ).05 o we kow that P( z >.5 ) <.05. Thi mea that either H0 : µ 5 i fale or H0 : µ 5 i true ad a evet, z.5, that ha P( z >.5 ) <.05 (oe chace i 0) ha happeed. A before, it i more reaoable to aume that the workig hypothei H0 : µ 5 i fale tha it i to aume that omethig that ha a probability of.05 ha happeed. So we reject H0 : µ 5 ad aume that Ha : µ 5 i true. We have a probability of of beig correct. That i, it would be correct 95% of the time. What i the advatage of what we did i Quetio 7 over what we did i Quetio ad 5? There really i't much of a advatage. The awer i Quetio ad 5 were more precie ad had a maller probability of beig icorrect. However, the advatage of the method i Quetio 7 i that the umber 1.96 ca be looked up oce ad for all ad be attached to a probability of.95 of beig correct (or, which i the ame thig, a probability of.05 of beig wrog i rejectig H0 ). By ettlig o a few key probability level (called "igificace level") of beig correct.90,.95,.99, etc. we reduce the pecial umber like 1.96 we have to look up or lear. We ca fid thoe few ad ue them without all the work we wet to i Quetio ad 5. Ay time I wat to ee if a H0 i correct with a 95% chace of beig right, I aume that the H0 i correct, calculate the value of z ad ee whether or ot it i true that P( z > 1.96). If it i, I reject H0 ad aume the alterative. If it i ot, I do ot reject H0 3

4 ad itead keep it a my workig hypothei. By comparig the z I calculate to 1.96, I ca make my deciio with a kow probability (.95) of beig correct. The Five Part of a Hypothei Tet Every hypothei tet ha five part. They are a follow: 1. H 0, a workig or ull hypothei. H a, the (logically) alterative hypothei to the workig hypothei 3. A tet tatitic, that i a tatitic computed with ample data combied with what the workig hypothei ay about a populatio parameter.. A rejectio regio, that i, a tatemet about the value of the tet tatitic that if it were true for the value of the tet tatitic would have uch a low probability of happeig that you would rather coclude that the workig hypothei, H 0, i correct tha to coclude that uch a low probability evet ha occurred. 5. A cocluio about the workig hypothei, H 0, o the bai of aumig it i true, calculatig the value of a tet tatitic, ad determiig if it fall ito a rejectio regio, that i, that thi would mea uch a low probability evet had happeed that it i better to reject the workig hypothei, H 0, tha to aume the low probability evet happeed. The cocluio i either i the form "reject H 0 " or "do ot reject H 0 ". It i ever i the form "accept H 0 " becaue the mot you ca prove by a hypothei tet i that the data you have are ot ufficiet to allow you to reject H Puttig together your awer to Quetio 1 through 7, tate the five part of the hypothei tet that you were performig i awerig them. Commet: H 0 : µ 5, the ull hypothei H a : µ 5, the alterative hypothei z y µ , the value of the tet tatitic (give H 0 : µ 5 ad the ample data) z > 1.96, the rejectio regio with a probability of.95 of leadig to the correct deciio Reject H 0 : µ 5, the formal cocluio to the hypothei tet. Sice we ca reject H 0 : µ 5 with a probability of.95 of beig correct, we do o ad aume itead that H a : µ 5 i correct. Thi i kow a performig a hypothei tet at the 95% igificace level ice we have a probability of.95 of makig the correct deciio (that i, we hould make the correct deciio 95% of the time uig thi procedure). Workheet 18 A Itroductio to Hypothei Tetig Thi i a cotiuatio of Dicuio Sheet 18. Pleae complete that ow if you have t already doe o. Thi correpod to 9.5 ad 9.6 i the textbook Pleae read thoe ectio. We will treat all hypothei tet for a mea a uig Studet t-ditributio ad a with σ ukow. Igore the part i the textbook about eparate form of hypothei tet for large ample with σ kow or ukow

5 ( ) P ( t > a).05 i Quetio 6 of Dicuio Sheet 18, fid a b uch that P ( t < b or t > b) P ( t > b).01 with ν 1 df. 1. Itead of the a uch that P t < a or t > a Commet: Agai we are coiderig a tadard ormal radom variable, z, that i, a ormally ditributed radom variable with mea 0 ad tadard deviatio 1. We ca look up probabilitie related to it i Table o page 109. Whe we look at P( z < b or z > b ), we are lookig at P[ ( z < b) ( z > b) ], the uio of two evet that are dijoit ice we ca't have z < b ad z > b true at the ame time, whatever b i. So P[ ( z < b) ( z > b) ] P( z < b) + P( z > b ). Sice the tadard ormal ditributio i ymmetric, we kow P( z < b) P( z > b), that i the probability of the part of the tail above b uder the curve for the deity fuctio i the ame a the probability of the part of the tail below b uder the curve for the deity fuctio. So P[ ( z < b) ( z > b) ] P( z < b) + P( z > b ) P( z > b) + P( z > b) P( z > b) Ad we wat P( z < b or z > b ).01 o we wat P( z > b ).01 which mea P( z > b) Ufortuately, we ca oly fid P( 0 < z < b) i Table. If P( z > b).005, what i P( 0 < z < b)? Well, if P( z > b).005 the P( z < b) 1 P( z > b) That mea that P( z < b).5 + P( 0 < z < b).9950 o P( 0 < z < b) So we look iide Table to fid a umber a cloe to.950 a we ca get. We ee that the etry for.57 i.99. The etry for.58 i.951. The value we wat i.950, which i betwee.00 ad.951. I fact, So we chooe a value halfway betwee.57 ad.58, that i,.575. That i, P( 0 < z <.575).950 which mea that P( z <.575).9950 which mea that P( z >.575).005. Sice the ormal ditributio i ymmetrical, P( z <.575) P( z >.575).005. So P ( z <.575) ( z >.575) ( ) + P( z >.575) [ ] P z <.575 By the way, we really oly kow that b i betwee.57 ad.58. To fid a more precie awer, we might ue liear iterpolatio. However, liear iterpolatio i ot correct with a ormal ditributio, o we will alway ue the value halfway betwee the two table value whe we fid two probabilitie that the deired probability lie betwee. I thi cae, ice the probability we wat i betwee.99 ad.951, which correpod repectively to.57 ad.58, we will ue the value halfway betwee, that i, z.575 ad ever bother with iterpolatio. [ ] i jut a mathematically more precie way of writig Thi i what we wat. P ( z < b) ( z > b) P( z < b or z > b ), o we have P( z <.575 or z >.575).01. What doe it mea to ay " z <.575 or z >.575 "? It mea that.575 < z <.575 i ot true. But.575 < z <.575 i the ame a ayig z <.575, o " z <.575 or z >.575 " mea that z <.575 i ot true, o that mea z >.575 i true. So we have P z <.575 or z >.575 wat i b.575. ( ) P z >.575 ( ).01 ad the b we 5

6 . If you ued thi regio i Quetio 1 a a rejectio regio for your hypothei tet of the H 0 i Quetio 1 of Dicuio Sheet 18 itead of the rejectio regio P t > a ( ) from Quetio 6 i Dicuio Sheet 18, what would you coclude ad what would be differet about the hypothei tet you had performed? Commet: We are till lookig at z y µ but ow or rejectio regio would be z >.575. Sice.5 >.575 we would reject H 0 : µ 5 a before, but we would kow that we had a probability of at leat.99 of beig correct (.01 of beig wrog). Thi would be a hypothei tet at the 99% igificace level. It part would be H 0 : µ 5, the ull hypothei H a : µ 5, the alterative hypothei z y µ , the value of the tet tatitic (give H 0 : µ 5 ad the ample data) z >.575, the rejectio regio with a probability of.95 of leadig to the correct deciio Reject H 0 : µ 5, the formal cocluio to the hypothei tet. A before, we would the aume the alterative, H a : µ 5, i correct. 3. Give the data i Quetio 1 from Dicuio Sheet 18, fid a 99% cofidece iterval for µ. How doe thi compare with your rejectio regio i Quetio 6 of Dicuio Sheet 18? Commet: For a mea, µ, ad a large ample, the form of a 95% cofidece iterval i: y z α < µ < y + z α which i thi cae would give 8 z < µ < 8 + z z.05 < µ < 8 + z < µ < < µ < < µ < So our 95% cofidece iterval i about ( 6.7,9.3). If our rejectio regio wa z > 1.96 that mea we would ot reject if z < That i, we would ot reject if 1.96 < z < < y µ < < y µ < 1.96 y 1.96 < µ < y < µ <

7 < µ < That i, we would ot reject H 0 : µ 5 if the hypotheized value of µ wa iide the cofidece iterval. The differece betwee a hypothei tet ad a cofidece iterval i that for a hypothei tet we aume a value for the parameter, ue it to calculate the tet tatitic ad compare that to the rejectio regio to make a deciio. For a cofidece iterval, we do't make a aumptio about the value of the parameter but calculate a iterval that it would likely be i. That mea hypothei tet are bet i we wat a ye/o awer while cofidece iterval are bet if we wat a umerical etimate. A 100(1-α)% Hypothei Tet for µ A (two-ided) hypothei tet for a populatio mea, µ, whe the populatio tadard deviatio, σ, i ukow but the ample ize i > 30 i a follow: 1. H 0 : µ µ 0, where µ 0 i ome give cotat. H a : µ µ 0, the alterative hypothei 3. A tet tatitic t y µ 0, where y,, ad come from ample data ad µ 0 come from H 0. ( ) α (, P t < t α,ν ) α ( ) ( P t > t α,ν ) α.. A rejectio regio, zt > t α,ν where P t > t α,ν P t > t α,ν or t < t α,ν 5. A cocluio : "reject H 0 " or "do ot reject H 0 ". ad o. Suppoe we have a radom ample of data about y of ize 5 that had a ample mea of y 8 ad a ample tadard deviatio of. We wat to kow whether or ot µ 5. Perform a hypothei tet at the 95% level to awer thi quetio. Commet: Whe we have < 30 the amplig ditributio for y would be a Studet' t-ditributio with 1 degree of freedom. Thi i epecially true if we wat to ue σ. Thi would lead to: 1. H 0 : µ µ 0, where µ 0 i ome give cotat. H a : µ µ 0, the alterative hypothei 3. A tet tatitic t y µ 0, where y,, ad come from ample data ad µ 0 come from H 0. (It called "t" becaue it will be compared to a Studet' t-ditributio, ot to a tadard ormal ditributio, z.). A rejectio regio, t > t α, 1 where P t > t α, 1 or t < t α P t > t, 1 α α., 1 5. A cocluio : "reject H 0 " or "do ot reject H 0 ". 7