Chapter 9. Key Ideas Hypothesis Test (Two Populations)

Transcription

1 Chapter 9 Key Idea Hypothei Tet (Two Populatio) Sectio 9-: Overview I Chapter 8, dicuio cetered aroud hypothei tet for the proportio, mea, ad tadard deviatio/variace of a igle populatio. However, ofte reearcher wat to compare two differet populatio. For example, urgeo may wat to try out a ew urgical techique for a certai ailmet, but they are ot ure if it will be better. To tet thi, they could take two ample of people. I oe ample, they could ue the exitig techique, ad i the other, they could ue the ew techique. By comparig urvival proportio of the two group, they could the determie whether the ew ample i ay better. I thi cae, the firt populatio i all people who would receive the exitig techique. The ecod populatio i all people who would receive the ew techique. By uig the method dicued i thi chapter, uch iferece ca be doe. Sectio 9-: Iferece About Two Proportio To ue the method decribed i thi ectio, we firt eed to rely o a few coditio that mut be met for everythig to work properly. Coditio. The proportio are take from two imple radom ample which are idepedet. Here, idepedet mea that obervatio from the firt populatio are ot related to, or paired with, obervatio from the ecod populatio.. For each of the two ample, there are at leat 5 uccee ad 5 failure. Notatio p = Populatio Proportio (Populatio ) p = Populatio Proportio (Populatio ) = Sample Size (Populatio ) = Sample Size (Populatio ) x = Number of Succee (Populatio ) x = Number of Succee (Populatio ) x x p = = Sample Proportio (Populatio ) p = = Sample Proportio (Populatio ) q p q = p = x x p = = Pooled Sample Proportio q = p The Tet The goal i to tet the hypothee give by: H 0 : p = p H : p p (or p > p or p < p ) I other word, we tet whether the proportio for each populatio are equal v. whether they are differet i ome way. Tet Statitic: ( p Z = p ) ( p p ) ( p p ), or Z = (For the tet tatitic, we aume H 0 i true, which mea p p = 0) The critical value ad P-Value come from the tadard ormal ditributio. Deciio are made i exactly the ame way a i Chapter 8.

2 Cloe to a electio, ballot iue #3 ad #4 are very cotroverial. Reearcher wat to ee whether there i a differece i the proportio of people who upport iue #3 ad thoe who upport iue #4. They radomly ample 00 total people. They ak 00 of thee people whether they upport iue #3, to which 56 ay ye. They ak the other 00 people whether they upport iue #4, to which 45 ay ye. I there a differece i the proportio of upporter for each iue? Tet thi with α = Solutio From the iformatio give, we ee that: = 00, x = 56, p = 0. 56, = 00, x = 45, p = x x p = = = = 0.505, q = H 0 : p = p H : p p ( p p ) Tet Statitic: Z = = = = (0.495) 0.505(0.495) P-Value Method Sice H ha a ig, we wat to fid area above Z =.56 ad below - Z = -.56 for the tadard ormal ditributio. From the Z-Table, thi area i (0.0594) = Now we compare thi area to α ad ee that 0.88 > Therefore, we do ot reject H 0. I other word, we coclude that there i ot eough evidece to claim that there i a differece i proportio of upporter for the two iue. Gloria, a hairdreer, claim that he i better tha a fellow hairdreer amed Jule. They decide to ru a hypothei tet to ee if thi i true. Out of 9 of Gloria cutomer, 87 are atified with their haircut. Out of 67 of Jule cutomer, 56 are atified. Ru the tet to ee if Gloria cutomer have a higher percetage of atifactio tha Jule cutomer (α = 0.05). Solutio From the iformatio give, we ee that: = 9, x = 87, p = , = 67, x = 56, p = x x p = = = = 0.899, q = H 0 : p = p H : p > p ( p p ) Tet Statitic: Z = = = = (0.0) 0.899(0.0)

3 P-Value Method Sice H ha a > ig, we wat to fid area above Z =.7 for the tadard ormal ditributio. From the Z-Table, thi area i = Now we compare thi area to α ad ee that 0.06 < Therefore, we reject H 0 ad coclude that Gloria ha a higher atifactio percetage tha Jule. Cofidece Iterval for p p Sometime, oe would like to etimate the differece betwee the proportio, rather tha jut eeig whether the proportio differ igificatly. To cotruct a cofidece iterval for p p, we have the followig: Poit Etimate: p p Critical Value: Z Stadard Error: α p q p q Thi give the followig cofidece iterval: p q p q p p ± Zα Coider the Gloria/Jule hairdreer example from the previou page. We had the followig quatitie: = 9, x = 87, p = , = 67, x = 56, p = Therefore, a 95% cofidece iterval would be: p q p q 0.946(0.054) p p ± Zα = ±.96 9 = 0.±.96(0.05) = 0.± ( 0.0, 0.) 0.836(0.64) 67 Notice that 0 i ot icluded i thi cofidece iterval. Due to thi fact, we could ay that there i a differece betwee the two proportio (i.e. we would reject H 0 i favor of H : p p ).

4 Sectio 9-3: Iferece About Two Mea: Idepedet Sample I imilar fahio to tetig for differece i proportio, oe may alo wih to tet for a differece i the mea of two populatio. I the iteret of time, we will coider the mot geeral cae, where the populatio tadard deviatio are ukow, ad o aumptio are made about them. Better hypothei tet exit whe thee value are both kow, or are ukow but aumed to be equal. Coult the textbook for more iformatio o thee tet. Agai, certai requiremet mut be met for the techique dicued i thi ectio to be theoretically oud. Coditio. σ ad σ are ukow ad o aumptio i made about the equality of σ ad σ.. The two ample are idepedet. 3. Both ample are imple radom ample. 4. Either both populatio are ormally ditributed or ad are both greater tha 30. Notatio µ = Populatio Mea (Populatio ) µ = Populatio Mea (Populatio ) = Sample Size (Populatio ) = Sample Size (Populatio ) x = Sample Mea (Populatio ) x = Sample Mea (Populatio ) = Sample Stadard Deviatio (Populatio ) = Sample Stadard Deviatio (Populatio ) Degree of Freedom = df = mi(, ) (o df i till, but here i the maller of the two ample ize) The Tet The goal i to tet the hypothee give by: H 0 : µ = µ H : µ µ (or µ > µ or µ < µ ) I other word, we tet whether the mea for each populatio are equal v. whether they are differet i ome way. Tet Statitic: ( x t = x ) ( ) ( x x ) µ µ, or t = (For the tet tatitic, we aume H 0 i true, which mea µ µ = 0) The critical value ad P-Value come from the Studet t ditributio with Degree of Freedom = mi(, ). Deciio are made i exactly the ame way a i Chapter 8. Reearcher wat to ee if the average moey pet per week by tourit i Chicago i le tha the average moey pet per week by tourit i New York City. They take a ample of 60 tourit from Chicago ad 56 tourit from NYC. Of the Chicago tourit, average pedig wa $635, with a tadard deviatio of $50. Of the NYC tourit, the average pedig wa $650, with a tadard deviatio of $30. Ru a hypothei tet with α = Solutio From the iformatio give, we ee that: = 60, x = 635, = 50, = 56, x = 650, = 30, df = mi(, ) = 56 = 55 H 0 : µ = µ H : µ < µ ( x x ) Tet Statitic: t = = = =

5 P-Value Method Sice H ha a < ig, we wat to fid area below t = for the t ditributio. Now, otice that the t-table doe ot allow oe to directly fid thi area. However, for df = 55 we ee that ha a area below of 0.05 ad ha a area below of Sice < t < -.673, the p-value will fall betwee 0.05 ad (ee picture) Now we compare thi area to α ad ee that 0.05 < p < 0.05 = α. Sice p < 0.05, we reject H 0. The evidece ugget that Chicago tourit pay le per week tha thoe i New York City. Oe quetio o everyoe mid i whether there i a differece i the average umber of pet owed by Columbu familie ad the average umber of pet owed by Clevelad familie. Reearcher et out to awer thi importat quetio. They ampled 35 Columbu familie ad 48 Clevelad familie. Of the Columbu familie, the average umber of pet wa.4, with a tadard deviatio of.4. Of the Clevelad familie, the average umber of pet wa.9, with a tadard deviatio of 0.9. Ru a hypothei tet (α = 0.05) to ee if there i a differece i the average umber of pet owed by familie i the two citie. Solutio From the iformatio give, we ee that: = 35, x =. 4, =.4, = 48, x =. 9, = 0.9, df = mi(, ) = 35 = 34 H 0 : µ = µ H : µ µ ( x x ) Tet Statitic: t = = = = P-Value Method Sice H ha a ig, the p-value will be the area above t =.85 ad below - t = -.85 for the t ditributio. Agai, otice that the t-table doe ot allow oe to directly fid thi area. However, for df = 34 we ee that.69 ha a two-tailed area of 0.0 ad.03 ha a two-tailed area of Sice.69 < t <.03, the p-value will fall betwee 0.05 ad 0.0. (ee picture) Now we compare thi area to α ad ee that p > 0.05 = α. Sice p > 0.05, we do ot reject H 0. There i ot ufficiet evidece to coclude that the average umber of pet i differet i the two citie.

6 Cofidece Iterval for p p A i the previou ectio, oe might like to etimate the differece betwee the mea, rather tha jut tetig for the differece. To cotruct a cofidece iterval for µ µ, we have the followig: Poit Etimate: x x Critical Value: t (df = mi(, ) ) Stadard Error: α Thi give the followig cofidece iterval: x x ± tα Coider the avg. umber of pet example from the previou page. We had the followig quatitie: = 35, x =. 4, =.4, = 48, x =. 9, = 0.9, df = mi(, ) = 35 = 34 Therefore, a 95% cofidece iterval would be:.4 x ± t =.4.9 ± x α = 0.5 ±.03(0.700) = 0.5 ± ( 0.049,.049) Notice that 0 i icluded i thi cofidece iterval. Due to thi fact, we could ay that there i ot a igificat differece betwee the two mea (i.e. we do ot reject H 0 i favor of H : µ µ, a i the example o the previou page).