SOLUTION: The 95% confidence interval for the population mean µ is x ± t 0.025; 49

Transcription

1 C Sprig 2011 Homework 7 olutio 1. Baed o a ample of 50 x-value havig mea ad tadard deviatio 4.26, fid a 95% cofidece iterval for the populatio mea. SOLUTION: The 95% cofidece iterval for the populatio mea µ i x ± t 0.025; 49 which i ± , or ± Thi iterval i (34.15, 36.57)., The value ued for t 0.025; 49 deped o the degree of freedom umber give i the table at had. Almot certaily there will ot be a lie for 49 degree of freedom. Your table will likely have ome of thee fact: t 0.025; 40 = t 0.025; 50 = t 0.025; 45 = t 0.025; 60 = A proper rule would ue the lie immediately below your umber of degree of freedom, here 49. If your table ha a lie for 45 degree of freedom, you d ue Thi would produce ± If your table ha o lie for 45 degree of freedom, but doe have a lie for 40 degree of freedom, ue Thi would produce ± It alo poible to ae t 0.025; 49 by iterpolatig. It hould be clear that oe of thee method i worthy of great cocer. The value ued here, , wa foud through Miitab. The commad i Calc Probability Ditributio t Ivere cumulative probability. Set the Degree of freedom umber to 49, ad et Iput cotat to (There i a illutratio with the olutio to problem 3.) Page 1 g2012

2 C Sprig 2011 Homework 7 olutio Of coure, the etire tak ca be doe with eae i Miitab. Do Stat Baic Statitic 1 Sample t Summarized data, ad the et up the pael a follow: Thi produce the followig output: Oe-Sample T N Mea StDev SE Mea 95% CI ( , ) 2 zα/2 σ 2. The formula i ued to elect a ample ize before doig a E experimet. The target error i pecified a E, ad thi repreet a propoed upper limit for the abolute error x µ. The probability of achievig thi limit i pecified a 1 α. The value ued for σ i a prior opiio o the tadard deviatio. Sometime the experimeter will requet a upper boud for H, the half-width of a cofidece iterval. For example, if the cofidece iterval tur out to be 78.2 ± 6.4 the H i 6.4. Suppoe that the experimeter pecifie H, decide to ue a 1 α cofidece iterval, ue σ a a kow value, ad will ue the cofidece iterval σ x ± zα/2. I term of H, α, ad σ, what value of would you recommed? SOLUTION: I the iterval x ± zα/2 The coditio i the zα/2 σ σ, it happe that zα/2 wat H, ad thi lead quickly to σ correpod to H. 2 zα/2 σ H. Page 2 g2012

3 C Sprig 2011 Homework 7 olutio Thi look like exactly the ame formula... ad it i. But ote * The experimeter who talk about the error boud E i uually goig to ak for 1 α of 0.70 or The dicuio i cofidece iterval term will almot certaily be akig for 1 α = * Both perpective are treatig the value of σ a kow. 3. Suppoe that a ample of 200 accout receivable etrie at a large mail-order buie had a mea price of $ ad a tadard deviatio of $1, Give a 95% cofidece iterval for the populatio mea. Be ure to tate ay aumptio that you ue. SOLUTION: The iterval i certaily x ± t 0.025; 1, which i here ± , which i about ± Thi i the iterval ($589.52, $1,102.88). The value i the two-ided 5% poit from the t ditributio with 199 degree of freedom; thi value wa obtaied from Miitab. If you are uig a tatitical table, it would certaily be reaoable to ue the value To get thi t poit from Miitab, do Calc Probability Ditributio t ad the et up the pael a follow: Page 3 g2012

4 C Sprig 2011 Homework 7 olutio Thi produce output Ivere Cumulative Ditributio Fuctio Studet' t ditributio with 199 DF P( X <= x ) x The oly aumptio required i that of radom amplig. The tadard deviatio of thi ample of poitive value greatly exceed the mea, o we do ot claim to be amplig from a ormal populatio. With = 200, the Cetral Limit theorem allow u to avoid aumptio. The Cetral Limit theorem oly let u make cocluio about the ormal ditributio of X, ad caot help u with the ditributio of (which deped o ormality). With a large ample ize we ca claim, through the Law of Average, that will be acceptably cloe to the true value of σ. If you ue Miitab through Stat Baic Statitic 1 Sample t Summarized data, you will get Oe-Sample T N Mea StDev SE Mea 95% CI ( , ) 4. Coider the cofidece iterval for µ baed o a ample of ize from a populatio which i aumed ormal. a. If = 10, how much loger (percetagewie) i a 95% cofidece iterval tha a 90% cofidece iterval? b. If = 20, how much loger (percetagewie) i a 95% cofidece iterval tha a 90% cofidece iterval? c. Fidig the limitig awer, a get very large, to the olutio to the quetio above. SOLUTION: The cofidece iterval i X ± tα / 2; 1, ad thu the legth of uch a cofidece iterval i 2 tα / 2; 1. Let divide the legth of a 95% cofidece iterval (uig α = 0.05) by the legth of a 90% cofidece iterval (uig α = 0.10). Thi will produce 2 t 2 t 0.025; ; 1 = t 0.025; 1 t 0.05; 1 Page 4 g2012

5 C Sprig 2011 Homework 7 olutio The ratio deped oly o the value from the t table! For part a, we ue = 10, ad fid the ratio to be % cofidece iterval i about 23% loger It look like the If you ued a prited table with three figure after the decimal poit, thi would be t give a 0.025; 9 = Thi dicrepacy i certaily ot worth t ; 9 worryig about. For part b, we ue = 20, ad fid the ratio to be 95% cofidece iterval i about 21% loger With = 20, the For part c, we are jut doig the ame calculatio for the lie of the t table. Equivaletly, we ca ue a ormal table to pull out the reult. The ratio i the Here the 95% iterval would be appraied a about 19% loger. 5. A ample of 612 voter were aked whether they would vote for cadidate Wolczak i the upcomig electio. Of thee, 344 aid that they iteded to vote for Wolczak. Fid the covetioal 95% cofidece iterval for the populatio parameter p (correpodig to the proportio who will vote for Wolczak). ( 1 pˆ ) pˆ SOLUTION: Thi covetio iterval i p ± z α/ , z α/2 = z = 1.96, ad = 612, the iterval i. Uig p = ± which compute a (about) ± Thi i the iterval ( , ). Thi could reaoably be rouded to (0.5228, ). Page 5 g2012

6 C Sprig 2011 Homework 7 olutio Miitab ca do thi a well. Ue Stat Baic Statitic 1 Proportio Summarized data. The fill the pael a follow: To get the covetioal iterval, you ll eed to do Optio Ue tet ad iterval baed o ormal ditributio. The output i thi: Tet ad CI for Oe Proportio Sample X N Sample p 95% CI ( , ) Uig the ormal approximatio. 6. For the ituatio of the previou problem, give the 95% cotiuity-corrected cofidece iterval for p. ( pˆ ) pˆ 1 1 SOLUTION: The iterval i p ± zα/2 +. We ca get thi from the 2 1 previou by extedig each edpoit by 2 = That i, itead of 1,224 uig ± , we would ue ± ( ), which i ± Thi iterval i ( , ). Thi could reaoably be rouded to (0.5220, ). With = 612, thi i a very delicate refiemet ad mot people would ot otice. Miitab doe ot give thi iterval, but it very eay to cotruct from the detail i problem 5. Jut ubtract from the lower ed, ad add to the upper ed. Page 6 g2012

7 C Sprig 2011 Homework 7 olutio 7. For the ituatio of the previou two problem, give the 95% Agreti-Coull cofidece iterval for p. SOLUTION: Fid firt p = The 95% iterval i the ± Thi i ± ad it ca be writte (0.5225, ). Miitab ca be tricked ito givig the Agreti-Coull cofidece iterval. Jut imagie that you had 346 Wolczak voter out of 616 urveyed. Cotiue to ue Optio Ue tet ad iterval baed o ormal ditributio. The Miitab output will be thi: Tet ad CI for Oe Proportio Sample X N Sample p 95% CI ( , ) Uig the ormal approximatio. 8. Hillary had a very buy week i the laboratory. She wa aked, over ad over agai, to produce 95% cofidece iterval. Over thi week, he gave 120 uch iterval. All 120 problem were idepedet of each other. What i the probability that eight or more of her iterval failed to cover the true populatio parameter? Page 7 g2012

8 C Sprig 2011 Homework 7 olutio SOLUTION: The probability that ay idividual iterval fail to cover the populatio parameter i Thu X, the umber of iterval i error, ca be regarded a a biomial radom variable with = 120 ad p = The P[ X 8 ] = 1 P[ X 7 ]. Miitab ca eaily upply thi awer. Ue Calc Probability Ditributio Biomial Cumulative probability. The Miitab output i Cumulative Ditributio Fuctio Biomial with = 120 ad p = 0.05 x P( X <= x ) The P[ X 8 ] = = Thi i about 25%. 9. Joe would like to give a cofidece iterval for the biomial parameter p. Ufortuately he i oly able to take oe obervatio. Thi obervatio he ll call X, ad the oly poible outcome are X = 0 ad X = 1. Joe ha decided to ue thee tatemet: If X = 0, he ll ay, I am percet cofidet that p i i the iterval [0, 0.80). If X = 1, he ll ay, I am percet cofidet that p i i the iterval (0.20, 1]. What umber hould go i the blak? HINT: You ll eed a careful defiitio of the cofidece coefficiet (the umber i the blak). Thi hould be the miimum value (over all p) of the probability that the iterval i correct. I ymbol, Cofidece coefficiet = mi P[ iterval iclude p p i true value] p SOLUTION: Suppoe that the true value i p 0. For ay p 0 with 0.20 < p 0 < 0.80, Joe i goig to make a correct tatemet with probability 1. Thu we will examie the iterval 0 p ad 0.80 p 0 1. Coider firt 0 p Joe will make a correct tatemet if ad oly if X = 0. If Joe get X = 0, he ll give the iterval a [0, 0.80), ad thi will cover p 0. If Joe get X = 1, he ll give the iterval a (0.20, 1], which will fail to cover p 0. Thu if 0 p , Joe will cover p 0 with probability P[ X = 0 ] = 1 p 0. The mallet value that thi ha over the iterval [0, 0.20] i The problem at the upper ed of [0, 1] i ymmetric, o we ll get the ame reult. Jut to ee the detail. Page 8 g2012

9 C Sprig 2011 Homework 7 olutio Coider 0.80 p 0 1. Joe will make a correct tatemet if ad oly if X = 1. If Joe get X = 1, he ll give the iterval a (0.20, 1], ad thi will cover p 0. If Joe get X = 0, he ll give the iterval a [0, 0.80), which will fail to cover p 0. Thu if 0.80 p 0 1, Joe will cover p 0 with probability P[ X = 1 ] = p 0. The mallet value that thi ha over the iterval [0.80, 1] i The cofidece coefficiet for thi problem i Joe ca ay that he givig a 80% cofidece iterval. By the way, the coverage fuctio i thi: P[ iterval cover true p 0 ] = 1 p0 if 0 p if 0.20 < p0 < 0.80 p0 if 0.80 p0 1 It hould be oted that thi i ot eve a cotiuou fuctio of p 0! 10. Suppoe that a ample of te value from a aumed ormal populatio ha mea x = 210 ad tadard deviatio = 32. The tadard 95% cofidece iterval for the populatio mea µ i x ± t 0.025; 9, which here i (187.1, 232.9). Thi iterval i baed o the t ditributio for X µ. It i alo true that X µ P t0.01;9 < < t0.04; 9 = 0.95 (*) a. Fid the value t 0.01; 9 ad t 0.04; 9. The firt you ca get from a prited table. The ecod ca be obtaied from Miitab Calc Probability Ditributio t Ivere cumulative probability. b. Give i umeric form the 95% cofidece iterval related to the probability tatemet (*) above. c. Give two reao why the tadard iterval i preferred over (*). Page 9 g2012

10 C Sprig 2011 Homework 7 olutio SOLUTION: For a, you ca fid t 0.01; 9 = from a prited table. Miitab will give t 0.04; 9 = You could alo ue Miitab to get t 0.01; 9 = The iterval requeted i b i X t0.04; 9, X + t0.01; 9. I umber, thi i , or ( , ). Thi ca be rouded to ( , ) or eve (190.0, 238.6). So what wrog with the iterval i b? For oe thig, the ample average x = 210 i ot at the ceter, ad may people fid thi aoyig. I additio, the iterval i b i lightly loger; it legth i = The legth of the covetioal iterval i = Shorter i better for cofidece iterval. Page 10 g2012