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1 114 Block 3 Erdeky == Begi 6.3 ============================================================== 1 / 8 / Correspodig Areas uder a ormal curve ad the stadard ormal curve are equal. Below: Area B = Area A. NORMAL 2 MU =50 σ=10 1 3:P(L<X<R) L=50 R=65 A ormal distributio has µ = 50 ad σ = 10. Fid P(50 < < 65). The traditioal method is to covert the values 50 ad 65 to their correspodig z scores 0 ad 1.5 usig the z-score formula. Net, use the areas of the stadard ormal table z(50) = 0 to fid A(0) =.5000, A(1.5) = =.4332 P(0 < z < 1.5) P(50 < < 65) =.4332 Page 332, 26b A ormal distributio A L R L R Give a ND with µ = 8.4 ad σ = 1.7. Let = the police respose time (i miutes). (b) Fid the probability that the police respod withi 5 miutes. That is fid P( < 5). The stadard ormal distributio µ = 50 µ = 0 σ = 10 σ = 1 1 = 50 z 1 = 0 2 = 65 z 2 = 1.5 P(50 < < 65) =.4332 P(0 < z < 1.5) =.4332 B = A B Z NORMAL 1 1 3:P(L<Z<R) L=0 R=1.5 µ z = σ z = z = 1.5 z(65) 1.5 INTERVAL É À NORMAL P( < 5) Draw the graph, showig itervals. P( < 5) z About 2% of the time, the respose time is less tha 5 miutes. 1 By table: z(5) = -2. A(-2) =.0228 P(z < -2).0228 P( < 5).0228

2 Page 333, 31d Give µ = 8. Whe σ is ot give, use the geeral rule: Let be the replacemet age of a TV set. Rage σ = = (d) Fid 0, the guarateed age of a TV, so that o more tha 10% of the sets are i the guaratee period. That is, fid 0 so P( < 0 ).10. INTERVAL NORMAL Draw the graph o paper z o =? P( < 6.08).10 Use the NORMAL program to fid the boudary of the kow Left Area..10 P( < 6.08).10 Coclusio: If the compay guaratees their sets for 6 years, they will ot have to replace more tha 10% of them. The Traditioal Method: Optioal Fid 0 so P( < 0 ).10. First fid z 0 so P(z < z 0 ).10 usig steps Draw left tail area C so that it cotais 10% of the distributio. Mark the boudary z 0. 2 Look for.1000 i the body of the SNC table is betwee.1003 ad Choose.1003 sice it is closer to The z-score that correspods to.1003 is z 0 = Net fid Use = zσ + µ to fid 0. 0 = So P( < 6) = 6. Page 332, 28 Give a ND with µ = 22 ad σ = 10. Let = the daytime high temperature. Fid P(29 < < 40). NORMAL 2 MU=22 σ= L=29 R=40 By table: z(29) =.7. z(40) = 1.8 A(.7) =.7580, A(1.8) = =.2061 Result: PROB =.2060 P(29 < < 40) z Coclusio: The temperature is betwee 29 ad 40 o about 21% of the days i Jauary days. There are approimately 6 days i Jauary whe the daytime high temperature is betwee 29 ad

3 == Begi 7.1 ===================================================================== A Parameter is a umber that is a descriptive measure of a populatio. µ ad σ are parameters. Page A Statistic is a umber that is a descriptive measure of a sample. ü ad s are statistics. Page A Samplig Distributio of ü is a probability distributio formed by all the meas ü of all the samples, all of size, from a populatio. This is called the ü distributio. See sheet 4. Page Cosider all K samples of size from a populatio : S 1, S 2, S 3,, S K. The probability distributio of all the meas: ü 1, ü 2, ü 3,, ü K is the ü distributio. ==7.2 =========================================================================== 4 The Mea of the ü Distributio is deoted µ ü. Page The Stadard Deviatio of the ü Distributio is deoted σ ü. Page 366 σ ü is also called The Stadard Error of the Mea ad deoted SE Mea. 6 Three Properties of the ü Distributio are See sheet 11. Page 366 σ a) If the populatio is ormal, the the ü distributio is ormal. b) µ ü = µ. c) σ ü = Cosider a populatio with µ = 100 ad σ = 20. Let ü deote the ü distributio for samples of size = a) If is ormal, the the ü distributio is also ormal. b) µ ü = 100. c) σ ü = = = The Cetral Limit Theorem states: a) The ü distributio from ay populatio is usually ormal whe > 29, σ b) µ ü = µ., ad c) σ ü = Page 366, 2b A ormal populatio has µ = 10.2 ad σ = 1.4. Five fish are caught. = 5. 3 See sheet 11. Page 368 Let ü be the mea (average) legth of the 5 fish. Fid P( 8 < ü < 12). The ü Distributio with = 5 The ü distributio has µ ü = 10.2 ad σ ü = = SE Mea. INTERVAL M=10.2 S=1.4 = 5 Draw the graph NORMAL 3 MU=10.2 σ=1.4 =5 1 3 L= 8 R=12. Result: PROB.9978 P( 8 < ü < 12) 99.8%. The probability is 99.8% that the average legth of the 5 fish will be betwee 8 ad 12 iches. Note: z( 8) z(12) A(-3.51) = 0 z = µ σ = = SE Mea = ü z 8 12 ZTXVALUE M=10.2 S=1.4 =5 Select 4. SE Mea =.626 = σ ü.

4 ======================= A Eample of a Samplig Distributio of ü ======================= Cosider the populatio : 1,1,1,1,1,1,2,2,2,3. N = 10, µ = 1.5, σ P() The Probability Distributio of The Distributio From the populatio, all samples of size = 3 are selected with replacemet. With replacemet meas that after each umber is selected, it is retured to populatio before the et umber is selected. There are1000 samples. 27 are differet samples. There are 7 differet values of ü The differet samples The ü distributio See sheet 10 Sample S ü P(S) 1,1, ,1,2 4/ ,1,3 5/ ,2,1 4/ ,2,2 5/ ,2, ,3,1 5/ ,3, ,3,3 7/ ,1,1 4/ ,1,2 5/ ,1, ,2,1 5/ ,2, ,2,3 7/ ,3, ,3,2 7/ ,3,3 8/ ,1,1 5/ ,1, ,1,3 7/ ,2, ,2,2 7/ ,2,3 8/ ,3,1 7/ ,3,2 8/ ,3, sum 1 ü P(ü) f P(1,1,1) = = P(1) =.216 P(1,1,2) = =.108 4/ P(4/3) = =.324 P(1,1,3) = =.036 5/ P(5/3) = =.270 Note that P(5/3) 6/27 because values of 5/3 are P(1,2,2) = =.054 7/ ot equally likely. 8/ sum 1 27 USE PROBDIST µ ü = 1.5, σ ü.387 Checkig the theory: σ ü = σ.671 = 3 1 4/3 5/3 2 7/3 8/3 3 The ü distributio.387 also, µ ü = µ ZTXVALUE M=1.5 S=.671 =3 Select 4. SE Mea.387 = σ ü. 4

5 Page 374, 8 Let be (the white blood cell cout) per (mm 3 of whole blood). is a ormal distributio with µ = 7500 ad σ = (a) Fid P( < 3500) for = 1, a sigle blood sample. Use INTERVAL to help make the graph. NORMAL 2 MU=7500 σ= R = 3500 Result: PROB = P( < 3500) 1.1% Note: z(3500) = σ = 1750 µ z = = = 2.29 σ (b) Fid P( ü < 3500) for = 2, for two blood samples. NORMAL 3 MU= 7500 σ= 1750 = R = 3500 Result: PROB = P(ü < 3500).0006 Note: z(3500) = Draw this o paper. µ µ ( ) z = = = = 3.23 σ σ (c) Fid P(ü < 3500) for = 3, for three blood samples. Use ZTXVALUE M=7500 S=1750 =2 1 NORMAL 3 MU=7500 σ =1750 =3 1 1 R = z Use INTERVAL.0006 SE Mea = ü z Use INTERVAL M=7500 S=1750 = SE Mea = M=7500 S=1750 =3 P(ü < 3500) =.004% Note: z(3500) = Use ZTXVALUE M=7500 S=1750 = ü z == 8.1 & 8.2 ====================================================================== These sectios develop the cocept of a Cofidece Iterval for µ. A cofidece iterval is a iterval that is estimated to cotai µ with a give level of cofidece or reliability. For eample, you ca say that µ is estimated to be i the iterval (2, 4) with 99% cofidece. 2 < µ < 4. 1 The Cofidece Level c is the reliability of the cofidece iterval. Page 401 c is writte as a percet. c = 95% or c = 99% are typical cofidece levels. c correspods to a cetral area uder a t curve foud iside the back cover of our book. A umber cv called a critical value determies the cetral area c. 5

6 2 The Critical Value cv,deoted z c or t c, is the right boudary of the cetral area c. Page 401 cv is foud i the t table i the book (iside back cover). Use c i the top row ad d.f. = 1 o the side. z c critical values are i the bottom row of the t table. t c critical values are i the body of the table. We will fid cv usig CV. CV.950 = c z z..95 = 1.96 CV 3 A c Cofidece Iterval for µ is a iterval cci that is estimated to cotai µ with a c level of cofidece. Eample: A 95%CI for µ is estimated to cotai µ with a 95% level of cofidece. If we say that we are 90% cofidet that the 90% CI cotais µ, we mea that if we have all of the 90% cofidece itervals of all the samples of size from the populatio, the µ would be i 90% of these itervals. t = 10 t.95 = out of 10 of the 90% C itervals will cotai µ. Oe out of 10 of the 90% C itervals will ot cotai µ. µ To fid a cofidece iterval for µ we eed: 1. a sample of size, 2 ü called A Poit Estimate of µ 3. c, 4. σ or s, 5. cv, 6. the formula (or Iterval fuctio i the TI) for the cci. σ 4 The cci for µ whe σ is give is cci = ± z c. This is the ZIterval. With σ,use z c. s 5 The cci for µ whe σ is ot give is cci = ± t c. This is the TIterval. With s, use t c. Note: If the sample is small, the the populatio must be ormal to use formula 4 or 5. Here ormal meas: moud shaped, uimodal, ad ot etremely skewed. A Small Sample has 29. A sample is small if = 2, 3,, 29. A Large Sample has 30. A sample is large if = 30, 31, 32,. 6

7 Page 405, 2 = 90, ü = 15.60, σ = Fid the 95% cofidece iterval for µ, her true average. Sice σ is give, a ZIterval is used. 7 put cursor o Calculate Í (Lower boud, Upper boud) 95% CI = (15.23, 15.97) We are 95% cofidet that her average time µ is betwee ad miutes. A 95% CI for µ ü Optioal Method By formula: cci = σ ± zc. CV 1:c c =.95 2:Z = z c %CI = 15.6 ± = (15.228, 15.97). The cv of the 95% ZIterval. 90 Keys for the lower boud: 15.6 ¹ y a90 ) Í Page 405, 2 cotiued E = =.372 = maimum error. SD 6 The Maimum Error of Estimate E of the cci is E = cv µ ü is called The Margi of Error. 1.8 I the problem above, E = The 95% CI would be ( , ) = (15.23, 15.97) ±.37 E E The ceter of the iterval is always ü ü Aother way to fid E, is to fid the cci, usig the method at the top of the page, the subtract ü from the upper boud. E = =.37 mi sec = 22.2sec. µ 7 Studet s t Distributio is give by t = usig ü of all samples, of size, from the populatio. s E c.95 E z ü Z The Studet s t distributio is used whe σ is ukow. The ormal distributio z is used if σ is kow. If is small, < 30, the the populatio must be ormal to use the t distributio. Page 413 7

8 A t distributio has µ = 0 ad σ = 1. Every sample size forms its ow t distributio. 3 See page 413 for eamples of t graphs for = 4 ad = 6. Whe =, (the t curve) = (the z curve). 8 The Number of Degrees of Freedom d.f. for a t iterval is the umber d.f. = 1. If = 10, the d.f. = 10 1 = 9. Also = d.f The equatio ( + y + z = 20) has 2 degrees of freedom. We are free to select values for two of the variables, say ad y. The, z must be computed. For eample if we select = 8 ad y = 7, the z must be 5. The equatio (a + b + c + d + e = 10) has 4 degrees of freedom. 416, put cursor o Calculate Í cv = E = 1.67 A 99% CI = (44.5, 47.8). 417, 3 8 put cursor o Calculate Í cv = E =.11 A 95% CI = (6.64, 6.86) == Begi 8.3 ===================================================================== r 9 The Poit Estimate ˆp for p i a biomial eperimet with trials ad r () successes is ˆp = = Page 426 The poit estimate for q is qˆ = 1 pˆ. p is the probability of success or the proportio of successes A sample = 800 studets was take from a populatio of studets ad give flu shots. Of the 800, 600 do ot get the flu. We estimate the probability p that ay studet i the that got a shot will ot get the flu 600 to be ˆp = = 75%. ˆp = P(will ot get the flu got a flu shot). The margi of error is p ˆp pq ˆ ˆ The Maimum Margi of Error E for a c level of cofidece is E = zc Page pˆ qˆ A c Cofidece Iterval for p is ˆp ± zc if ˆp > 5 ad ˆq > , 5d A sample of size = 800 studets is take from a populatio of studets ad give flu shots. Of the 800, 600 do ot get the flu. Fid a 99% CI for p. A 71% 75% 79% We are 99% cofidet that (71%, 79%) is oe of the itervals that cotais the actual value of p. That is, if all studets got a shot, betwee 71% ad 79% would ot get the flu. We say this with 99% cofidece. 75% ± 4% will ot get the flu

9 , 4 = 188, r = = 66. Fid a 90% CI for p. A A 90% CI = (29%, 41%) The margi of error is is 5.7% 6% 29% 35% 41% 35% of the books sold at a local bookstore are murder mysteries. This estimate has a margi of error of 6%. (ofte stated as ± 6%) == Begi 8.4 ===================================================================== zc s 10 The Sample Size for a cci for µ with maimum error of estimate E is = rouded up. E s zc s The formula for above is derived from E = zc E = zc s = square each side E Page 442, 2 Fid the sample size eeded, to form a 90% CI for µ, for a maimum error of estimate E =.5. SAMPSIZE After diggig up at least 30 roots, s is foud to be 8.94 i. IF,for eample, ü of the 865 plats is 10 iches, the our 90%CI with E =.5, would be (9.5, 10.5) iches. The formula method : For a 90% CI the cv = z c = foud o the bottom row of the t-table uder c = Coclusio: We eed the root legths of 865 plats i order to be 90% cofidet that the mea of the 865 measuremets will be withi.5 iches of µ, the mea of the populatio of root legths i YNP. 2 zc s = E = rouds up to 866. Keys: Í.5 Note: SAMPSIZE uses the eact value of z c, gettig ad roudig up to = 865. == Begi 8.5 ===================================================================== 11 A way to tell if two populatios 1 ad 2 are differet is to eamie the differece i µ 1 ad µ 2 i.e. (µ 1 µ 2 ) or the differece i p 1 ad p 2 i.e. (p 1 p 2 ). If both sides of a cci for (µ 1 µ 2 ) are egative, the we are c cofidet (µ 1 µ 2 ) < 0 (µ 1 < µ 2 ). If both sides of a cci for (µ 1 µ 2 ) are positive, the we are c cofidet (µ 1 µ 2 ) > 0 (µ 1 > µ 2 ). If a cci for (µ 1 µ 2 ) has opposite sigs, the we caot say which mea is larger. The same applies to all of the above if we replace µ with p. A c CI for (µ 1 µ 2 ) whe σ 1 ad σ 2 are kow is σ1 σ2 (1 2 ) ± zc +. Page If 1 ad 2 are both < 30, the the populatios must be ormal to use this formula. A 2-Sample Z Iterval. If 1 ad 2 are both > 29, the the populatios do ot eed to be ormal

10 , 9 Populatios 1 ad 2 have σ 1 = 1.9 ad σ 2 = 2.3. Fid a 95% CI for (µ 1 µ 2 ). 1 before. 2 after. Samples are draw with 1 = 167, ü 1 = 5.2 ad 2 = 125 ad ü 2 = Both sides of the CI are egative we are 95% cofidet that (µ 1 < µ 2 ) (µ 2 > µ 1 ) A 95% CI for (µ 1 µ 2 ) = (-2.1, -1.1). The mea umber of fish caught/day µ 2 after the fire is greater tha the mea umber of fish caught/day µ 1 before the fire A c CI for (µ 1 µ 2 ) whe σ 1 ad σ 2 are ukow is foud usig (2-Samp TIt ). Page , 10 Populatios 1 ad 2 are ormal ad have ukow σ 1 ad σ 2. Fid a 90% CI for (µ 1 µ 2 ). Samples are draw with 1 = 15, ü 1 = 19.65, s 1 = 1.86 ad 2 = 14, ü 2 = 6.59, s 2 = 1.91 ü 1 ü 2 = wie 0. A 90% CI for (µ 1 µ 2 ) = (11.9, 14.3). µ 1 is the mea brai activity of people that drak ½ liter of wie before sleepig. µ 2 is the mea brai activity of people that drak o wie before sleepig. Both sides of the CI are positive we are 90% cofidet that (µ 1 > µ 2 ). (The mea brai activity of the drikig group) is greater tha (the mea brai activity o-drikig group) pˆ 1qˆ 1 pˆ 2qˆ 2 A c CI for (p 1 p 2 ) for two idepedet biomial eperimets is (pˆ ˆ 1 p 2 ) ± z c +. Page Where all of ˆ 1p 1, ˆ 1q 1, ˆ 2p 2, ˆ 2q 2, > 5 A 2-Propositio Z-Iterval , 11 Group 1 watched a comedy before sleep with 1 = 175, ad r 1 = 1 = 49 had bad dreams. B Group 2 watched othig before sleep with 2 = 180, ad r 2 = 2 = 63 had bad dreams. Fid a 95% CI for p 1 p 2. A 95% CI = (-17%, 3%). Sice the CI cotais opposite sigs, we caot say that a comedy before sleep reduced bad dreams. == Ed Block 3 ==================================================================== 10

11 Samplig Distributios of ü for 4 Differet Populatios ad 3 Sample Sizes Note that as becomes larger, the ü distributios ted to get closer to ormal. Origial Populatio Samplig Distributio Samplig Distributio Samplig Distributio Uiform of ü for = 2 of ü for = 5 of ü for = 30 ormal Origial Populatio Samplig Distributio Samplig Distributio Samplig Distributio V Shaped of ü for = 2 of ü for = 5 of ü for = 30 ormal Origial Populatio Samplig Distributio Samplig Distributio Samplig Distributio J Shaped of ü for = 2 of ü for = 5 of ü for = 30 ormal Origial Populatio Samplig Distributio Samplig Distributio Samplig Distributio of ü for = 2 of ü for = 5 of ü for = 30 Normal Note that all distributios here are ormal. ormal 11

12 114 Block 3 Laguage Test Statemets Erdeky Here are the actual 18 test items as they will appear o Test L3. Te of these will be selected for the test but you are required to be able to do all of them. The test item is i this fot. The aswer is writte i this fot or i a bo Complete: Correspodig Areas uder a ormal curve ad the stadard ormal curve are equal. 2 Defie: A Statistic A Statistic is a umber that is a descriptive measure of a sample. 3 Complete: A samplig distributio of ü is a probability distributio formed by all the meas ü of all the samples, all of size, from a populatio. 4 Complete: The Mea of The ü Distributio is deoted µ ü. 5 Complete: The Stadard Deviatio of The ü Distributio is deoted σ ü. 6 Complete: Three Properties of The ü Distributio are a) If the populatio is ormal, the the ü distributio is ormal. b) µ ü = µ. c) σ ü = σ. 7 Complete: The Cetral Limit Theorem states: a) The ü distributio from ay populatio is usually ormal whe > 29, b) µ ü = µ. c) σ ü = σ. 1 Defie: The Cofidece Level c The Cofidece Level c is the reliability of the cofidece iterval. 2 Defie: The Critical Value cv The Critical Value cv, deoted z c or t c, is the right boudary of the cetral area c. 3 Defie: A c Cofidece Iterval for µ A c Cofidece Iterval for µ is a iterval cci that is estimated to cotai µ with a c level of cofidece. σ 4 Complete: The cci for µ whe σ is give is cci = ± z c s 5 Complete: The cci for µ whe σ is ot give is cci = ± t c 12

13 SD 6 Complete: The Maimum Error E of Estimate of the cci is E = cv 7 Complete: 1) Studet s t Distributio is used whe σ is ukow. 2) If < 30, the the populatio must be ormal to use the t distributio. 8 Complete: The Number of Degrees of Freedom for a t distributio is d.f. = 1. 9 Complete: The Poit Estimate ˆp for p i a biomial eperimet with trials ad r successes is r ˆp = = 10 Complete: The sample size for a cci with maimum error E is = zc s E 2 rouded up 11 To tell if two populatios 1 ad 2 are differet, eamie the differece i µ 1 ad µ 2 i.e. (µ 1 µ 2 ) or the differece i p 1 ad p 2 i.e. ( p 1 p 2 ) If both sides of a cci for (µ 1 µ 2 ) are egative, the we are c cofidet that µ 1 < µ 2. If both sides of a cci for (µ 1 µ 2 ) are positive, the we are c cofidet that µ 1 > µ 2. If a cci for (µ 1 µ 2 ) has opposite sigs, the we caot say which mea is larger. 13

14 114 Block 3 Laguage Test Statemets Erdeky You are required to be able to do everythig below. 1 Complete: Correspodig Areas uder curve ad curve are equal. 2 Defie: A Statistic 3 Complete: A samplig distributio of ü is 4 Complete: The Mea of The ü Distributio is deoted 5 Complete: The Stadard Deviatio of The ü Distributio is deoted 6 Complete: Three Properties of The ü Distributio are a) b) c) 7 Complete: The Cetral Limit Theorem states: 1 Defie: The Cofidece Level c 2 Defie: The Critical Value cv 3 Defie: A c Cofidece Iterval for µ 4 Complete: The cci for µ whe σ is give is cci = 5 Complete: The cci for µ whe σ is ot give is cci = 6 Complete: The Maimum Error E of Estimate of the cci is E = 7 Complete: 1) Studet s t Distributio is used whe 2) If < 30, the 8 Complete: The Number of Degrees of Freedom for a t distributio is d.f. = 9 Complete: The Poit Estimate ˆp for p i a biomial eperimet with trials ad r successes is 10 Complete: The sample size for a cci with maimum error E is = 11 To tell if two populatios 1 ad 2 are differet, eamie the differece i µ 1 ad µ 2 i.e. (µ 1 µ 2 ) or the differece i p 1 ad p 2 i.e. ( p 1 p 2 ) If both sides of a cci for (µ 1 µ 2 ) are egative, the we are c cofidet that If both sides of a cci for (µ 1 µ 2 ) are positive, the we are c cofidet that If a cci for (µ 1 µ 2 ) has opposite sigs, the 14

15 114 B3 Problems , 1 NORMAL Note: I a cotiuous distributio P(3 < < 6) = P(3 6). P(3 < < 6) 53% 330, 9 NORMAL P( > 90) 75% 330, 11 Fid z o so P(z < z o ) =.06. NORMAL z o 331, 19 NORMAL z z o P(z < ) , 29b NORMAL 2 z.10 o 45 If the batteries are guarateed for 35 moths, the compay will oly have to replace about 10.6% of them , 31 NORMAL 2 b c , 33 a b c d e

16 , 35a NORMAL % b NORMAL o 90 Guaratee a chip for 81 moths ad there is a 99% probability that it will last beyod the guaratee period. Guaratee the chip for 84 moths ad there is a 5% probability that it will malfuctio i the guaratee period. 334, 35 (c) $50,000, = $2,622,110 is the epected loss to the compay. 334, 35 (d) Profit = 3,000,000 2,622,110 = $377,890. (The books aswer is the result of roudig.) P(( > 20 ad ( > 15)) P( > 20) 335,39a P( > 20 > 15) = = see below % P( > 15) P( > 15) = , 1a NORMAL b P(15 < ü < 17) 34% ü c I the (a) distributio, 15 ad 17 have z-scores of 0 ad 1. I the (b) distributio, 15 ad 17 have z-scores of 0 ad There is more area betwee 0 ad σ ü = 1.75 i b ad is smaller tha σ ü = 2 i a. b has a arrower distributio, so the area betwee 15 ad 17 is bigger , 11 is the mothly percet retur of the mutual fud. The fud has stocks from 250 compaies. This is a large sample of all of the compaies i the world. Because is a large sample of averages, it is a ü distributio that is ormal. For, µ = 1.6 ad σ =.9. (Both are percets.) (b) = 6. This sample is small, = 6, but it is from a ormal distributio so its distributio ü is ormal. NORMAL ü ü (c) is doe like (b) ecept = 24. P(1 < ü < 2) = 81%

17 , 13 b NORMAL P(ü < 6%) =.014 With a probability this small, it is ulikely that the mea was really 10.8%. The market is weaker tha 10.8%. NORMAL P(ü > 16%) =.009 It is very ulikely that the mea was really 10.8%. The market is heatig up ad the yield is more tha 10.8% , 19 (a) If the total weight w of 45 m 3 is 9500g, the the average weight ü of 1 m 3 is NORMAL ü (c) NORMAL 9500 P(w < 9500) = P < ü P(9500 < w < 12000) = P < < 97% , 21 (a) If the total distace d i 5 years is 90 ft, the the average distace d i 1 year is 90 5 (b) If the total distace d i 5 years is 80 ft, the the average distace d i 1 year is 80 5 = 18 ft. = 16 ft. a P( d > 90) = P( d > 18) 25% b P( d < 80) = P( d < 16) 25% the book rouds z to.68 d c P( 80 < d < 90) = P(16 < d < 18) 50% d

18 , 1 7 σ is kow, so a ZIterval is correct. 408, , , , 9 a Eter data ito L1 b 7 e 411,11c 7 99%CI = (49.92, 52.40) , 1 CV 420, 3 CV

19 , 5 Eter the 9 data values ito L1. 8 σ is ukow, so a t iterval is correct , 7 421, 9 421, 11 b c , 1 A 433, 3 A 434, 5 A 435, 15 A 436, 17 A ,1 SAMPSIZE 442, 5 1 = 56 s = SAMPSIZE = 64 We eed 64 more weights to be 90% cofidet that the average weight of the 120 players will be withi 4 pouds of the actual average of all of the players , 15 1 = 167 s = 3.8. E =.5 (30 secods =.5 miutes) SAMPSIZE =217. Get 217 more customers to be 99% cofidet that the mea of the 384 times will be o more tha 30 secods from the true mea

20 , 1 Eter data ito L1 ad L2 0 The TI gives better results. The CI for (µ 1 µ 2 ) has opposite sigs, so we caot say with 90% cofidece which mea is larger , 3 b 0 The TI gives better results. Both sides of the CI for (µ 1 µ 2 ) are positive. We are 85% cofidet that (µ 1 > µ 2 ) , 5 b 0 The TI gives better results. Both sides of the CI for (µ 1 µ 2 ) are egative. We are 90% cofidet that (µ 1 < µ 2 ) , 7 B 461, 11 B Both sides of the CI for (p 1 p 2 ) are positive. We are 99% cofidet that (p 1 > p 2 )

Inferential Statistics and Probability a Holistic Approach. Inference Process. Inference Process. Chapter 8 Slides. Maurice Geraghty,

Inferential Statistics and Probability a Holistic Approach. Inference Process. Inference Process. Chapter 8 Slides. Maurice Geraghty, Iferetial Statistics ad Probability a Holistic Approach Chapter 8 Poit Estimatio ad Cofidece Itervals This Course Material by Maurice Geraghty is licesed uder a Creative Commos Attributio-ShareAlike 4.0