OPIM 5103 Stats Prof. J. Stallaert Solutions to Homework Set 2 CHAPTER 6

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1 CHAPTER Computig Z-scores ad usig the table i the book yields: (a) P(34 < X < 50) = P( 1.33 < Z < 0) = = P(X < 30) + P(X > 60) = P(Z < 1.67) + P(Z > 0.83) = ( ) = P(X > A) = 0.80 P(Z < 0.84) 0.20 (d) Z 0.84 A = (12) = thousad miles or 39,920 miles The smaller stadard deviatio makes the Z-values larger. (a) P(34 < X < 50) = P( 1.60 < Z < 0) = = P(X < 30) + P(X > 60) = P(Z < 2.00) + P(Z > 1.00) = ( ) = A = (10) = 41.6 thousad miles or 41,600 miles 6.9 Computig Z-scores ad usig the table i the book yields: (a) P(12 < X < 15) = P(-0.15 < Z < 1.35) = P(X < 10) = P(Z < -1.15) = P(X lower < X < X upper ) = 0.95 P(Z < 1.96) = ad P(Z < 1.96) = Z 1.96 X low er 5 Z 1.96 X 5 upper X lower = ad X upper = A Computig Z-scores ad usig the table i the book yields: (a) P(X < 91) = P(Z < 2.25) = P(65 < X < 89) = P( 1.00 < Z < 2.00) = = P(X > A) = 0.05 P(Z < 1.645) = A 73 Z A = (8) = 86.16% 8 (d) Optio 1: P(X > A) = 0.10 P(Z < 1.28) Z Sice your score of 81% o this exam represets a Z-score of 1.00, which is below the miimum Z-score of 1.28, you will ot ear a A grade o the exam uder this gradig optio Optio 2: Z Sice your score of 68% o this exam represets a Z-score of 2.00, which is well above the miimum Z-score of 1.28, you will ear a A grade o the exam uder this gradig optio. You should prefer Optio 2.

2 6.11 Computig Z-scores ad usig the table i the book yields: (a) P(X < 180) = P(Z < 1.50) = P(180 < X < 300) = P( 1.50 < Z < 1.50) = = P(110 < X < 180) = P( 3.25 < Z < 1.50) = = (d) P(X < A) = 0.01 P(Z < 2.33) = 0.01 A = (40) = secods 6.13 (a) Computig Z-scores ad usig the table i the book yields: P(21.99 < X < 22.00) = P( 2.4 < Z < 0.4) = P(21.99 < X < 22.01) = P( 2.4 < Z < 1.6) = P(X > A) = 0.02 Z = 2.05 A = (d) (a) P(21.99 < X < 22.00) = P( 3.0 < Z < 0.5) = P(21.99 < X < 22.01) = P( 3.0 < Z < 2) = P(X > A) = 0.02 Z = 2.05 A = (a) P(0.75 < X < 0.753) = P( 0.75 < Z < 0) = P(0.74 < X < 0.75) = P( 3.25 < Z < 0.75) = = P(X > 0.76) = P(Z > 1.75) = = (d) P(X < 0.74) = P(Z < 3.25) = (e) P(X < A) = P(Z < 1.48) = 0.07 A = (0.004) = (a) P(1.90 < X < 2.00) = P( 2.00 < Z < 0) = P(1.90 < X < 2.10) = P( 2.00 < Z < 2.00) = = P(X < 1.90) + P(X > 2.10) = 1 P(1.90 < X < 2.10) = (d) P(X > A) = P( Z > 2.33) = 0.99 A = (0.05) = (e) P(A < X < B) = P( 2.58 < Z < 2.58) = 0.99 A = (0.05) = B = (0.05) =

3 7.21 (a) = X CHAPTER 7 P(7.8 < X < 8.2) = P( 0.50 < Z < 0.50) = = P(7.5 < X < 8.0) = P( 1.25 < Z < 0) = = = 0. 2 X 100 P(7.8 < X < 8.2) = P( 1.00 < Z < 1.00) = = (d) With the sample size icreasig from = 25 to = 100, more sample meas will be closer to the distributio mea. The stadard error of the samplig distributio of size 100 is much smaller tha that of size 25, so the likelihood that the sample mea will fall withi 0.2 miutes of the mea is much higher for samples of size 100 (probability = ) tha for samples of size 25 (probability = ) (a) p = 15/50 = 0.30 p = 0.40(0.60) 50 = (a) P(0.50 < p < 0.60) = P(0 < Z < 2.83) = P( < Z < 1.645) = 0.90 p = (0.0354) = p = (0.0354) = P(p > 0.65) = P (Z > 4.24) = virtually zero (d) If = 200, P(p > 0.60) = P (Z > 2.83) = = If = 1000, P(p > 0.55) = P (Z > 3.16) = = More tha 60% correct i a sample of 200 is more likely tha more tha 55% correct i a sample of (a) p 0.56, p = 100 = P(p < 0.5) = P (Z < ) = p 0.56, p = 500 = P(p < 0.5) = P (Z < ) = Icreasig the sample size by a factor of 5 decreases the stadard error by a factor of. The samplig distributio of the proportio becomes more cocetrated aroud the true proportio of 0.56 ad, hece, the probability i becomes smaller tha that i (a).

4 P = 0.98, p 500 = (a) P(p > 0.99) = P(Z > ) = P(0.97 < p < 0.99) = P( < Z < ) = P(p < 0.97) = P(Z < ) = = , = X X (a) 20 = PHStat output: Probability for X <= X Value -10 Z Value P(X<=-10) 1 P( X < -10) = P(Z < ) = PHStat output: Probability for a Rage From X Value -20 To X Value 0 Z Value for Z Value for P(X<=-20) P(X<=0) P(-20<=X<=0) P(-20 < X < 0) = P(3.7868< Z < ) = PHStat output: Probability for X > X Value -30 Z Value P(X>-30) P( X > -30) = P(Z > ) = (a) = 10 P(X < 0) = P(Z < ) = P(10 < X < 20) = P( < Z < ) = P(X> 10) = P(Z > ) = (d) 10 = 24.03, = X X = 5 4 P( X < 0) = P(Z < ) = (e) P(10 < X < 20) = P( < Z < ) = (f) P( X > 10) = P(Z < ) = (g) Sice the sample mea of retur of a sample of treasury bod is distributed closer tha the retur of a sigle treasury bod to the populatio mea, the probabilities i (a) ad are greater tha those i (d) ad (e) while the probability i is smaller tha that i (f).

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6 CHAPTER (a) X Z Sice the value of 1.0 is icluded i the iterval, there is o reaso to believe that the mea is differet from 1.0 gallo. No. Sice is kow ad = 50, from the Cetral Limit Theorem, we may assume that the samplig distributio of X is approximately ormal. (d) The reduced cofidece level arrows the width of the cofidece iterval. X Z Sice the value of 1.0 is still icluded i the iterval, there is o reaso to believe that the mea is differet from 1.0 gallo (a) Usig the Excel spreadsheet ( std dev kow ): Sample size 64 Sample Mea 350 Kow Std Dev 100 Cofidece level 0.95 Poit Estimate 350 Alpha 0.05 Margi of error Iterval Hece: No. The maufacturer caot support a claim that the bulbs last a average 400 hours. Based o the data from the sample, a mea of 400 hours would represet a distace of 4 stadard deviatios above the sample mea of 350 hours. No. Sice is kow ad = 64, from the Cetral Limit Theorem, we may assume that the samplig distributio of X is approximately ormal. (d) The cofidece iterval is arrower based o a process stadard deviatio of 80 hours rather tha the origial assumptio of 100 hours. (a) X Z Based o the smaller stadard deviatio, a mea of 400 hours would represet a distace of 5 stadard deviatios above the sample mea of 350 hours. No, the maufacturer caot support a claim that the bulbs have a mea life of 400 hours.

7 8.18 Oe eeds to fid the sample mea ad sample stadard deviatio of the data set first by eterig the data i Excel a usig the fuctios AVERAGE ad STDDEV. Usig the Excel spreadsheet ( std dev ukow ) the yields: (a) Sample size 6 Sample Mea Sample Std Dev Cofidece level 0.95 Poit Estimate Alpha 0.05 Margi of error Iterval Hece: You ca be 95% cofidet that the populatio mea price for two tickets with olie service charges, large popcor, ad two medium soft driks is somewhere betwee $31.93 ad $ Oe eeds to fid the sample mea ad sample stadard deviatio of the data set first by eterig the data i Excel a usig the fuctios AVERAGE ad STDDEV. Usig the Excel spreadsheet ( std dev ukow ) the yields: Sample size 16 Sample Mea Sample Std Dev Cofidece level 0.95 Poit Estimate Alpha 0.05 Margi of error Iterval (a) You ca be 95% cofidet that the populatio mea miles per gallo of 2009 family sedas priced uder $20,000 is somewhere betwee ad Sice the 95% cofidet iterval for populatio mea miles per gallo of 2007 small SUVs does ot overlap with that for the populatio mea miles per gallo of 2007 family sedas, we are 95% cofidet that the populatio mea miles per gallo of 2007 small SUVs is lower tha that of 2007 family sedas.

8 8.30 (a) Usig the Excel spreadsheet ( proportios ) we get: Sample size 500 Sample proportio 0.52 Cofidece level 0.95 Poit estimate 0.52 Alpha 0.05 Margi of error Iterval (d) Or: X 260 p(1 p) 0.52(1 0.52) p = 0.52 p Z Hece: Sice the 95% cofidece iterval cotais 0.50, you caot claim that more tha half of all U.S. workers have egotiated a pay raise. X 2600 p(1 p) 0.52(1 0.52) (a) p = 0.52 p Z Hece: Sice the lower limit of the 95% cofidece iterval is greater tha 0.50, you ca claim that more tha half of all U.S. workers have egotiated a pay raise. The larger the sample size, the arrow is the cofidece iterval holdig everythig else costat Usig the Excel spreadsheet ( proportios ) we get: Sample size 1000 Sample proportio 0.76 Cofidece level 0.95 Poit estimate 0.76 Alpha Margi of error Iterval Or: X 760 p = p(1 p) 0.76(1 0.76) p Z

9 CHAPTER H 0 : 2.8 feet. The mea legth of steel bars produced is at least 2.8 feet ad the productio equipmet does ot eed immediate adjustmet. H 1 : < 2.8 feet. The mea legth of steel bars produced is less tha 2.8 feet ad the productio equipmet does eed immediate adjustmet. From the hypo.xls spreadsheet (oe sample meas): INPUT Sample mea 2.73 Sample Std. Dev. 0.2 Hypothesized mea 2.8 Sample size 25 OUTPUT t-value P-value (two-tailed test) P-value (oe-tailed test) Decisio: Sice p value = is less tha = 0.05, reject H 0. There is eough evidece to coclude the productio equipmet eeds adjustmet. The probability of obtaiig a sample whose mea is 2.73 feet or less whe the ull hypothesis is true is (a) H 0 : 0.5 H 1 : < 0.5 From the hypo.xls spreadsheet (oe sample proportio): INPUT Sample proportio 0.42 Hypothesized proportio 0.5 Sample size 500 OUTPUT Std. Dev. (ormal approx.) Z-value P-value (two-tailed test) P-value (oe-tailed test) ASSUMPTIONS p (1-p) Coclusio: p-value of forces us to reject the ull hypothesis. H 0 : 0.5 H 1 : < 0.5 INPUT Sample proportio 0.42

10 Hypothesized proportio 0.5 Sample size 100 OUTPUT Std. Dev. (ormal approx.) 0.05 Z-value -1.6 P-value (two-tailed test) P-value (oe-tailed test) ASSUMPTIONS p (1-p) Coclusio: p-value of meas that there is ot eough evidece to reject the ull hypothesis. The larger the sample size, the smaller the samplig error. Eve though the sample proportio is the same value at 0.42 i (a) ad, the test statistic is more egative while the p-value is smaller i (a) compared to because of the larger sample size i (a) (a) H 0 : < 0.5 H 1 : > 0.5 From the hypo.xls spreadsheet 9oe sample proportio): INPUT Sample proportio 0.52 Hypothesized proportio 0.5 Sample size 1132 OUTPUT Std. Dev. (ormal approx.) Z-value P-value (two-tailed test) P-value (oe-tailed test) ASSUMPTIONS p (1-p) Decisio: Sice p-value = > 0.05, do ot reject H 0. There is ot eough evidece to show that more tha half of all mobile phoe users are ow usig the mobile Iteret. The claim by the author is ivalid. INPUT Sample proportio 0.53 Hypothesized proportio 0.5 Sample size 1132 OUTPUT Std. Dev. (ormal approx.) Z-value

11 (d) P-value (two-tailed test) P-value (oe-tailed test) ASSUMPTIONS Np (1-p) H 0 : 0.5 H 1 : > 0.5 Decisio rule: If p-value < 0.05, reject H 0. p-value = Decisio: Sice p-value = < 0.05, reject H 0. There is eough evidece to show that more tha half of all mobile phoe users are ow usig the mobile Iteret. The claim by the author is valid. With the same level of sigificace, 53% is cosidered as far eough above 50% to reject the ull hypothesis of o more tha half of all mobile phoe users are ow usig the mobile Iteret while 52% is ot cosidered as far eough above 0.5.