Practice Problems for Test 3 Note: these problems oly cover CIs ad hypothesis testig You are also resposible for kowig the samplig distributio of the sample meas, ad the Cetral Limit Theorem Review all the exercises we did i class, ad the homework problems also (Please let me kow if I have ay mistakes o this review sheet extra credit) 1 From a radom sample of days i a recet year, the closig stock prices of Hasbro had a mea of $1931ad stadard deviatio of $237 Use t* = 168 for 90%, ad t* = 203 for 95% cofidece a Costruct the 90% ad 95% cofidece itervals for the populatio mea Iterpret the CIs 90% CI: Coditios: SRS, ad >30 * s x ± t = 19 31± 168 2 37 = ( 18 64319, 977) Or if you have a TI-83 use STAT TESTS 8:T-Iterval We are 90% cofidet that the mea closig stock prices of Hasbro is betwee $1864 ad $1998 95% CI: Coditios: SRS, ad >30 * s x ± t = 19 31± 2 03 2 37 = ( 18508, 20112 ) Or if you have a TI-83 use STAT TESTS 8:T-Iterval We are 95% cofidet that the mea closig stock prices of Hasbro is betwee $1851 ad $2011 c Which iterval is wider? What are the margis of errors? The 95% CI is wider Ad it is suppose to be wider, because higher cofidece level gives wider CI itervals Margi of error for 90% cofidece: 168 2 37 = 0 66 Margi of error for 95% cofidece: 2 03 2 37 = 0802
2 You radomly select 20 mortgage istitutios ad determie the curret mortgage iterest rate at each The sample mea rate is 693% with a sample stadard deviatio of 042% Fid the 99% cofidece iterval for the populatio mea mortgage iterest rate Assume the iterest rates are approximately ormally distributed Coditios: SRS, populatio is ormally distributed Both satistfied t* =2861 * s x ± t = 6 93 ± 2 861 0 42 = ( 6 66, 7 20 ) 20 We are 99% cofidet that the mea mortgage iterest rate for ALL mortgage istitutios is betwee 666% ad 720% 3 A biologist reports a CI of (21cm, 35cm) whe estimatig the mea height of a sample of seedligs Fid the sample mea ad the margi of error The poit estimator, the sample mea is exactly i the middle of the CI The margi of error, m, is the half of the iterval m m (-------------------------- -------------------------) 21 cm x 35 cm 21 + 35 Thus, the sample mea is: x = = 2 8 cm 2 Ad the margi of error is: m = 28 cm 21 cm = 07 cm 4 A radom sample of airfare prices i dollars for a oe-way ticket from New York to Housto is give: 288, 290, 292, 295, 298, 300, 305, 305, 306, 307, 314, 320, 322, 326, 327 Assumig that the prices of tickets are ormally distributed, fid ad iterpret the 90% CI for the mea price of all tickets from New York to Housto (The sample mea is $30634, ad the sample stadard deviatio is $1297, t* = 1761) Coditios: SRS, ormally distributed populatio: both are satisfied If you have a TI-83, 84, you ca eter the list ad the use STAT TESTS 8: TIterval ad highlight Data
* s 12 97 90% CI: x ± t = 306 34 ± 1761 = ( 300 43, 312 23 ) 15 We are 90% cofidet that the mea price of ALL airfare prices for a oe-way ticket from New York to Housto is betwee $30043 ad $31223 5 The followig table show is from a survey of radomly selected 900 US adults Who are the more dagerous drivers? Teeagers 565 People over 75 297 No opiio 38 a Costruct a 95% CI for the proportio of adults who thik that teeagers are the more dagerous drivers What s the margi of error? Coditios: SRS, x + 2 p$ = + = 565 + 2 4 900 + 4 = 0 627 p$ ± z * p$( 1 p$) + 4 = 0 627 ± 196 0 627( 1 0 627) 904 = ( 0596, 0 659 ) We are 95% cofidet that the proportio of ALL adults who thik that teeagers are the more dagerous drivers is betwee 596% ad 659% The margi of error: 196 0 627( 1 0 627) 904 = 0 0315 3% b Costruct a 95% CI for the proportio of adults who thik that people over 75 are the more dagerous drivers What is the margi of error? Coditios: SRS, x + 2 p$ = + = 297 + 2 4 900 + 4 = 0 331 * p$( 1 p$) 0 331( 1 0 331) p$ ± z = 0 331 ± 196 = ( 0 298, 0 359 ) + 4 904 We are 95% cofidet that the proportio of ALL adults who thik that teeagers are the more dagerous drivers is betwee 298% ad 359%
The margi of error: 196 0 331( 1 0 331) 904 = 0 0307 3% 6 You are ruig a political campaig ad wish to estimate, with 99% cofidece, the proportio of registered voters who will vote for your cadidate Your estimate must be accurate withi 3% of the true populatio Fid the miimum sample size eeded if a o prelimiary estimatio is available Whe o prelimiary estimatio is available, use p $ = 05 2 2 * z = m p p = 2 576 $( 1 $) ( 05 )( 1 05 ) = 1843271 0 03 Thus, the miimum sample size required is 1844 b a prelimiary estimatio gives p $ = 0 31 2 2 * z = m p p = 2 576 $( 1 $) ( 0 31)( 1 0 31) = 1577102 0 03 Thus, the miimum sample size required is 1578 7 A used car dealer says that the mea price of a 2002 Ford F-150 Super Cab is $18,800 You suspect this claim is too high You fid that a radom sample of 14 similar vehicles has a mea price of $18,250 ad a stadard error of $1,250 Is there eough evidece to reject the dealer s claim at α=005? Assume the populatio is ormally distributed Is your coclusio the same at the 10% level? Ca you reach the same coclusio by usig a CI? H0: µ = 18800 : µ < 18800 H a Coditios: SRS, ormal distributio Both satisfied t = x µ 0 1820 18800 = = 1646 1250 s 14 The p-value is 00618 Sice the p-value is greater tha 005, the sigificace level, we caot reject the ull hypothesis At the 5% level, we do t have eough evidece to reject the claim that the mea price of 2002 Ford F-150 Super Cab is $18,800
At the 10% level, sice the p-value is less tha 10%, we ca reject the ull hypothesis We have eough evidece to reject the claim that the mea price of 2002 Ford F-150 Super Cab is $18,800 No, we ca t use a CI to test the claim because it s ot a two-tailed test (We do t have a i the alterative hypothesis) 8 A compay that makes cola driks states that the mea caffeie cotet per oe 12-ouce bottle of cola is 40 milligrams You work as a quality cotrol maager ad are asked to test this claim Durig your tests, you fid that a radom sample of 30 12-ouce bottles of cola has a mea caffeie cotet of 392 milligrams with stadard deviatio of 75 milligrams Assume that the caffeie cotet is ormally distributed At α = 1% level, ca you reject the compay s claim? Ca you reach the same coclusio by usig a CI? H0: µ = 40 : µ 40 H a Coditios: SRS, the populatio is ormally distributed t = x µ 0 39 2 40 = = 0584 7 5 s 30 p-value: 0559 Coclusio: sice the p-value is about 56%, we have o evidece to reject the ull hypothesis We do t have eough evidece to reject the claim that the mea caffeie level of all 12-ouce bottle of cola is 40 milligrams Yes, we ca use a 99% cofidece iterval to test the claims sice we have a two-sided test Use t* = 275 * s x ± t = 39 2 ± 2 75 7 5 = ( 3543, 42 97 ) 30 Sice the 99% CI does cotai the hypothesized value, 40, we caot reject the ull hypothesis at the 1% level
9 A scietist estimates that the mea itroge dioxide level i West Lodo is 28 parts per billio You believe that the mea itroge dioxide level is higher So, you determie the itroge dioxide levels for radomly selected days The results (i parts per billio) are listed below At α = 10%, ca you support the scietist s estimate? Ca you reach the same coclusio by usig a CI? 27, 29, 53, 31, 16, 47, 22, 17, 13, 46, 99, 15 20, 17, 28, 10, 14, 9, 35, 29, 32, 67, 24, 31, 43, 29, 12, 39, 65, 94, 12, 27, 13, 16, 40, 62 (The sample mea is 3286, ad the sample stadard deviatio is 2213) Hypotheses: H 0: µ = 28 H a : µ > 28 Radom sample,ad > 30 If you have a Ti-83, 84, you ca eter the list ad the use STAT TESTS 2: T-test, ad highlight Data t = x µ 0 32 86 28 = = 1318 22 12 s p-value: 0098 Coclusio: sice p-value is less tha α (0098 < 10), strictly speakig the result is statistically sigificat That is, we ca reject the ull hypothesis, although it s really a borderlie case sice 0098 010 The decisio is i your hads This would be my coclusio: We have little or o evidece agaist the ull hypothesis We have little or o evidece agaist the claim that the mea itroge dioxide level i West Lodo is 28 parts per billio Further ivestigatio is ecessary No, we ca t use a CI iterval here to test the claim because it s ot a two-tailed test (we do t have i the alterative hypothesis) 10 A medical researcher claims that 20% of adults i the US are allergic to a medicatio You thik that less tha 20% are allergic to a medicatio I a radom sample of 100 adults, 15% say they have such a allergy Test the researcher s claim at α = 1% level Ca you reach the same coclusio usig the appropriate cofidece iterval?
Hypotheses: H p 0: = 0 20 H : p < 0 20 a Coditios: SRS, p0 = 100( 0 20) = 20 > 10, ( 1 p0) = 100( 1 0 20) = 80 > 10, all satistfied Test statistic: z = p p$ p ( 1 p ) 0 0 0 = 015 0 20 0 20( 1 0 20) 100 = 125 (You ca use your calculator, STAT TESTS 5:1-PropZTest Use x = p$ = 100( 015 ) = 15 ) p-value: 01056 Coclusio: Sice p-value > 10%, we have o evidece agaist the ull hypothesis We ca t reject it Based o our sample, we do t have eough evidece to reject the medical researcher s claim that 20% of adults i the US are allergic to a medicatio No, we ca t use a CI iterval here to test the claim because it s ot a two-tailed test (we do t have i the alterative hypothesis) 11 USA TODAY reports that 5% of US adults have see a extraterrestrial beig You decide to test this claim ad ask a radom sample of 250 US adults whether they have ever see a extraterrestrial beig Of those surveyed, 8% reply yes At α = 5%, is there eough evidece to reject the claim? Ca you reach the same coclusio usig the appropriate cofidece iterval? Hypotheses: H p 0: = 0 05 H : p 0 05 a Coditios: SRS, p0 = 250( 0 05) = 12 5 > 10, ( 1 p0 ) = 250( 1 0 05) = 237 5 > 10, all satisfied Test statistic: z = p p$ p ( 1 p ) 0 0 0 = 0 08 0 05 0 05( 1 0 05) 250 = 2176 (You ca use your calculator, STAT TESTS 5:1-PropZTest Use x = p$ = 250( 0 08) = 20 ) p-value: 0029 (If you wat to fid the p-value usig the z-table, fid the value for z = 2176, ad multiply if with 2 because it s a two-tailed test)
Coclusio: Sice p-value < 5%, we ca reject the ull hypothesis We have good evidece agaist the USA TODAY s claim that 5% of US adults have see a extraterrestrial beig Yes, we ca use a 95% CI iterval (the cofidece level ad the α must add up to 100%) because it s a two-tailed test (we have i our alterative hypothesis) 20 + 2 Usig the adjusted sample proportio: p $ = 250 + 4 = 0 087 The 95% CI: p$ ± z * p$( 1 p$) + 4 = 087 ± 196 087 ( 1 087 ) 250 + 4 = ( 0 052, 0121 ) TI-83: A: 1-PropZIt with x = 22, = 254 Sice the claimed value, 005 is NOT iside the 95% CI, at the 5% level we ca reject the ull hypothesis claimed by USA TODAY that 5% of US adults have see a extraterrestrial beig Thus, we ca reach the same coclusio what we got usig the hypothesis test