s 1. C. The formula for the cofidece iterval for a populatio mea is: x t, which was based o the sample Mea. So, x is guarateed to be i the iterval you form.. D. Use the rule : p-value <alpha, reject H 0. The P-value is greater tha the sigificace level (=.10), so we ca coclude the data do ot provide sufficiet evidece to reject the ull hypothesis (H 0 ). Fail to reject H 0. 3. A. The formula for cofidece iterval is: x t s where (t s ) is the margi of error. Other thigs beig equal, the margi of error of a cofidece iterval icreases as the sample size decreases. So, whe the sample size decreases, the legth of the cofidece iterval will become bigger. 4. B. Similar to the previous questio. Other thigs beig equal, the margi of error of a cofidece iterval decreases as the cofidece level (t -score) decreases. So, the legth of the cofidece iterval will become smaller whe the cofidece level decreases. 5. D. From the results of the previous two questios, we kow that whe the sample size icreases, the cofidece iterval will be smaller. However, it will become bigger as the cofidece level icreases. Therefore, we caot coclude how the cofidece iterval will be i this questio, sice we do t have eough iformatio to determie whether the chage i sample size or the cofidece level is more ifluetial here. 6. C. The samplig distributio of a statistic is the distributio of values take by the statistic i all possible samples of the same size from the same populatio. The mea of the samplig distributio of x is the populatio mea. 7. A. Sice the P-value is a probability, so it must be betwee 0 ad 1. 8. B. For 95% cofidece, z = 1.96. For a margi of error of 0.5, we have z* s = m = 1.96*10 =1536.6. So, the sample size should be 1537. (Always 0.5 roud up to the ext higher whole umber whe fidig ). 9. C. Take the 93% ad chage it to a decimal (0.93). Take 1 -.93 = 0.07 (this is the area i the tails). Divide this umber i half (0.035.) Look i the middle of the table for the etry 0.035. This correspods to - z=-1.81. Thus, z=1.81. I short look up (1-0.93)/=.035 i the middle of the table ad z is the absolute value of the z-score.
10. D. The samplig distributio of x is the distributio of values take by x i all possible samples of the same size from the same populatio. 11. B. Because we ifer coclusios about the populatio from data o selected Idividuals (all sample). 1. a. F. I a very large umber of samples, 95% of the cofidece itervals would cotai the populatio mea. If the edpoits of the CI are give, use the term cofidece, ot probability. b. T. The defiitio of cofidece iterval. We are 95% cofidece that the ukow lies betwee (1.15, 4.0). c. F. The ceter of each iterval is at x, ad therefore varies from sample to sample. So, whe 100 itervals calculated the same way, we ca expect 100 of them to capture their ow sample mea. Not oly 95% of them. d. F. This setece states that idividuals (all America households) is i that iterval. This is wrog. CI made statemets about ot idividuals. e. T. I a very large umber of samples, 95% of the cofidece itervals would cotai the populatio mea. f. T. The ceter of each iterval is at x, ad therefore varies from sample to sample. So, whe 100 itervals calculated the same way, we ca expect 100 of them to capture the sample mea. 13. C. Use the rule : p-value <alpha, reject H 0. Our reasoable alpha levels are.10,.05, ad.01. We reject H 0 at all these levels, so III is true. II is ot true because there is ot a iterval i HT. I is true because the defiitio of the p-value is the probability that you would see a result this extreme if the ull were true. This p-value is so low that the probability of gettig a sample like this if H 0 were true is ulikely. 14. B. A parameter is a umber that describes the populatio. A statistic is a umber that describes the sample. 15. B This problem is a questio about the samplig distributio of the sample meas. The amout of moey eared i tips is a quatitative variable. The sample mea has a Normal distributio with mea equal to 10 ad stadard error equal to.5. Draw the picture. 35 1310 z 7.09 The probability greater tha 7.09 is a very small umber almost zero..5 35
16. C. A parameter is a umber that describes the populatio. So here, the parameter should be the average umber of jelly beas i all packages made, which is 375. 373 375 17. A. z= =-.61. Look the table A, the probability of beig less tha -.61 is.709. 8/ 6 18. C. Sice the umber of jelly beas follows the ormal distributio, we ca use the z table. 19. C. Accordig the cetral limit theory, whe is large, the samplig distributio of the sample mea x is approximately ormal. That is, x ~,. s 0. B. The formula for the cofidece iterval for a populatio mea is: x t.however, is large, so we ca use the z istead of the t. x z s. x =7.1. For 95% cofidece, z = 1.96. 5 So the cofidece iterval is 7.1 1.96* 00 =7.1.69=(6.41, 7.79) 1. D. The defiitio of cofidece iterval. We are 95% cofidet that the ukow populatio mea work hours lies betwee 6.8 ad 7.38. A is wrog because it was the term probability whe the umbers are give. B is wrog because it talks about idividuals rather tha the populatio mea. C is wrog because of it estimates the average i our sample. A CI estimates the average i our populatio.. B. The estimate ( x i this case) is our guess for the value of the ukow parameter (). So, we eed to calculate the margi of error shows how accurate we believe our guess is, based o the variability of the estimate. That s why we have 95% cofidece i our iterval, istead of 100%. 3. C. From the coclusio of questio 4 ad 5, we kow that the cofidece iterval will become arrower whe the size icreases ad the cofidet level decreases. 4. B. We will reject the ull whe the p-value is smaller tha the sigificace level. The p-value of this test is 0.044, which is smaller tha the levels at.10,.05, but larger tha.01. So we reject the H 0 whe =0.10 ad.05, but fail to reject the ull whe =.01. 5. a. T. Sice the populatio has a ormal distributio, we ca use the Normal table for the probability that oe perso is more tha 00 lbs. b. T. Sice the populatio has a ormal distributio, the samplig distributio of x is ormal. So, we ca use the z table. c. T. Sice the populatio has a ormal distributio, the samplig distributio of x is also ormal. So, we ca use the z table.
d. F. The distributio is ot Normal because the 68,95,99.7% rule does ot apply. The sample size is quite small (1), so the CLT does ot apply. So, we ca t use the z table. e. F. The sample size is quite small (10), So the CLT does ot apply. So, we ca t use the z table. f. T. Accordig to the CLT, whe we draw a SRS of size from ay populatio with mea ad fiite stadard deviatio. Whe is large, the samplig distributio of the sample mea x is approximately ormal. x ~N(, ). Here, the sample size is large, so we ca apply the CLT. Therefore, we ca use the Z table to fid the probability. p ( 1 g. F. ~N( p, ) whe values of, p satisfyig p15 ad (1-15. However, p 10 =0* =4150, therefore, you ca t use Normal table here. 50 h. T. ~N( p, p ( 1 ) whe values of, p satisfyig p15 ad (1-15. 1000 p =91* =18.. (1-=91*(4000/5000) 7.8. 5000 The rule ca be applied. So we ca use the z table here. 6. C ~N( p, p ( 1 ) whe values of, p satisfyig p15 ad (1-15. p =0.*100=0> 15 ad (1-=100*0.8 = 80 15. So, ~N( 0., ~N( 0., 0.04 ) 0.(0.8) 100 7. B Use the samplig distributio of the sample proportio that you used above ad the z-score. 0.4 0.0 z 1.0 Look up 1.00 i the table. 0.8413 is listed i the table. This is the proportio 0.04 less tha, we wat the proportio greater tha so we take 1-0.8413=0.1587. 8. C ~N( p, p ( 1 ) whe values of, p satisfyig p15 ad (1-15. p =0.05*300=160 > 15 ad (1-=300*0.95=3040 15. So, ~N( 0.05, ~N( 0.05, 0.00385) ) 0.05(0.95) 300 )
p ( 1 9. D ~N( p, ) whe values of, p satisfyig p15 ad (1-15. p =0.06*100=6 < 15 So, the distributio of the sample proportio is ukow. p ( 1 30. C ~N( p, ) whe values of, p satisfyig p15 ad (1-15. p =0.06*300=18 > 15 ad (1-=300*0.94=8 15. So, ~N( 0.06, 0.06(0.94) 300 ) ~N( 0.06, 0.013711). 31. A Use the samplig distributio of the sample proportio that you used i problem 5 ad the 0.10 0.06 z-score. z.90 Look up.90 i the table. P(z<.90) = 0.9981. We wat the 0.013711 probability that our sample proportio is greater tha 0.10 so we take 1-0.9981=0.0019. ( z) p(1 (1.96) (.04)(.96) 3. B. = = m.01 So should be 1476 =1475.1. 33. B. ˆp == X =40/800=0.05 estimate of the stadard error of ˆp = (1 ) = 0.05*(0.95) 800 From the z table, we fid the value of z to be 1.960. =.0077. So the CI is ˆp z*se =.05 (1.960)(.0077)= (.035,.065) 34. G. The populatio proportio p is ukow, ad that s why we wat to estimate it by the sample proportio. E. P-hat is the sample proportio. =X/=87/350=.8 A. Sice the researchers are testig to determie if more tha half of all reported scams victimize the elderly, the p 0 should be 0.5. C. From the sample size, we record the cout X of success ( here ifers the victim over 65 years old). The X should be 87. D. The total sample size is 350.
35. E. Sice the P-value is a probability, so it must be betwee 0 ad 1. 36. C Note that we are takig 50 idepedet observatios from the distributio. This is a large eough sample size to use the Cetral Limit Theorem for x-bar. Hece x-bar is distributed approximately ormal with mea 48 ad stadard error 5/ (50) 37. A Here, we are samplig 100 idepedet observatios from a populatio with proportio p = 0.75. Therefore, the samplig distributio of the sample proportio has mea 0.75 ad stadard error ((0.75*0.5)/100) = 0.0433. Now, sice p = 75 > 15 ad (1- = 5 > 15, we ca coclude that the samplig distributio of the sample proportio is approximate ormally distributed with mea 0.75 ad stadard error 0.0433. Now, we ca compute the z-score ad get z = (0.65 0.75) / 0.0433 = -.31. The, we use a z-table to fid P(Z < -.31) = 0.0104. 38. B Note that we are takig a radom sample of 143 adult males. Sice this sample size is larger tha 30, we ca use the Cetral Limit Theorem ad coclude that average weight is approximately ormally distributed eve though we do ot kow the distributio of the weight of adult males i Alachua Couty. So the average weight is distributed ormally with mea 190 ad stadard error 0/ (143) = 1.67484. Now, we ca compute the z-score ad get z = (193 190) / 1.67484 = 1.79. Fially, we ow use the z-table to fid P(Z > 1.79) = 0.0367. 39. D We are ot give the distributio of the populatio we are samplig from ad our sample size is oly 16 (<30). Hece, we caot give a approximate probability here. 40. B Here, the umber of hits is a biomial radom variable with = 400 ad probability of a hit p = 0.3. Therefore, the mea of the samplig distributio of the proportio of hits has mea 0.3 ad stadard error 0.09. 41. C Rule: p-value < alpha Reject Ho. At alpha = 0.10, reject Ho. At alpha = 0.05, reject Ho. At alpha = 0.01, fail to reject Ho. 4. D (1 ) z 0.8467*(1.8467) 0.8467 1.96 150 0.8467 0.058 43. B For a cofidece iterval for the populatio proportio, we must assume that the data comes from a radom sample, that we have categorical data, p> 15 ad (1-> 15. 44. C The size of the populatio does ot affect how accurate the results are. The size of the sample affects how accurate a predictio we ca make.
45. C is the icorrect statemet. The cofidece iterval is suppose to estimate the populatio proportio ot the sample proportio. A is just givig the cofidece iterval that is o.k. B is talkig about estimatig the populatio proportio with the cofidece iterval that is correct. D is estimatig the complemet of ot i favor of gu cotrol i favor of gu cotrol. 46. B p ˆ = 17 > 15, but (1 ) = 3 < 15. Therefore, we ca compute the cofidece iterval usig the large sample formula if we add successes ad failures. The 17 0 se.791 4 1 (.7917)(.083) 0 4.089 ad the resultig 95% cofidece iterval is p ˆ 1.96( se).79 1.96.083 (.69,.954). 47. C The oly correct statemet is the first oe --The large-sample cofidece iterval formula for proportios is valid if phat 15 ad (1-phat) 15. The large sample cofidece iterval oly cotai the true value a certai percetage of the time. A 95% CI will cotai the value 95% of the time. You add successes ad failures. 48. B First, we use a calculator to fid that the sample stadard deviatio s = 83.55. The se s 83.55 37.36. 5 49. A The 95% cofidece iterval for the populatio mea is x t. 05 se. I this particular problem, we have x 566.0 se 37.36 Usig df = 1 = 4, we look up (i a table) that t.05 =.776. The our cofidece iterval is x t se 566.0.77637.36 (46.3, 669.7).. 05 50. C Assumptios for the cofidece iterval for the mea are as follows: data is quatitative,
radom sample, data comes from a ormal distributio. Oly statemet (c) is true. 51. C Accordig to the Cetral Limit Theorem for large the samplig distributio of sample mea is Normal. 5. B The stadard error is (0.65 * 0.35)/ 1000= 0.01508 53. B ~N( p, p ( 1 ) whe values of, p satisfyig p15 ad (1-15., p satisfyig 1000*0.6565 ad 1000*0.35 15. So, ~N( 0.65, ~N( 0.65,0.01508) 0.65(1 0.65) 1000 ) 54. B The Z- value for this (0.7 0.65)/ 0.01508 = 3.3 Now P( Z> 3.3) = 1 P(Z 3.3) = 0.0005 55. C p = 0.5 = the probability of gettig heads whe you flip a ubiased(fair) coi You eed to have p> 15 ad p> 15. This happes whe = 50. (50*.5=5 ad 50*(1-.5) = 5) 56. B The hypothesis was ot rejected at level alpha=.05.so p value was higher tha 0.05 ad so higher tha 0.05 as well. So the test will agai fail to reject the ull hypothesis at level =0.05. 57. A The hypothesis was rejected at level=0.01.so, p value was less tha 0.01 ad so less tha 0.10 as well. Hece the test will agai reject the ull hypothesis at level=0.10. 58. C The hypothesis was rejected at level=0.10.so p value was less tha 0.10.But that might be more tha 0.05 or might be less tha 0.05 which we do t kow from above iformatio. Therefore we do t kow what will happe for the test at level=0.05. 59. D They wat to show that more whales tur away tha usual with the extra souds emitted. p(1 0.4(1 0.4) 60. D Solutio: p-hat is 4/5=0.4615. se 0 0. 067937. 5 p0 0.4615 0.4 Thus, z 0. 905 se 0 0.067937 The probability shaded greater tha 0.905 is (1-0.8186) =0.1814. p-value = 0.1814. p-value is ot less tha alpha So, we fail to reject Ho.
61. B The assumptios of the hypothesis test for the proportio are the data must be categorical, data must come from a radom sample, p> 15 ad (1-> 15. 6. B The bottles are of iterest ot UF studets.