STA 2023 Practice Questions Exam 2 Chapter 7- sec 9.2. Case parameter estimator standard error Estimate of standard error

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1 STA 2023 Practice Questios Exam 2 Chapter 7- sec 9.2 Formulas Give o the test: Case parameter estimator stadard error Estimate of stadard error Samplig Distributio oe mea x s t (-1) oe p ( 1 p) CI: prop. p pˆ ST: pˆ(1 pˆ) p 1 p ) ( 0 0 z Memorize these Formulas: Geeral Format for Cofidece Iterval: estimator +/- (t or z) est. stadard error Geeral Format of Test Statistic: estimator # from Ho ( t or z) estimate of stderr Determiig sample size for estimatig proportios ad meas. Practice Problems o ext page.

7 28. A auto isurace compay has 32,000 cliets, ad 5% of their cliets submitted a claim i the past year. We will take a sample 3,200 cliets, ad determie how may of them have submitted a claim i the past year. What is the samplig distributio of ˆp? a) ˆp ~ N(3200, 0.2) b) ˆp ~N(160, 152) c) ˆp ~N(0.05, ) d) Ca ot be determied Questios Suppose 20 doors come to a blood drive. Assume that the blood doors are ot related i ay way, so that we ca cosider them idepedet. The probability that the door has type-o blood is 0.06, which is costat from door to door. Let X = the umber of doors that have type-o blood. 29. For a sample of 100 doors, what is the samplig distributio of the sample proportio? a) ˆp ~ Biomial (100, 0.06) b) ˆp ~Normal (0.06, ) c) ˆp ~Normal(6, 2.37) d) Ca ot be determied 30. For a sample of 300 doors, what is the samplig distributio of the sample proportio? a) ˆp ~ Biomial (200, 0.06) b) ˆp ~Normal (12, 3.359) c) ˆp ~Normal(0.06, ) d) Ca ot be determied 31. For the sample of 300 doors, what is the probability that the sample proportio is greater tha 0.10? a) b) c) d) The executives at Sadbachia, Ic. havig recetly solved their widget crises, have aother major problem with oe of their products. May cities are sedig complaits that their mahole covers are defective ad people are fallig ito the sewers. Sadbachia, Ic. is pretty sure that oly 4% of their mahole covers are defective, but they would like to do a study to cofirm this umber. They are hopig to costruct a 95% cofidece iterval to get withi 0.01 of the true proportio of defective mahole covers. How may mahole covers eed to be tested? a) 8 b) 1476 c) 9604 d) 9605

12 Questios We kow that 65% of all Americas prefer chocolate over vailla ice cream. Suppose that 1000 people were radomly selected. 52. The stadard error of the sample proportio is a) b) c) d) The Samplig Distributio of the sample proportio is a) Biomial ( 1000, 0.65) b) Normal( 0.65, ) c) Normal(10000,0.65) d) Noe of the above 54. What is the probability that our sample will have more tha 70% of people prefer chocolate ice cream? a) b) c) 0.70 d) oe of the above 55. We are doig a experimet where we record the umber of heads whe we get whe we flip a ubiased coi may times. For what sample sizes below would the samplig distributio of the sample proportio be approximately ormally distributed? a) 5 b) 28 c) 50 d) All of the above e) Noe of the above 56. For a test with the ull hypothesis Ho: p = 0.5 vs. the alterative Ha: p > 0.5, the ull hypothesis was ot rejected at level alpha=.05. Das wats to perform the same test at level alpha=.025. What will be his coclusio? a) Reject H 0. b) Fail to Reject H 0. c) No coclusio ca be made. d) Reject Ha. 57. The ull hypothesis Ho: p=.5 agaist the alterative Ha: p>.5 was rejected at level alpha=0.01. Nate wats to kow what the test will result at level alpha=0.10. What will be his coclusio? e) Reject H 0. f) Fail to Reject H 0. g) No coclusio ca be made. h) Reject Ha.

13 58. A ull hypothesis was rejected at level alpha=0.10.what will be the result of the test at level alpha=0.05? a) Reject Ho b) Fail to Reject Ho c) No coclusio ca be made d) Reject Ha Questios Commercial fisherme workig i certai parts of the Atlatic Ocea sometimes fid their efforts beig hidered by the presece of whales. Ideally, they would like to scare away the whales without frighteig the fish. Oe of the strategies beig experimeted with is to trasmit uderwater the souds of a killer whale. O the 52 occasios that that techique has bee tried, it worked 24 times (that is, the whales left the area immediately). Experiece has show, though, that 40% of all whales sighted ear fishig boats leave o their ow accord, ayway, probably just to get away from the oise of the boat. 59. What would a reasoable hypothesis test be: a) Ho: p=0.4 versus Ha: p = 0.46 b) Ho: p=0.46 versus Ha: p > 0.46 c) Ho: p=0.46 versus Ha: p 0.46 d) Ho: p=0.4 versus Ha: p > Suppose you wat to test Ho: p=0.4 versus Ha: p > 0.40 at the 0.05 level of sigificace. What would your coclusio be? a) Reject Ho. b) Accept Ho. c) Accept Ha. d) Fail to reject Ho. 61. Which of the followig are the assumptios that must be satisfied i order to be able to coduct a sigificace test for p? I The data is obtaied from a radom sample II The variable is categorical III The variable is quatitative IV The populatio size is large V The populatio is ormally distributed VI The sample size is sufficietly large VII The samplig distributio of pˆ is approximately ormal a) I, IV, ad VII b) I, II, ad VII c) I, III, ad VI d) I, IV, V ad VI

14 ANSWERS 1. C. The formula for the cofidece iterval for a populatio mea is: x t s, which was based o the sample Mea. So, x is guarateed to be i the iterval you form. 2. 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 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 B. For 95% cofidece, z = For a margi of error of 0.5, we have 2 2 z* s = m = 1.96*10 = So, the sample size should be (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 = 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 This correspods to - z= Thus, z=1.81. I short look up (1-0.93)/2=.035 i the middle of the table ad z is the absolute value of the z- score.

15 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). 12. 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, 4.20). 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 usual 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 2.5. Draw the picture z 7.09 The probability greater tha 7.09 is a very small umber almost zero

16 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 A. z= =-.61. Look the table A, the probability of beig less tha -.61 is / 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 20. 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 = So the cofidece iterval is * 200 =7.1.69=(6.41, 7.79) 21. D. The defiitio of cofidece iterval. We are 95% cofidet that the ukow populatio mea work hours lies betwee 6.82 ad 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. 22. 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%. 23. 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. 24. 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 = a. T. Sice the populatio has a ormal distributio, we ca use the Normal table for the probability that oe perso is more tha 200 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.

18 p ( 1 p) 29. D pˆ ~N( p, ) whe values of, p satisfyig p 15 ad (1-p) 15. p =0.06*100=6 < 15 So, the distributio of the sample proportio is ukow. p ( 1 p) 30. C pˆ ~N( p, ) whe values of, p satisfyig p 15 ad (1-p) 15. p =0.06*300=18 > 15 ad (1-p)=300*0.94= So, pˆ ~N( 0.06, 0.06(0.94) 300 ) pˆ ~N( 0.06, ). 31. A Use the samplig distributio of the sample proportio that you used i problem 52 ad the z-score. z 2.90 Look up 2.90 i the table. P(z< 2.90) = We wat the probability that our sample proportio is greater tha 0.10 so we take = ( z) p(1 p) (1.96) (.04)(.96) 32. B. = = 2 2 m.01 So should be 1476 = B. ˆp == X =40/800=0.05 estimate of the stadard error of ˆp = pˆ(1 pˆ) = 0.05*(0.95) 800 From the z table, we fid the value of z to be = So the CI is ˆp z*se pˆ =.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. pˆ =X/=287/350=.82 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 287. D. The total sample size is 350.

20 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 pˆ ) = 3 < 15. Therefore, we ca compute the cofidece iterval usig the large sample formula if we add 2 successes ad 2 failures. The 17 2 pˆ pˆ 1 pˆ se (.7917)(.2083) ad the resultig 95% cofidece iterval is p ˆ 1.96( se) (.629,.954). 47. C The oly correct statemet is the first oe --The large-sample cofidece iterval formula for proportios is valid if p 15 ad (1-p) 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 2 successes ad 2 failures. 48. B First, we use a calculator to fid that the sample stadard deviatio s = The se s A The 95% cofidece iterval for the populatio mea is x t. 025 se. I this particular problem, we have x se Usig df = 1 = 4, we look up (i a table) that t.025 = The our cofidece iterval is x t se (462.3, 669.7) 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.

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