Lab2-AnswerKey-WithMinitabOutput

Lab2-AnswerKey-WithMinitabOutput - In apoll of 800 1. ;...

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Unformatted text preview: In apoll of 800 1. ; 0,08” a D C r7ng /? randomly selected adults, 8.5% said they are in favor of the cloning of human beings. How many of these 800 adults feel that cloning should be allowed? Let this represent the number of successes in your sample data. Recall: # of sucCesses = n*p A 6 g X=80o$0.0€r F=§7¢§Z x:ée Assume that all of the necessary assumptions/conditions (page 493 - plausible independence, randomization, 10% condition, success/failure condition) have been met for approximating the distribution of sample proportions (I; ~vaiues) with a normal distribution ' Use Minitab to. find a 95% confidence interval for the proportion of all adults who feel that cloning should be allowed. Click on STAT, Basic Statistics, then 1-Proportion. Since you have summary results (not the individual data), click on the Summarized Data button. Enter the number of successes (# of events = 68) and number of trials (11 = 800) into the appropriate fields. Click the Options button to choose the appropriate confidence level (95%). Choose NOT EQUAL for the Alternative and click the check box for Use Test and Interval Based on a Normal Distribution. Click OK ie. he confidence interval results should now be in the session window. Did you go 0.065675, 0.104325) ‘ Interpret the meaning of your confidence interval in the context of the given scenario. T 4,05 gafzr/OM 4i/V‘S rfi‘r‘flom Sam-Fit. a? 890 aduli—s) WGL Qr‘e. 95,75 wwpa'éflzvx7" 5 OMQ\,A\r\e~r‘e tut-h. éeéyb 0m"? MDfilZ 0’? QJL (Maui-L; 00:. \‘q Qawor 35‘? damt‘nj a? ‘AMVWQR befiflfis‘. A newspaper re orter has seen this Gallup poll information and mites an article with the following headline: ore Than 7.5° f Adults Favor Cloning”. Write an appropriate null and alternative hypothesis to determine w ether or ' is appropriate for a newspaper to run the article. (2 Yak-H—JVM c4 HA1 Use Minitab to test the hypothesis you wrote in question #3. Q Click on STAT, Basic Statistics, then 1-Proportion. Click on the Summarize and chew . F?” M guard-Hm *5] ea 0C; Onto 5. Based on your 21pr test result (p—value 1" 0.141), is it appropriate for .1 - newspaper to run the article? Use a 10% level of significance. Give a reason, not just a YES ‘1 4 swer. -)\Io gimme: 4dr“- ‘p—nVeAvm. 1's lm' lucr- ‘l-lmux 0L, +l4¢r¢ "5' “1—01.— en: musk 5+c‘i’i'x’s-1er/ fiVt'fi’cflt/‘z. JP: Fad.le, We my.“ HyFleqf-S‘Lr (H0)- THU§)"H/\c-v~e_ X'S Ale'é—l—r‘aij canOfiM @vl'a/mm 'l’b S'MP br‘l' +1214» mli—WM’J‘M- laypa—Urxfl/lfigl 6. The 95% confidence interval found previously is (0.066, 0.104) The width of the interval is 0.104 ~ 0.066 = 0.038 Would a 90% confidence interval have the same width, be wider, or ban the 95% interval? fl) 0|) louder Laval: 9% COA‘G’Jenaa —% A— (\ Od'f‘fi‘l v.3 ind-ex Vail 9 0- (0'019 J HUMMLN‘RU 5": Con-Q2239.“ —-+ A. Wl‘ddzr Mel-tr‘v‘ad 7. Knowing that the width of the 95% confidence interval is 0.03 8, find the margin of error of the interval. j—lfic MMji'n b'e'af‘f‘bf‘ {‘5’ Vs: 09744:: k4)\‘a(+{\ Mt; M «J m E =- 3;me _ miflia. 0"“? ' ' 9+E—u——'9——r (14.5-: 0.01? meal, “5&6 6.50% 8. What would happen to the width of the 95% confidence interval if thgfl'aisters had doubled the sample size to 1600 adults and found the same proportion of successes in this larger sample? _ r“ orch {y ¥VVLS “M983 51.22... leeKAQOE ‘ the new confidence interval would have a width of less than 0.038 MM {A 5-? esp—r as r ‘—? C. the new confidence interval would have a width of more than 0.03 8 r as Uflffi \n‘ 6" a} ' n arr-awed“ («tire-4“! 9. Write a conclusion statement for the hypothesis test you designed in question #3. Remember to include your p-value (0.141), a correct decision about whether or not there is enough evidence to support HA, and to restate the alternative hypothesis (HA) in words. NFL“ Gt —l>—V°Jl‘k£ 6‘? OHM“ %X\cho.Ma\e ell aim el__L_)_cj-_s / pPaV't’épe. $117“an elm (“43k Mes/[jamae ‘l—o (ignol‘xgc “fl/(fl MOCEFWGLW 7- 5—2, Q? a“ gaunt“ 48A; (Juan-Mm 43 kt at r\ s: . 10. The p-value from the hypothesis test was found to be 0.141. What would happen to t fine if till) Gallup pollsters had increased the sample size to 1600 adults and found the same proportion of A. the new confidence interval would have the same width of 0.038 successes in this larger sample? A—gg‘um {719 Q} J .3 cagn ‘WL an at? ct, A. the new p—value would remain equal to 0.141 ' l‘(\ Weaotgi 49:4 a, S WW 5‘ 12¢(n B. the new p»value would be larger than 0.141 (p Q\-L\<Q "Hm; i—cg-ir ® the new p-value would be smaller than 0.141 M QL’Cle'S'vzf a... bf} ef‘. 'Tl\ WJ—l’lqa P_ fa pw-thne. \ms legal ofja'l—SMQ/kafi _ M... Z‘ ff?" 2.. ii. In this lab, you failed to reject Ho (7.5% oigeople favor cloning) — thus failing to support HA. What if more than 7.5% of people really do favor cloning (i.e. HA really is true . If this were the case, which type of error did you make? Type I or e II 5.90 or e; we to «flare/Ag; "We, is».an a 5.1/3" 1, pk ' (,Q Stream“ W.a'lc¥+ b ' “510.1512. (7 2 .13. Ina study of red! green color blindness, 1300 Alachua county residents and 1100 Columbia county residents are randomly selected and tested. Among the Alachua county residents, 120 have red] green color blindness. Among the Columbia county residents, 140 have red! green color blindness. Assume that all of the necessary assumptions/conditions (page 560) have been satisfied for using a 2 proportion z—interval and a 2-proportion hypothesis test to compare the populations of Alachua county and Columbia county residents. A Success” 1‘5 an M 3’ Wm? M9 WVAAA red 3 We“ 0" 5“ HW‘tdQnsf-f 12. Is there a difference in the percentage of red/ green color blindness between he residents of these two count'eg? Write an appropriate null and alternative hypothesis. If you use subscripts of 1 and 2, be sure to efine which groups the 1’s represent and which group the 2’s represent. 3 7“ I i7 “TM; (‘9’ W‘i'l’éfi if? Aiatiqxw tank-“41‘s H0 P. P2 \ F‘s-\Fdfi-Si'i‘" “Me: Gore. WM/flf‘cfih 4’9"" HA: ‘17! # i7: is \fa a9. 62 w'l-ou‘i at? Fflz’flqg Fretbat'i'rén ‘9’ch dbiul‘eifira MVfl'l'?’ "-- 13. Use Minna to test the hypothe ' you wrote in question #12, as well as to create a 99% confidence interval for describing the difference between the populations. Click on STAT, Basic Statistics, then 2-Proportions. Click on the Summarized Data button. Enter 120 for the first # of events and 1300 for the number of trials. Enter 140 for the second # of events and 1100 for the number of trials. Click the Options button & enter 99 for the confidence level and 0.0 for the “Test Difference” value. Choose “not equal” for the alternative hypothesis. Do NOT check the checkbox for pooling. Click OK twice. The hypothesis test results should now be in the session window. w \3 - Did y’op g@ & p-value ’1 $5 , P _ [£29 g :15}; (Mfg; i"— in UN) I Q. I A A -\’ *-o.n391?§, ‘ 0 Pt 2. o , D923} :10- !27‘27 (Q: 14. Based on the p-value ‘of the hypothesis test, should you reject the null hypothesis (H0), or should you not reject the null hypothesis? Assume a 5% level of significance. 9-wes'fg- @A‘ 04=o_o,_§'_' - g ‘ (T‘hcszJa‘tmL pg gmaii er] a. +0 (“a not H b, €14 f or-t— ' - if b‘i’L‘V- fps—i" F15 Units are 5 ah'sdhml lg 15. Write a conclusron statement for the hypothesrs be! t you d Slgned 1n question #12. Remember to g 1'5 ‘ ' include your p-value, a correct decision about whether or not there is enough evidence to support HA, and to restate the alternative hypothesis (HA) in words. ngii’A & ?—V4\M=. ,9? b. 0257} +03% SquKQ priming/:5 Very SULNH e—Vl'apQAci— weer himth is (L a°\'-izg®f‘cflc.e-_. in like. yef‘rfic—nf “‘9 A\“J?\u~ac CGKhW“F\- Vex-Sear Co’lumqiu'a. (Lona-i7 refit-Janus KARO SWQQM‘“ 4N)“ VertI/Socmn Cnidar- Biffiéfncxsz 16. Afier rounding, the confidence interval that you found in question #13 should be (-0.068, 0002). Interpret the meaning of this confidence interval in the context of the given scenario. FMJm “i475, Sawfic. 69ka we. Cd“: 9 9?,fi CO n—€,~¢fen+" fics+——ie/\c Wmfl'f'cj’fi— 5"? Ebb. (5(\O—c}(\\fi6k__ comm-l7 h:5f.ei/cn+f Lulu, are Fccfizprcan Lea/6r“ .bimJ 1'5 heal—M4 men 0. 3‘70 and J, 87., ifimer‘ 4&1 “ %g cg“ aver; rc__ '9‘? 9B C29 hm‘i- (“CS 'LJ. W A P \T ,7 3 Kevin Kasper's Answer Key % . “£37 1° Test and CI for One P1032506 4/— Sample X N Sample p 95% CI 1 68 000 0.005000 (0.065675, 0.104325) Using the normal approximation. Test and CI for One Proportion é———-—-" Test of p = 0.075 vs p > 0.075 Sample X N Sample p 1 68 800 0.085000 Using the noiTEl/agproximation. 08S” 137,0. ‘ Test and CI for Two Proportions Sample X N Sample p r» __ \ 0 1 120 1300 0.092308 F“ ’F‘ " Tch 2 140 1100 0.127273 2 Mom Difference = p (l) — p (2) Estimate for difference: :799% CI for difference: Test for difference = Wat test: \ //V Iflfier e. WVS'V adv“ —0.0349650 (—0.0680951, 0 (vs not m 0 : ) = 9.007 ._— ._—. Hat. 5? 0,575" HA ‘- J? > (3-975” ...
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This note was uploaded on 11/22/2010 for the course STA 2023 taught by Professor Ripol during the Spring '08 term at University of Florida.

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Lab2-AnswerKey-WithMinitabOutput - In apoll of 800 1. ;...

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