This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 1Proportion. 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 SamFit. a? 890 aduli—s) WGL Qr‘e. 95,75 wwpa'éﬂzvx7" 5 OMQ\,A\r\e~r‘e tuth. éeéyb 0m"? MDﬁlZ 0’? QJL (MauiL; 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 YakH—JVM c4 HA1 Use Minitab to test the hypothesis you wrote in question #3. Q
Click on STAT, Basic Statistics, then 1Proportion. Click on the Summarize and chew .
F?” M guardHm *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 ‘llmux 0L, +l4¢r¢ "5' “1—01.—
en: musk 5+c‘i’i'x’s1er/ ﬁVt'ﬁ’cﬂt/‘z. JP: Fad.le, We my.“ HyFleqfS‘Lr (H0)
THU§)"H/\cv~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 index Vail 9 0
(0'019 J HUMMLN‘RU 5": ConQ2239.“ —+ A. Wl‘ddzr Meltr‘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 arrawed“ («tire4“!
9. Write a conclusion statement for the hypothesis test you designed in question #3. Remember to include
your pvalue (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; (JuanMm 43 kt at r\ s: .
10. The pvalue 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—cgir
® the new pvalue would be smaller than 0.141 M QL’Cle'S'vzf a... bf} ef‘. 'Tl\ WJ—l’lqa
P_ fa pwthne. \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 2proportion 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‘tdQnsff
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\FdﬁSi'i‘" “Me: Gore. WM/ﬂf‘cﬁh 4’9""
HA: ‘17! # i7: is \fa a9.
62 w'lou‘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 2Proportions. 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 [email protected] & pvalue ’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 pvalue ‘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. 9wes'fg @A‘ 04=o_o,_§'_'  g ‘
(T‘hcszJa‘tmL pg gmaii er] a. +0 (“a not H b, €14 f ort— '  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 pvalue, 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\ VexSear Co’lumqiu'a. (Lonai7 reﬁtJanus 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__ comml7 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? > (3975” ...
View
Full Document
 Spring '08
 Ripol
 Statistics, Null hypothesis, Statistical hypothesis testing, new confidence interval

Click to edit the document details