Answers HW-7 MATH2300-18Spring2011

Answers HW-7 MATH2300-18Spring2011 - MATH 2300 Home work 07...

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Unformatted text preview: MATH 2300 Home work # 07 Answers Is the observed sample proportion unusual? A 1) Based on previous studies, researchers believe that 6% of children are born with a gene that is 1) linked to a certain childhood disease. If the researchers test 950 newborns for the presence of this gene, would it be unlikely for them to find fewer than 25 children with the gene? Answer by calculating the appropriate z—score. YES, Z = -4.41 5 0“” en emvo‘t ‘J-Qm \AaS 0‘ 3“,.» Pl 1% $l5'\;v '. kg," 16... B) No, 2 = —4.41 6.1 ; u, '6 C) NO,Z=-0.005 Mean 2 ? : 0 tag D) No,z=—6.59 _____._____. E) Yes,z=-6.59 5- e, :- JM 9) +03 “90’ ° ‘3‘“) = 0 007403 V‘ 359 Pepi/Julio.” ?Tp\’)m\!w : p P ,— 62 :o'ol: Calcytfik ~2— scam, Sanvle 3:59 = “‘350 :2: "lg/5579* am, 2 “ ”632. 01:07:”? gauplg QTDVMXLGAZ £5— 5S0 (ll/its {5 unllkmlg . baton/(Se, ‘2 Value ‘5 leSS Mu ' 3 ' ’t 1,, it” Amsuu Chow/23 A l g a Volt/u shout; be -» $32 Find the mean/standard error of the sampling distribution of the proportion. A 2) A candy company claims that its jelly bean mix contains 21% blue jelly beans. Suppose that the 2) candies are packaged at random in small bags containing about 400 jelly beans. Describe the sampling distribution model of the proportion of blue jelly beans in a bag mean = 21%; standard error = 2.0% xx i (go a. low on V ) mean = 21%; standard error = 0.8% 30‘ \9\:.~a d S i C) mean = 79%; standard error = 0.8% 3“ ”9U l)” 9‘” o A D) mean: 79%; standard error=2.0% we”. : P 2 0 Zt ; 2.} / E) There is not enough information to describe the distribution. . 1r 2\ I 02! W?“\V\Lo-« Pmt’w’me =l>=2"/.‘°’Z~‘ 5 Q rail“ 0 :10 (+00 ) 30w \2 5? e. x V\: 00 o 0203. 9 E + 2 2. 032 Provide an appropriate response. 3) Which of the following are true: 3) B x I) the population distribution is approximately normal for large enough n x 11) the sampling distribution of ; is approximately normal for any n \/ III) regardless of the population distribution, the mean of the sampling distribution of § equals the population mean A) II only ‘lll only C) I only D) all of these 7) True or False: A sampling distribution is a distribution for a statistic. Provide an appropriate response. 7) True, 8) In one region, the September energy consumption levels for single—family homes had a mean of 8) 3) 1050 kWh and a standard deviation of 218 kWh. Describe the center and spread of the sampling distribution of the sample mean for a random sample of 50 single—family homes from this region. A) center = 1050, spread = 92.49 B) center = 1050, spread = 61.66 _ , C) center = 1050, spread = 4.36 55w \3\ W83 d is Ah“: bu \n o.» a f; M @center = 1050, spread = 30.83 E) center = 1050, spread = 218 Savflfil W‘ C9“ ”£1050 Mann =/(A: (OYO Stag-g :1— ____:30‘823. “:60 I64” J50 9) The gestation time for humans has a mean of 266 days and a standard deviation of 25 days. If 100 women are randomly selected, find the probability that they have a mean re an between 266 days and 268days. A) 0.2119 B) 0.7881 C) 0.5517 @2881 : 25 :1 if ‘1 “:3? :rzwitz“) on i be 0.. W“ 0‘0““ \) c 9 (o < z < 0.8) 37W SomYV. 03 um Home» ; r ?(2 40’8)’ Péég .. :5: M f i a “0mm5/A’J ZLL :: O'ZSSI a b-g fl 9)CD 4) In one region, the September energy consumption levels for single-family homes had a mean of 4) B 1050 kWh and a standard deviation of 218 kWh. If 50 different homes are randomly selected, find the o robabili J -. eir mean ener consum tion level for Se utember is - eater than 1075 kWh. A) 0.0438 C) 0.2910 D) 0.4562 E) 0.4180 wa‘ax‘w “(An =/‘4 :: H350 Papalslmn S'FD. f. V“: 2'8 F (i > 107,15) : P 1;: >IO?{-\IOSO 7 5'¢ 30' :23 EMMY“ 855:. : v\ = 50 :{362 > 08!) Le’& '71 be “AL Mam enevaj COMSMM? kfiO-A \Q¢L\ &u( 50 howeg‘ «w L: 1.. PC%CO”E') ° M" a a s3 anknl \zmt Mame...) 'x = "0 ‘0 “t?*"""‘“"‘ *W (1 WM‘ = o' 2030 Arstvshak‘. On N '. 30. Wm“ : fl T IOSD 5.9.. = \2: £525.. = gonna, / 13 [so / Find the mean/standard error of the sampling distribution of the proportion. 3 5) A realtor has been told that 43% of homeowners in a City prefer to have a finished basement. She 5) surveys a group of 300 homeowners randomly chosen from her client list. Describe the sampling distribution model of the ro ortion of homeowners in this sample who prefer a finished bastflgng, r r \' gm 0" A) mean = 57%; standard error = 1.4% 55"“ \3‘ ""6 A" \r'v’” w B) mean = 43%; standard error = 1.4% ' Saw pin. 9”?“ km" C) mean = 57%; standard error = 2.9% o ‘ 3 / m : 3 :‘ i @mean = 43%; standard error = 2.9% N (a C P 4’ 4t— E) There is not enough information to describe the distribution. 3 , t : ”)6 -f) :JC” 1’3 6‘ bi + 3) 3 co P ; (f3 '/ t 0 ' 4‘. .2 M __ 0 0 , o 2858 . 7") r . . 300 :2'858X" 2.3/‘ Provide an appropriate response. . E 6) As the sample size increases, which of the following are true: 6) v’ I) the sample mean tends to fall closer to the population mean x 11) the sampling distribution of the sample mean ; approximates the population distribution \/ III) the standard error decreases r . M ‘- 3:5 “CCU-VAN” 0“ A) allofthese L5 “AL 56‘ Vb“ 96¢ mcrca ’ d C2fl'\xx\ L‘u—dk Md?“ ) 55wp‘h46 AlS‘n‘le*ibA 08 B) II only M \ l 3"“ h ”“1" ~ 1...». “b: a WWW C) In only V l “WY” .453».sz at. D)Ionly . (x ’ v? \ E bothIandIII ’ ’ ll. v3 ’ 1‘ 9%. 0‘5 9am?L¢_1~ 2 m 508 s & 5‘3. tnCwasas. S4. Azermszs. .'. 11: (S )‘YML4 10) Assume that blood pressure readings have a mean of 120 and a standard deviation of 8. If 100 10) g 2 people are randomly selected, find the probabilig that their mean blood Eressure will be geater was A) 0.8819 B) 0.9938 @00062 D) 0.8615 :120 4:: 8' Pfiw”) V\ :/00 Le x 31 bt M MCPW‘ bloc; Q‘VCSSuN. Orr Saw‘iu, : P (‘2 7 Z'QD en la . 08 too P ? z‘,o(‘£<2-S) W‘Ae {$5 5LT) 7L 0‘?me 0 2'5 b9 0 man angled” m s l— o. 3638 (,9. Ma : meant/L4,“: '20 ‘ 0.0062 Set :' 2:: ;:O‘& J v! J Ian 11) Assume that the heights of adult Caucasian women have a mean of 63.6 inches and a standard 11) C deviation of 2.5 inches. If 100 women are randomly selected, find the gflfijflflaflheyham mean height greater than 63.0 inches. A) 0.8989 B) 0.0082 @0 9918 D) 0.2881 V“; 2'5 6 L V1 , [00 3 ; 11’” >63;':§ > w LA ( s e. Wine—WA i :?(2>-29 It km jaWYLL ml “’0 “”4" \ C) l ;l/P(%<’2+) «21* 0 f as an, i avvmww" :1' 0.0032 0 novw‘ A‘S'xrll’xeio‘ “‘KL 3 0'3918’ weamr/A: 63‘ MM; 5 0,, :Z:2;£: 0.25» F J7?» 12) Which of the following is true about the sam lin distribution of the sam le mean? ' 12) A /@ The mean of the sampling distribution is always it. X B) The shape of the sampling distribution is W. (Nov) alum”) X C) The standard deviation of the sampling distribution is always a. 4< 5 X-M A m A AEVWAW" 06 1N” ‘J' ‘5 V719? Sarawak cifis‘iwwu"-°* ‘3 X D) All of the above are true. (“lie 3. gabvé at, n r at v) :3 V477 13) Why is the Central Limit Theorem so important to the study of sampling distributions? 13) m X A) It allows us to disregard the size of the sample selected when the population is not normal. x B) It allows us to disregard the size of the population we are sampling from. x C) It allows us to disregard the shape of the sampling distribution when the size of the population is large. /@ It allows us to disregard the shape of the population when n is large. (in :5 M swath stable) Provide an appropriate response. 14) If the proportion of American adults who believe that America is ready for a woman 14) president is 0.70, what are the mean and standard error for the proportion of 129931.333 believe that America is readyigrgmwoman presidentfgr a samplepf size 1000? popu\a\§ow prmlwcv: 03"? gawplimb Am’wibulwv 05 M Saw-Vb. P'r’b VOYV-GV‘ mean 2 P: 6'? 6. e. :’ Q’P) 31093;?) ‘3 O~O\44\—8 55mph 3‘32 = Vt: (000 15) If the proportion of American adults who feel that increases in gasoline prices have caused 15) financial hardship for their family is 63%, what are the mean and standard deviation for the number of people who feel that increases in gasoline prices have caused their family financial hardship for a random sample of 100? P 3 (33/: 0'63 gawvlwxg $5¥flt§u§t06 06 M anw- Yb. YW?“Y \».¢_ mean’ : V—‘sTO‘ 63 5,3. 2 PM?) 2W) ‘1 otoqsze VI:{OO ...
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