{"id":40055,"date":"2024-04-26T23:05:52","date_gmt":"2024-04-26T23:05:52","guid":{"rendered":"http:\/\/localhost\/branding\/statistic-project-sampling\/"},"modified":"2024-04-26T23:05:52","modified_gmt":"2024-04-26T23:05:52","slug":"statistic-project-sampling","status":"publish","type":"post","link":"https:\/\/sheilathewriter.com\/blog\/statistic-project-sampling\/","title":{"rendered":"Statistic Project, Sampling"},"content":{"rendered":"

Student\u2019s Name<\/p>\n

Instructor<\/p>\n

Subject<\/p>\n

Date of Submission<\/p>\n

Statistic Project, Sampling<\/p>\n

11. Population = 828 claims<\/p>\n

a.) SRS \u2013 sample 85 each with 215 fields<\/p>\n

Claims1142257<\/p>\n

Error43210<\/p>\n

fiyi438220<\/p>\n

we estimate mean(Y) with mean(y)<\/p>\n

n= \u2211fi = 10, \u2211fiyi=37<\/p>\n

mean(y) = \u2211fiyi\/\u2211fi = 37\/10 = 3.7<\/p>\n

The standard error mean(y) is\u03b4y= [var mean(y)]1\/2= [S2\/n( 1 – f)]1\/2<\/p>\n

S2= 1\/(n-1) \u2211[yi- mean(y)]2<\/p>\n

S2= 1\/9 * (523 \u2013 372\/10) = 42.9<\/p>\n

Estimated \u03b4y=0.45289429<\/p>\n

b.) Estimated (Y) = Ny= 3.7 * 828 = 3063.8<\/p>\n

standard error of Estimated Y = N \u03b4y = 828 * 0.45289429<\/p>\n

c.) Sample \u2013 18275 fields<\/p>\n

pop = 178,020 fields<\/p>\n

Number of errors in sample = 10<\/p>\n

Mean(y) = 3.7<\/p>\n

14. <\/p>\n

a.) School1234<\/p>\n

Smokers female729457511800<\/p>\n

Smokers<\/p>\n

Smokers105627<\/p>\n

n= \u2211fi = 46, \u2211fiyi= 1495, mean(y)= 32.5<\/p>\n

S2=1\/(n-1)[\u2211fiyi2 – \u2211(fiyi)2\/\u2211fi]<\/p>\n

=1\/99[ 52525 – 14952\/46] = 39.7727273<\/p>\n

b.) The (1-\u03b1)100% confidence interval of mean(y) is:<\/p>\n

mean(y) \u00b1Z \u03b1\/2* S\/n1\/2[(N- n\/N]1\/2<\/p>\n

\u2211f = 46, \u2211fiyi=1495, \u2211(fiyi)2, N=2550, n=100 , S=6.30656224.<\/p>\n

32.5 \u00b1 Z 0.025* (6.365224\/1001\/2)[(2550 – 100)\/2550]1\/2<\/p>\n

32.5 \u00b1 1.96* (6.365224\/1001\/2)[(2550 – 100)\/2550]1\/2 = 33.72287683<\/p>\n

c.) The 100 (1 – \u03b1)% C.I for the population total<\/p>\n

Nmean(y)\u00b1Z \u03b1\/2* Ns\/n1\/2 *[(N-n) \/N]1\/2<\/p>\n

2550 \u00b1 1.96 * (255 * 6.30656224) \/(100)1\/2) [(2550 – 100)\/2550]1\/2= 2709<\/p>\n

16.<\/p>\n

> school1=read.table(“c:\/xyz\/data.csv”,sep=”,”,header=T)<\/p>\n

> school1<\/p>\n

returnf<\/p>\n

1 1<\/p>\n

2 1<\/p>\n

3 1<\/p>\n

4 0<\/p>\n

5 1<\/p>\n

6 9<\/p>\n

7 1<\/p>\n

8 1<\/p>\n

9 0<\/p>\n

10 0<\/p>\n

11 1<\/p>\n

12 0<\/p>\n

13 0<\/p>\n

14 1<\/p>\n

15 0<\/p>\n

16 1<\/p>\n

17 0<\/p>\n

18 0<\/p>\n

19 1<\/p>\n

20 0<\/p>\n

21 0<\/p>\n

22 9<\/p>\n

23 0<\/p>\n

24 0<\/p>\n

25 0<\/p>\n

26 1<\/p>\n

27 0<\/p>\n

28 0<\/p>\n

29 0<\/p>\n

30 1<\/p>\n

31 1<\/p>\n

32 0<\/p>\n

33 1<\/p>\n

34 1<\/p>\n

35 0<\/p>\n

36 0<\/p>\n

37 1<\/p>\n

38 1<\/p>\n

39 1<\/p>\n

40 1<\/p>\n

a.) >sum (school1)<\/p>\n

[1] 37<\/p>\n

Percentage of parents who returned the forms: 37\/78 *100 =47.44%<\/p>\n

>sum (read.table(“c:\/xyz\/school2.csv”,sep=”,”,header=T))<\/p>\n

[1] 37<\/p>\n

Percentage of parents who returned the forms: 37\/238 *100 =15.54%<\/p>\n

>sum (read.table(“c:\/xyz\/school3.csv”,sep=”,”,header=T))<\/p>\n

[1] 31<\/p>\n

Percentage of parents who returned the forms: 31\/261 *100 =11.88%<\/p>\n

>sum (read.table(“c:\/xyz\/school4.csv”,sep=”,”,header=T))<\/p>\n

[1] 18<\/p>\n

Percentage of parents who returned the forms: 18\/174 *100 =10.34%<\/p>\n

>sum (read.table(“c:\/xyz\/school5.csv”,sep=”,”,header=T))<\/p>\n

[1] 48<\/p>\n

Percentage of parents who returned the forms: 48\/236 *100 =20.34%<\/p>\n

>sum (read.table(“c:\/xyz\/school6.csv”,sep=”,”,header=T))<\/p>\n

[1] 22<\/p>\n

Percentage of parents who returned the forms: 22\/188 *100 =11.70%<\/p>\n

>sum (read.table(“c:\/xyz\/school7.csv”,sep=”,”,header=T))<\/p>\n

[1] 24<\/p>\n

Percentage of parents who returned the forms: 24\/113 *100 =21.24%<\/p>\n

>sum (read.table(“c:\/xyz\/school8.csv”,sep=”,”,header=T))<\/p>\n

[1] 84<\/p>\n

Percentage of parents who returned the forms: 84\/170 *100 = 49.41%<\/p>\n

>sum (read.table(“c:\/xyz\/school9.csv”,sep=”,”,header=T))<\/p>\n

[1] 50<\/p>\n

Percentage of parents who returned the forms: 50\/296 *100 =16.89%<\/p>\n

>sum (read.table(“c:\/xyz\/school10.csv”,sep=”,”,header=T))<\/p>\n

[1] 43<\/p>\n

Percentage of parents who returned the forms: 43\/207 *100 =20.77%<\/p>\n

c.) > sum (read.table(“c:\/xyz\/consent.csv”,sep=”,”,header=T))<\/p>\n

[1] 339<\/p>\n

Percentage of parents who returned the forms: 339\/9962 *100 =3.40%<\/p>\n

0.95 *339= 322<\/p>\n

b.)<\/p>\n

The procedure is as follows:<\/p>\n

\u2022 The weights wi are the inverses of the selection probabilities \u03c8i.<\/p>\n

\u2022 The weighted estimator of the population total is 1st \u03c8 = \u2211witi.<\/p>\n

\u2022 We calculate \u03c8 (estimate) for each.<\/p>\n

Sample n=18275 pop N=178020Var(Y) =(N2S2\/n)*(N-n)\/n<\/p>\n

9.)<\/p>\n

Procedure<\/p>\n

– Suppose the number of samples, n is greater than 1 and we sample with replacement.<\/p>\n

-This implies \u03c0i = 1\u2212 (1 \u2212 \u03c8i)n<\/p>\n

-The probability that an item i is selected on the first draw is the same as the probability that item i is selected on any other draw.<\/p>\n

-Sampling with replacement gives us n independent estimates of the population total, one for each unit in sample.<\/p>\n

-We average these n estimates.<\/p>\n

-Estimated variance is variance of the estimates divided by n<\/p>\n

-N = 52 classes of states in the USA<\/p>\n

– Mi students in class i (i = 1 to 52)<\/p>\n

– Values of Mi range from 1 to 3142.<\/p>\n

-We want a sample of 10 states.<\/p>\n

-In this case \u03c8i=Mi\/3142<\/p>\n

units size Cumulative size Y=Population 1<\/p>\n

2<\/p>\n

3<\/p>\n

4<\/p>\n

5<\/p>\n

6<\/p>\n

7<\/p>\n

8<\/p>\n

9<\/p>\n

10<\/p>\n

.<\/p>\n

.<\/p>\n

.<\/p>\n

52 67<\/p>\n

25<\/p>\n

15<\/p>\n

75<\/p>\n

58<\/p>\n

63<\/p>\n

8<\/p>\n

3<\/p>\n

1<\/p>\n

67<\/p>\n

.<\/p>\n

.<\/p>\n

.<\/p>\n

159 67<\/p>\n

92<\/p>\n

107<\/p>\n

182<\/p>\n

240<\/p>\n

303<\/p>\n

311<\/p>\n

314<\/p>\n

315<\/p>\n

382<\/p>\n

.<\/p>\n

.<\/p>\n

.<\/p>\n

3142 Select a random number R between 1 and (TN) =52 by using random number table. <\/p>\n

4137511<\/p>\n

587766<\/p>\n

3832368<\/p>\n

2394253<\/p>\n

30895356<\/p>\n

3464675<\/p>\n

3279116<\/p>\n

690884<\/p>\n

585221<\/p>\n

13482716<\/p>\n

464736<\/p>\n

If Ti-1\u2264R\u2264Ti, then the ith unit is selected with probability Xi\/52,<\/p>\n

i = 1, 2,\u2026, 52. <\/p>\n

Repeat the procedure 10 times to get a sample of size 10. <\/p>\n

First Draw: Draw a random number between 1 and 3142.<\/p>\n

Suppose it\u2019s 167<\/p>\n

T3\u2264132\u2264T4, Unit Y is selected and Y4 = 2394253 enters in the sample.<\/p>\n

2. Second Draw: Draw a random number between 1 and 64 <\/p>\n

Suppose it is 308 <\/p>\n

T6< 38 < T7 , Unit 7 is selected and Y7 = 3279116<\/p>\n

Enters in the sample and so on. <\/p>\n

This procedure is repeated till the sample of required size is obtained. <\/p>\n

10.) <\/p>\n

units size Cumulative size Y=Population<\/p>\n

1<\/p>\n

2<\/p>\n

3<\/p>\n

4<\/p>\n

5<\/p>\n

6<\/p>\n

7<\/p>\n

8<\/p>\n

9<\/p>\n

10<\/p>\n

67<\/p>\n

25<\/p>\n

15<\/p>\n

75<\/p>\n

58<\/p>\n

63<\/p>\n

8<\/p>\n

3<\/p>\n

1<\/p>\n

67<\/p>\n

67<\/p>\n

92<\/p>\n

107<\/p>\n

182<\/p>\n

240<\/p>\n

303<\/p>\n

311<\/p>\n

314<\/p>\n

315<\/p>\n

382<\/p>\n

4137511<\/p>\n

587766<\/p>\n

3832368<\/p>\n

2394253<\/p>\n

30895356<\/p>\n

3464675<\/p>\n

3279116<\/p>\n

690884<\/p>\n

585221<\/p>\n

13482716<\/p>\n

Works Cited<\/p>\n

Chambers, John M.\u00a0Software for Data Analysis: Programming with R. Berlin: Springer New York, 2008. Print.<\/p>\n

Gardener, Mark.\u00a0Beginning R: The Statistical Programming Language. Indianapolis: John Wiley & Sons, 2012. Print.<\/p>\n

Gentleman, Robert.\u00a0R Programming for Bioinformatics. Boca Raton: CRC Press, 2009. Print.<\/p>\n

Matloff, Norman S.\u00a0The Art of R Programming: Tour of Statistical Software Design. San Francisco: No Starch Press, 2011. Print.<\/p>\n","protected":false},"excerpt":{"rendered":"

Student\u2019s Name Instructor Subject Date of Submission Statistic Project, Sampling 11. Population = 828 claims a.) SRS \u2013 sample 85<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[],"tags":[],"class_list":["post-40055","post","type-post","status-publish","format-standard","hentry"],"yoast_head":"\nStatistic Project, Sampling - sheilathewriter<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/sheilathewriter.com\/blog\/statistic-project-sampling\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Statistic Project, Sampling - sheilathewriter\" \/>\n<meta property=\"og:description\" content=\"Student\u2019s Name Instructor Subject Date of Submission Statistic Project, Sampling 11. Population = 828 claims a.) 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