SIMPLE RANDOM SA 36 shading in Figure 2.4. List the sample data. Use the sample
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Question
SIMPLE RANDOM SA 36 shading in Figure 2.4. List the sample data. Use the sample to estimate (b) Repeat part (a), selecting another sample of size 10 by simple sampling (without replacement) and making new estimates. Indie positions of the units of the samples on the sketch. (c) Give the inclusion probability for the unit in the upper left-hand com many possible samples are there? What is the probability of selectin the sample you obtained in part (a)? 2. A simple random sample of 10 households is selected from a population 100 households. The numbers of people in the sample households are 2,5. 4, 4, 3, 2, 5, 2, 3 (a) Estimate the total number of people in the population. Estimate the variance of your estimator. (b) Estimate the mean number of people per household and estimate the vari. ance of that estimator. 3, Consider a small population of nits, labeled 1, 2, 3 with respective values 3, 1, 0, 3. Consider a simple random sampling design with a sample si n E3 For your convenience, several parts of the following may be combined into a single tablet (a) Give the valu of the population parameters t, and a3. List every C that it is sample of size na3 For each sample, what is the probability the one selected? e mons that the sample ean is unbiased for the population and determine whether the sample median is for population median the m the ng us in amp ing lect le per aking d mi what (a) list o Ho What (c) (d) Estin Esti (e) How 6. Repeat (a) UsExplanation / Answer
Given,
Sample size(n)=10
Population(N)=100
sample values 2,5,1,4,4,3,2,5,2,3
A.a) to find,
population and variance of estimator
sample mean(X)= (2+5+1+4+4+3+2+5+2+3)/10
sample mean(X)=3.1
Population total=100*3.1
Population total =310
SD for sample values 2,5,1,4,4,3,2,5,2,3 is
SD(sample)=1.37
Standard error of mean=S/sqrt(n)
SE=1.37/sqrt(10)
SE=0.43
Variance=0.43*0.43=0.19
Hence variance of estimate is 0.19
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