Question

Simple random sampling question:

The revenue (Y ) for 1,000 food producing establishments in a country may be described by an exponential distribution with parameter 1/(12*10 ^ 6)  , use seed=1999.

a. Generate the 1000 revenues and retain the integer portion.

b. Obtain a histogram for the data in (a).

c. Select a simple random sample of size 30 from this population (use seed = 2008).

d. Using this sample, estimate the average and total revenue and the associated variance estimates. (use SAS or R)

e. Construct 95% confidence intervals for the mean and total revenue of the population. (use SAS or R)

f. For this type of distribution, why is a simple random sampling not the best design to yield a precise estimate? Suppose there was an auxiliary information prior to the sampling design, propose an alternate sampling design that can lead to a more precise estimate. Explain how you could do this

Answer 1

ANSWER:-

set.seed(1999)
x = rexp(1000,rate = 1/(12*10 ^6))
x= as.integer(x)

4e-08 - 3e-08 2e-08 1e-08 0е+00 8e+07

histogram(x)

set.seed(2008)
sampl = sample(x,30,replace = F)

avg = mean(sampl)
avg
[1] 13628350
variance = var(sampl)
variance
[1] 1.88232* 10^14

But for exponentially distributed data,

mean = 1/lambda = 12*10 ^6 = 12000000

variance = 1/lambda^2 = 1.44*10^14

The 95% confidence interval is obtained as;

quantile(sampl,c(0.025,0.975))
2.5% 97.5%
956513 42051242

The sample doesn't effective capture the mean and standard deviations for exponential distributiuons. This is because data is skewed and a simple random sampling is preffered for uniform or symmetrical data.