Question

Suppose that a population of 540 elementary units is divided into 3 strata and the samples selected from the strata are presented in the table below. Stratum 1 Stratum 2 Stratum 3 X 11= 6 X21 = 9 X31 = 20 X12 = 10 X22 = 18 X32 = 26 X13 = 2 X23 = 12 X33 = 16 X14 = 4 X24 = 13 X34 = 26 X15 = 8 X35 = 22 X16 =12 a) Estimate the mean of the population. b) Estimate the variance of the estimator. c) Find a 95 % confidence interval about the mean. d) By considering all of them as a simple random sample, estimate the standard error. e) Compare the variance from the stratified sampling with the variance from the simple random sample and comment on your answers.

Answer 1

a)

$\mu=\frac{\sum x_i}{N}=\frac{6+9+20+10+18+26+2+12+16+4+13+26+8+22+12}{15}=13.6$

b)

$Var(X)=\frac{\sum(x_i-\mu)^2}{N}=$

$=\frac{7.6^2+4.6^2+6.4^2+3.6^2+4.4^2+12.4^2+11.6^2+1.6^2+2.4^2+9.6^2+0.6^2+12.4^2+5.6^2+8.4^2+1.6^2}{15}=$

$=\frac{799.6}{15}=53.31$

c)

$\mu\pm Z_{0.95}\frac{\sigma}{\sqrt{n}}=13.6\pm 1.96\cdot \frac{\sqrt{53.31}}{\sqrt{15}}=13.6\pm 3.845$

confidence interval:

$13.6-3.845<\mu<13.6+3.845$

$9.755<\mu<17.445$

d)

$SE=\sigma/\sqrt{n}=\sqrt{53.31/15}=1.885$

e)

the variance from the simple random sample:

$Var_1(X)=\frac{\sum(x_i-\mu)^2}{N-1}=\frac{799.6}{14}=57.11$

$Var_1(X)-Var(X)=57.11-53.31=3.8$