Suppose that 5000 sales invoices are separated into four strata. Stratum 1 contains 50 invoices,
stratum 2 contains 500 invoices, stratum 3 contains 1000 invoices, and stratum 4 contains 3450
invoices. A sample of 500 sales invoices is needed.
a) what type of sampling should you do? why?
A stratified random sampling is suitable for this study. This is because the population of invoices is
heterogenous which results in increase in the variance of the estimators. When the population is
divided into homogeneous strata and stratified random sampling is adopted a reduction in the
variance of the estimators is attained.
b) Explain how you would carry out the sampling according to the method stated in (a)
The size of the population is 50 + 500 + 1000 + 3450 = 5000
The size of the sample is 500
The sampling fraction is 500/5000 = 0.10 = 10%
From stratum 1, a simple random sample of size 5 (10% of 50) is chosen.
From stratum 2, a simple random sample of size 50 (10% of 500) is chosen.
From stratum 3, a simple random sample of size 100 (10% of 1000) is chosen.
From stratum 4, a simple random sample of size 345 (10% of 3450) is chosen.
Thus a stratified random sample of size 5 + 50 + 100 + 345 = 500 is obtained.
c) Why is the sampling in (a) not simple random sampling?
In simple random sampling all possible samples of size 500 are given the same probability of beind
selected. In stratified random sample, this not the case.
For example the probability of selecting 500 invoices from stratum 4 is 0. Only 345 invoices are
selected from stratum 4.
The NBC hit comedy Friends was TiVo's most popular show during the week of April 18-24, 2004.
According to the Nielsen ratings, 29.7% of TiVo owners in the United States either recorded Friends
or watched it live ("Prime - Time Nielsen Ratings," USA Today, April 28, 2004, p. 3D)
Suppose you select a random sample of 50 TiVo owners.
Let X be the number of TiVo owners who watched or recorded out of the 50 selected. Then X follows
the binomial distribution with parameters n = 50 and p = 29.7/100 = 0.297
The mean of X is np = 50*0.297 = 14.85 and the variance is np(1-p) = 50*0.297*(1-0.297)=10.440.
The standard deviation is sqrt(10.440)=3.231.
Since n>30, the distribution of X can be approximated by the normal distribution with mean 14.95
and standard deviation 3.231.
Z = (X-14.95)/3.231 follows the standard normal distribution.
a)What is the probability that more than half the people in the sample watched or recorded Friends?
half of 50 is 25
P(X>25) is required.
The Z value for X = 25 is (25-14.95)/3.231 = 3.141
P(Z>3.141) = 1-P(Z<3.141) = 1-0.999159=0.000841
b)What is the probability that less than 25% of the people in the sample watched or recorded
Friends?
25% of 50 is 12.5
P(X<12.5) is required.
The Z value for X = 12.5 is (12.5-14.95)/3.231 = -0.727
P(Z<-0.727) = 0.234
c)If a random sample of size 500 is taken, how does this change your answers to (a) and (b)?
mean of X is np = 500*0.297 =149.5
Standard deviation of x is sqrt(500*0.297*(1-0.297)) = 10.217
part(a) becomes
half of 50 is 25
P(X>25) is required.
The Z value for X = 25 is (25-149.5)/10.217 = -12.087
P(Z>-12.087) = 1-P(Z<-12.087) = 1-0 = 1
Part(b) becomes
25% of 50 is 12.5
P(X<12.5) is required.
The Z value for X = 12.5 is (12.5-149.5)/10.217 = -13.311
P(Z<-13.311) = 0