Research Methodology - Dr. Krishan K. Pandey

Sampling Fundamentals 173
We now illustrate the use of this formula by an example.
Illustration 4
A market research survey in which 64 consumers were contacted states that 64 per cent of all
consumers of a certain product were motivated by the product’s advertising. Find the confidence
limits for the proportion of consumers motivated by advertising in the population, given a confidence
level equal to 0.95.
Solution: The given information can be written as under:
n = 64
p = 64% or .64
q = 1 – p = 1 – .64 = .36
and the standard variate (z) for 95 per cent confidence is 1.96 (as per the normal curve area table).
Thus, 95 per cent confidence interval for the proportion of consumers motivated by advertising in
the population is:
p ± z⋅
= . 64 ± 196 .
pq
n
b gb g
= . 64 ± 196 . . 06
b gb g
064 . 036 .
64
= . 64 ± . 1176
Thus, lower confidence limit is 52.24%
upper confidence limit is 75.76%
For the sake of convenience, we can summarise the formulae which give confidence intevals
while estimating population mean bg μ and the population proportion bg $p as shown in the following
table.
Table 8.3: Summarising Important Formulae Concerning Estimation
Estimating population mean
p
X ± z⋅
n
σ
bg μ when we know σ p
Estimating population mean
s
X ± z⋅
n
σ
bg μ when we do not know σ p
In case of infinite In case of finite population *
population
σ p
X ± z ⋅ ×
n
σ s
X ± z⋅
×
n
N − n
N − 1
N − n
N − 1
○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○ ○
Contd.