Microsoft Word research method fm doc

Estimating a proportion
• Estimate how big the proportion might be (P) 
• Choose the margin of error you will allow in the estimate of the proportion (say ± w)
• Choose the level of confidence that the proportion in the whole population is indeed 
between (p-w) and (p+w). We can never be 100% sure. Do you want to be 95% sure? 
• The minimum sample size required, for a very large population (N>10,000) is:
n = Z
2
p(1-p) / w
2
 
Example 1 (
Prevalence of diarrhoea) 
a) p = 0.26 , w = 0.03 , Z = 1.96 ( i.e., for a 95% C.I.) 
n = (1.96)
2
(.26 
× .74) / (.03)
2
= 821.25
≈ 822 
 
Thus, the study should include at least 822 subjects. 
 
b) If the above sample is to be taken from a relatively small population (say N = 3000), the 
required minimum sample will be obtained from the above estimate by making some 
adjustment. 

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Research methodology 
49
821.25 / (1+ (821.25/3000)) = 644.7
≈ 645 subjects 
Example 2 
 
A hospital administrator wishes to know what proportion of discharged patients are unhappy 
with the care received during hospitalization. If 95% Confidence interval is desired to estimate 
the proportion within 5%, how large a sample should be drawn?
n = Z
2
p(1-p)/w
2
=(1.96)
2
(.5
×.5)/(.05)
2
=384.2 
≈ 385
patients
 
N.B.
If you don’t have any information about P, take it as 50% and get the maximum value of 
PQ which is 1/4 (i.e., 25%). 
 
Estimating a mean 
The same approach is used but with SE =
σ / √n 
The required (minimum) sample size for a very large population is given by :