Question

A research agency conducted a survey on randomly selected 2000 Malaysian adults aged 18 and over regarding weight issue. These adults were asked if they want to lose weight and 63% said yes.
i) Construct a 97% confidence interval for the corresponding population proportion.
ii) Find the minimum sample size if the research agency wants to be sure that the margin of error is within 0.04 of the population proportion for a 97% confidence interval. The preliminary estimate for the proportion of adults who wants to lose weight is 0.75.

Answer 1

Proportion of adult who want to loose weight= p= 63%=0.63

q= proportion of adult who don't want to loose weight= 1-p= 1-0.63= 0.37

i) 97% confidence interval for population proportion is given by

p ± z(alpha= 3%)* standard error ( p)

= 0.63± 1.88*√(p*q/n)= 0.63±1.88*√(0.63*0.37)/2000=

0.63±0.0203=( 0. 6097,0.6503)≈(0.61,0.65)

So, 0.61<P<0.65

ii) margin of error for 97% confidence interval is given by

Z( alpha= 3%)*standard error (P)= 1.88* √(p*q/n)

Here, p= 0.75 and q= 1-0.75= 0.25

So , margin of error= 0.04

Or, 1.88*√(0.75*0.25/n)= 0.04

Or, n = 1.88^2*0.75*0.25/0.04= 414.1875≈414

So, the minimum sample size= 414