Question

A researcher wishes to estimate the proportion of adults who have high-speed internet access. What size sample should be obtained if she wishes the estimate to be within 0.03 95% confidence ifa) she uses a previous estimate of 0.36?b) she does not use any prior estimates?

Answer 1

Answer: a) 984   b) 1068

Step-by-step explanation:

When the prior estimate of the population proportion(p) is available .

Then the formula to find the sample size  :-

$n=p(1-p)(\dfrac{z^*}{E})^2$

, where E = margin of error

and z* = Critical z-value .

a) p= 0.36

E= 0.03

Critical value for 95% confidence level = z*= 1.96

Required sample size=$n= 0.36(1-0.36)(\dfrac{1.960}{0.03})^2$

$n= 0.36(0.64)(65.3333333333)^2$

$n=(0.2304)(4268.44444444)=983.4496\approx984$

Hence, the required sample size is 984.

b) When the prior estimate of the population proportion is unavailable .

Then we use formula to find the sample size  :-

$n= 0.25(\dfrac{z^*}{E})^2$

, where E = margin of error

and z* = Critical z-value

Put E= 0.03 and z*= 1.960

Required sample size =$n= 0.25(\dfrac{1.960}{0.03})^2$

$n= 0.25(65.3333333333)^2$

$n= 0.25(4268.44444444)=1067.11111111\approx1068$  

Hence, the required sample size is 1068.