Question

A prior study estimated as 34%. The analysts would like to conduct a second study on the same topic with a margin of error, E, of 0.027 and a confidence level of 90% (z*-score 1.645). What is the minimum sample size that should be used so the estimate of will be within the required margin of error of the population proportion? n = (1 – ) • 17 23 430 833

Answer 1

Answer: 833

Step-by-step explanation:

When the prior population proportion (p) is known , then the formula to find the minimum sample size is given by :-

$n=p(1-p)(\dfrac{z_{c}}{E})^2$

where, $z_{c}$ is the z-score for confidence level (c) and   E = the margin of error .

Given : A prior study estimated as 34%.

i.e. p= 0.34

Confidence level = 0.90

Critical z-score for 90% confidence level : $z_{c}1.645$

Margin of error : E= 0.027

then , the sample size = $n=0.34(1-0.34)(\dfrac{1.645}{0.027})^2$

Simplify ,

$n=832.9657\approx833$

Hence, the minimum sample size=833

Answer 2

The minimum sample size required for a test with a confidence interval of $100(1 - \alpha )%$ with a z-score of $z_{ \alpha /2}$ and margin of error of E and a population proportion of p is given by:

$n= \frac{p(1-p)z_{ \alpha /2}^2 \alpha }{E^2}$

Given p = 34% = 0.34, E = 0.027, $z_{ \alpha /2}=1.645$

Therefore, 

$n= \frac{0.34(1-0.34)(1.645)^2}{0.027^2} \\ \\ = \frac{0.34(0.66)(2.706025)}{0.000729} = \frac{0.60723201}{0.000729} \\ \\ =832.97=833$