Question

If we wanted to create a new 90% confidence interval from a different sample for the proportion of those with a two on one date while keeping the margin of error at 0.05, what would the needed sample size be? Assume you have no prior knowledge of the proportion.

Answer 1

Answer: 271

Step-by-step explanation:

The formula we use to find the sample size is given by :-

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

, where $z_{\alpha/2}$ is the two-tailed z-value for significance level of $(\alpha)$

p = prior estimation of the proportion

E = Margin of error.

If prior estimation of the proportion is unknown, then we take p= 0.5 , the formula becomes

$n=0.5(1-0.5)(\dfrac{z_{\alpha/2}}{E})^2$

$n=0.25(\dfrac{z_{\alpha/2}}{E})^2$

Given :   Margin of error : E= 0.05

Confidence level = 90%

Significance level $\alpha=1-0.90=0.10$

Using z-value table , Two-tailed z-value for significance level of $0.10$

$z_{\alpha/2}=1.645$

Then, the required sample size would be :

$n=0.25(\dfrac{1.645}{0.05})^2$

Simplify,

$n=270.6025\approx271$

Hence, the required minimum sample size =271