Answer:
[tex]n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11[/tex]
And rounded up we have that n=1068
Step-by-step explanation:
We have the following info given:
[tex] Confidence= 0.95[/tex] the confidence level desired
[tex] ME =0.03[/tex] represent the margin of error desired
The margin of error for the proportion interval is given by this formula:
[tex] ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex] (a)
The confidence level is 95% or 0.95, the significance is [tex]\alpha=0.05[/tex] and the critical value for this case using the normal standard distribution would be [tex] z_{\alpha/2}=1.96[/tex]
Since we don't have prior information we can use [tex]\hat p= 0.5[/tex] as an unbiased estimator
Also we know that [tex]ME =\pm 0.03[/tex] and we are interested in order to find the value of n, if we solve n from equation (a) we got:
[tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex] (b)
And replacing into equation (b) the values from part a we got:
[tex]n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11[/tex]
And rounded up we have that n=1068
