Respuesta :
Answer:
a) [tex]n=\frac{0.32(1-0.32)}{(\frac{0.03}{1.96})^2}=928.81[/tex]
And rounded up we have that n=929
b) [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:
Part a
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)
And on this case we have 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)
For a confidence of 95% we have that the significance is [tex]\alpha=0.05[/tex] and the critical value would be:
[tex] z = 1.96[/tex]
And replacing into equation (b) the values from part a we got:
[tex]n=\frac{0.32(1-0.32)}{(\frac{0.03}{1.96})^2}=928.81[/tex]
And rounded up we have that n=929
Part b
For this case since we don't have prior info we can use as estimator for the true proportion the value [tex]\hat p=0.5[/tex] and replacing 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
