Respuesta :
Answer:
a) [tex]n=\frac{0.76(1-0.76)}{(\frac{0.01}{1.64})^2}=4905.83[/tex] Â
And rounded up we have that n=4906
b) For this case since we don't have prior info we need to use as estimatro for the proportion [tex]\hat p =0.5[/tex]
[tex]n=\frac{0.5(1-0.5)}{(\frac{0.01}{1.64})^2}=6724[/tex] Â
And rounded up we have that n=6724
Step-by-step explanation:
We need to remember that the confidence interval for the true proportion is given by : Â
[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]
Part a
The estimated proportion for this case is [tex]\hat p =0.76[/tex]
Our interval is at 90% of confidence, and the significance level is given by [tex]\alpha=1-0.90=0.1[/tex] and [tex]\alpha/2 =0.05[/tex]. The critical values for this case are:
[tex]z_{\alpha/2}=-1.64, t_{1-\alpha/2}=1.64[/tex]
The margin of error for the proportion interval is given by this formula: Â
[tex] ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex] Â Â (a) Â
The margin of error desired is given [tex]ME =\pm 0.01[/tex] and we are interested in order to find the value of n, if we solve n from equation (a) we got: Â
[tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex] Â (b) Â
Replacing we got:
[tex]n=\frac{0.76(1-0.76)}{(\frac{0.01}{1.64})^2}=4905.83[/tex] Â
And rounded up we have that n=4906
Part b
For this case since we don't have prior info we need to use as estimatro for the proportion [tex]\hat p =0.5[/tex]
[tex]n=\frac{0.5(1-0.5)}{(\frac{0.01}{1.64})^2}=6724[/tex] Â
And rounded up we have that n=6724
