Respuesta :
Answer:
[tex]n=\frac{0.04(1-0.04)}{(\frac{0.03}{1.64})^2}=114.76[/tex]
And rounded up we have that n=115
Step-by-step explanation:
In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by [tex]\alpha=1-0.90=0.1[/tex] and [tex]\alpha/2 =0.05[/tex]. And the critical value would be given by:
[tex]z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64[/tex]
Solution to the problem
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 proportion of defectives is estimated as: [tex]\hat p=0.04[/tex]. And on this case we have that the margin of error is [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{0.04(1-0.04)}{(\frac{0.03}{1.64})^2}=114.76[/tex]
And rounded up we have that n=115
