Respuesta :
Answer:
(a) The sample size required is 1034.
(b) The sample size required is 1040.
Step-by-step explanation:
The confidence interval for population proportion is:
[tex]CI=\hat p\pm z_{ \alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n} }[/tex]
The margin of error is:
[tex]MOE=z_{ \alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n} }[/tex]
Given:
[tex]MOE= 0.04\\Confidence\ level =0.99[/tex]
The critical value of z for 99% confidence level is:
[tex]z_{\alpha /2}=z_{0.01/2}=z_{0.005}=2.58[/tex] *Use a standard normal table.
(a)
Compute the sample size required if the previous estimate is [tex]\hat p = 0.54\\[/tex] as follows:
[tex]MOE=z_{ \alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n} }\\0.04=2.58\times \sqrt{\frac{0.54(1-0.54)}{n} }\\n=\frac{(2.58)^{2}\times 0.54\times (1-0.54)}{(0.04)^{2}} \\=1033.4061\\\approx1034[/tex]
Thus, the sample size required is 1034.
(b)
Compute the sample size required if there was no previous estimate as follows:
Assume that the estimated proportion be 0.50.
[tex]MOE=z_{ \alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n} }\\0.04=2.58\times \sqrt{\frac{0.50(1-0.50)}{n} }\\n=\frac{(2.58)^{2}\times 0.50\times (1-0.50)}{(0.04)^{2}} \\=1040.0625\\\approx1040[/tex]
Thus, the sample size required is 1040.
