Respuesta :
Answer:
The sample size required is 910.
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]\hat p = 0.16\\MOE= 0.02\\Confidence\ level =0.90[/tex]
The critical value of z for 90% confidence level is:
[tex]z_{\alpha /2}=z_{0.10/2}=z_{0.05}=1.645[/tex] *Use a standard normal table.
Compute the sample size required as follows:
[tex]MOE=z_{ \alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n} }\\0.02=1.645\times \sqrt{\frac{0.16(1-0.16)}{n} }\\n=\frac{(1.645)^{2}\times 0.16\times (1-0.16)}{(0.02)^{2}} \\=909.2244\\\approx910[/tex]
Thus, the sample size required is 910.
