Question

The estimate of the population proportion should be within plus or minus 0.02, with a 90% level of confidence. The best estimate of the population proportion is 0.16. How large a sample is required

Answer 1

Answer:

The sample size required is 910.

Step-by-step explanation:

The confidence interval for population proportion is:

$CI=\hat p\pm z_{ \alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n} }$

The margin of error is:

$MOE=z_{ \alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n} }$

Given:

$\hat p = 0.16\\MOE= 0.02\\Confidence\ level =0.90$

The critical value of z for 90% confidence level is:

$z_{\alpha /2}=z_{0.10/2}=z_{0.05}=1.645$ *Use a standard normal table.

Compute the sample size required as follows:

$MOE=z_{ \alpha /2}\sqrt{\frac{\hat p(1-\hat p)}{n} }\\0.02=1.645\times \sqrt{\frac{0.16(1-0.16)}{n} }\\n=\frac{(1.645)^{2}\times 0.16\times (1-0.16)}{(0.02)^{2}} \\=909.2244\\\approx910$

Thus, the sample size required is 910.