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

The sample size required is 910.

Explanation:

The confidence interval for population proportion is:

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

The margin of error is:

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

Thus, the sample size required is 910.