Question

A researcher wishes to estimate the proportion of adults who have​ high-speed Internet access. What size sample should be obtained if she wishes the estimate to be within 0.040.04 with 9999​% confidence if ​(a) she uses a previous estimate of 0.540.54​? ​(b) she does not use any prior​ estimates?

Answer 1

(a) The sample size required is 1034.

(b) The sample size required is 1040.

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:

`MOE= 0.04\\Confidence\ level =0.99`

The critical value of z for 99% confidence level is:

`z_(\alpha /2)=z_(0.01/2)=z_(0.005)=2.58` *Use a standard normal table.

(a)

Compute the sample size required if the previous estimate is
`\hat p = 0.54\\` as follows:

`MOE=z_( \alpha /2)\sqrt{(\hat p(1-\hat p))/(n) }\\0.04=2.58* \sqrt{(0.54(1-0.54))/(n) }\\n=((2.58)^(2)* 0.54* (1-0.54))/((0.04)^(2)) \\=1033.4061\\\approx1034`

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.

`MOE=z_( \alpha /2)\sqrt{(\hat p(1-\hat p))/(n) }\\0.04=2.58* \sqrt{(0.50(1-0.50))/(n) }\\n=((2.58)^(2)* 0.50* (1-0.50))/((0.04)^(2)) \\=1040.0625\\\approx1040`

Thus, the sample size required is 1040.