Question 1060911
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 0.05 with 99​% confidence if
​(a) she uses a previous estimate of 0.36​?
​(b) she does not use any prior​ estimates?
<pre><b><font face = “Century Gothic” size = 4 color = "indigo">
The ESTIMATED SAMPLE PROPORTION should be calculated using the following formula: n = p&#0770q&#0770{{{matrix(1,2, "*", (Z[c]/E)^2)}}}
<b>(a)</b>
Now, with a PRELIMINARY ESTIMATE of .36, p&#0770 = .36, and q&#0770 = 1 - p&#0770 = 1 - .36 = .64
Since this is a 99% CONFIDENCE INTERVAL, then significance level is {{{matrix(1,3, .01/2, or, .005)}}}, and so, {{{Z[critical] = 2.575}}}
E (Margin of Error) = .05
Therefore, n = p&#0770q&#0770{{{matrix(1,2, "*", (Z[c]/E)^2)}}} becomes: {{{highlight_green(matrix(1,16, n, "=", (.36)(.64), "*", (2.575/.05)^2, "=", .2304*(51.5)^2, "=", ".2304(2,652.25)", "=", 611.0784, ",", rounded, up, to, highlight(matrix(1,2, 612, adults))))}}}</pre>
<pre><b>(b)</b>
Now, since NO PRELIMINARY ESTIMATE WAS GIVEN/WAS AVAILABLE, then .5(50%) should be used for p&#0770, or for the ASSUMED proportion
With p&#0770 being .5, q&#0770 = 1 - p&#0770 = 1 - .5 = .5
Since this is a 99% CONFIDENCE INTERVAL, then significance level is {{{matrix(1,3, .01/2, or, .005)}}}, and so, {{{Z[critical] = 2.575}}}
E (Margin of Error) = .05
Therefore, n = p&#0770q&#0770{{{matrix(1,2, "*", (Z[c]/E)^2)}}} becomes: {{{highlight_green(matrix(1,16, n, "=", (.5)(.5), "*", (2.575/.05)^2, "=", .25*(51.5)^2, "=", ".25(2,652.25)", "=", 663.0625, ",", rounded, up, to, highlight(matrix(1,2, 664, adults))))}}}


When I used STATDISK to calculate the ESTIMATED SAMPLE PROPORTION, I got the same result: a) 612 and b) 664 adults.</font></b></pre>

This problem looks a lot like one from Chamberlain College of Nursing. I tutor a lot of nursing students who attend that college. Is that where you attend?