Question 1062598
A researcher wants to estimate with 99% confidence, the proportion of people who have a computer desktop.  A previous study showed that 30% of those interviewed had a computer desktop at home.  The researcher wants to be accurate within 3% of the true proportion.  What is the minimum sample size to be needed?
<pre>See Probability-and-statistics/1060911 below. 
Do it the same way. You only need a) though.
Pay attention to the different values. For example, in yours, Margin of Error (E) is: 3% (.03) instead of 5% (.05).


A researcher wishes to estimate the proportion of adults who have&#8203; high-speed Internet access. What size sample should be obtained if she wishes the estimate to be within 0.05 with 99&#8203;% confidence if
&#8203;(a) she uses a previous estimate of 0.36&#8203;?
&#8203;(b) she does not use any prior&#8203; 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?