The margin of error is inversely proportional to the square root of the sample size. We know that for a certain sample size n, the margin of error is +/- E, thus:
[tex]E=\frac{k}{\sqrt{n}}[/tex]Where k is the constant of proportionality.
We need to know the sample size N that produces a margin of error of +/-9/10E:
[tex]\frac{9}{10}E=\frac{k}{\sqrt{N}}[/tex]Substituting the value of E from the first equation:
[tex]\frac{9}{10}\frac{k}{\sqrt{n}}=\frac{k}{\sqrt{N}}[/tex]Simplifying by k:
[tex]\frac{9}{10}\frac{1}{\sqrt{n}}=\frac{1}{\sqrt{N}}[/tex]Squaring and solving for N:
[tex]\begin{gathered} \frac{81}{100}\frac{1}{n}=\frac{1}{N} \\ \\ N=\frac{100}{81}n \end{gathered}[/tex]We need to calculate the percentage of increase of n to get N. Subtracting n:
[tex]N-n=\frac{19}{81}n[/tex]Dividing by n:
[tex]\frac{N-n}{n}=\frac{19}{81}[/tex]The sample size must be increased by 19/81 = 0.2346.
It's equivalent to 23% (rounding to the nearest percent)
Answer. 23%
