A leading dental journal claims that 68% of young adults do not brush their teeth regularly. A dental hygienist wants to conduct a survey to verify this with her young adult patients, and wants to do so with a margin of error (ME) of ± 5%. What minimum sample size, rounded to the nearest whole person, does the hygienist need to use? Hint: ME equals plus or minus two times square root of start fraction p hat left parenthesis one minus p hat right parenthesis over N end fraction end square root, where p hat is the yes proportion in the sample and N is the sample size N ≥ 175 N ≥ 348 N ≥ 366 N ≥ 400

To calculate the minimum sample size needed to conduct a survey to verify the claim that 68% of young adults do not brush their teeth regularly.

The sample size for a proportion estimation is given by
$n= (\frac{z_{ \alpha /2}\sqrt{p(1-p)}}{ME} )^2$
where: $z_{ \alpha /2}$ is the z-score of the confidence level and p is the proportion being tested.

Setting our confidence level to be 95%, then $z_{ \alpha /2}$ = 1.96

Thus,
$n= (\frac{1.96\times\sqrt{0.68(1-0.68)}}{0.05} )^2 \\ = (\frac{1.96\times \sqrt{0.2176} }{0.05})^2= (\frac{1.96(0.466476)}{0.05})^2 \\ = (\frac{0.914293}{0.05} )^2=(18.2859)^2=335$
i.e. N ≥ 335.

Setting our confidence level to be 90%, then $z_{ \alpha /2}$ = 1.645

Thus,
$n=
 (\frac{1.645\times\sqrt{0.68(1-0.68)}}{0.05} )^2 \\ = (\frac{1.645\times 
\sqrt{0.2176} }{0.05})^2= (\frac{1.645(0.466476)}{0.05})^2 \\ = 
(\frac{0.767353}{0.05} )^2=(15.3471)^2=236$
i.e. N ≥ 236.

Setting our confidence level to be 99%, then $z_{ \alpha /2}$ = 2.58

Thus,
$n=
 (\frac{2.58\times\sqrt{0.68(1-0.68)}}{0.05} )^2 \\ = (\frac{2.58\times 
\sqrt{0.2176} }{0.05})^2= (\frac{2.58(0.466476)}{0.05})^2 \\ = 
(\frac{1.20351}{0.05} )^2=(24.0702)^2=580$
i.e. N ≥ 335.

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