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
[tex]n= (\frac{z_{ \alpha /2}\sqrt{p(1-p)}}{ME} )^2[/tex]
where: [tex]z_{ \alpha /2}[/tex] is the z-score of the confidence level and p is the proportion being tested.
Setting our confidence level to be 95%, then [tex]z_{ \alpha /2}[/tex] = 1.96
Thus,
[tex]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[/tex]
i.e. N ≥ 335.
Setting our confidence level to be 90%, then [tex]z_{ \alpha /2}[/tex] = 1.645
Thus,
[tex]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[/tex]
i.e. N ≥ 236.
Setting our confidence level to be 99%, then [tex]z_{ \alpha /2}[/tex] = 2.58
Thus,
[tex]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[/tex]
i.e. N ≥ 335.
