Question

Cilantro tastes like soap to some people. This soap taste is inherited through the olfactory receptor gene OR6A2. About 14% of the population has this gene.
You want to estimate the proportion of Americans who have this gene. How large a sample must you test, with a 3% margin of error and 95% confidence, to estimate the proportion of people who carry the OR6A2 gene?

Answer 1

Answer:

Sample size of at least 514 Americans is required.

Step-by-step explanation:

We are given that the soap taste is inherited through the olfactory receptor gene OR6A2. About 14% of the population has this gene.

Let p = % of population having this gene = 0.14

Also, Margin of error = 3%

        Confidence level = 95%

Margin of error formula = $Z_\frac{\alpha}{2} * \frac{\sigma}{\sqrt{n} }$

where, $Z_\frac{\alpha}{2}$ = At 5% level of significance z score has value of 1.96

             $\sigma$ = $\sqrt{p(1-p)}$ = $\sqrt{0.14 * 0.86}$ = 0.347

So, Margin of error = $1.96* \frac{\sqrt{0.14*0.86} }{\sqrt{n} }$

            $\sqrt{n}$ = $\frac{1.96*0.347}{0.03}$

             n = $22.671^{2}$ = 513.95 ≈ 514

Therefore, sample must be of 514 Americans .

Answer 2

Answer:

You must test a sample size of at least 514 Americans.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence interval $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$ is the margin of error.

95% confidence interval

So $\alpha = 0.05$, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$, so $z = 1.96$.

We have that:

$M = 0.03, \pi = 0.14$

So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.03 = 1.96\sqrt{\frac{0.14*0.86}{n}}$

$0.03\sqrt{n} = 0.68$

$\sqrt{n} = 22.67$

$n = 513.92$

You must test a sample size of at least 514 Americans.