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

Sample size of at least 514 Americans is required.

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_(\alpha)/(2) * (\sigma)/(√(n) )`

where,
`Z_(\alpha)/(2)` = At 5% level of significance z score has value of 1.96

`\sigma` =
`√(p(1-p))` =
`√(0.14 * 0.86)` = 0.347

So, Margin of error =
`1.96* (√(0.14*0.86) )/(√(n) )`

`√(n)` =
`(1.96*0.347)/(0.03)`

n =
`22.671^(2)` = 513.95 ≈ 514

Therefore, sample must be of 514 Americans .

Answer 2

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

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{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

`M = z\sqrt{(\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 - (0.05)/(2) = 0.975`, so
`z = 1.96`.

We have that:

`M = 0.03, \pi = 0.14`

So

`M = z\sqrt{(\pi(1-\pi))/(n)}`

`0.03 = 1.96\sqrt{(0.14*0.86)/(n)}`

`0.03√(n) = 0.68`

`√(n) = 22.67`

`n = 513.92`

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