Question

The consumption of asparagus results in the release of methanethiol and S-methyl thioesters metabolites in the urine. To some people, these metabolites have a very distinctive odor, but others cannot detect the odor ("asparagus anosmia"). How common is asparagus anosmia?

In a genetic study aiming to find the genetic markers supporting the phenotype, researchers contacted 6909 participants in a large scientific cohort and found that 4161 had asparagus anosmia. Assuming that the cohort is representative of the American adult population, obtain a 90% confidence interval for the true population of Americans with asparagus anosmia. Use technology, for instance, the 1-PropZInt function in the TI graphing calculator or the Statistics / Proportion / 1-sample function (Confidence Interval tab) on CrunchIt!.

Answer 1

The 90% confidence interval for the true population of Americans with asparagus anosmia is (0.5926, 0.6120).

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence level of
`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)`.

Researchers contacted 6909 participants in a large scientific cohort and found that 4161 had asparagus anosmia.

This means that
`n = 6909, \pi = (4161)/(6909) = 0.6023`

90% confidence level

So
`\alpha = 0.1`, z is the value of Z that has a pvalue of
`1 - (0.1)/(2) = 0.95`, so
`Z = 1.645`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.6023 - 1.645\sqrt{(0.6023*0.3977)/(6906)} = 0.5926`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.6023 + 1.645\sqrt{(0.6023*0.3977)/(6906)} = 0.6120`

The 90% confidence interval for the true population of Americans with asparagus anosmia is (0.5926, 0.6120).