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!.

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

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

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

This means that [tex]n = 6909, \pi = \frac{4161}{6909} = 0.6023[/tex]

90% confidence level

So [tex]\alpha = 0.1[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.1}{2} = 0.95[/tex], so [tex]Z = 1.645[/tex].

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6023 - 1.645\sqrt{\frac{0.6023*0.3977}{6906}} = 0.5926[/tex]

The upper limit of this interval is:

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

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