Sulfur compounds cause "off‑odors" in wine, so winemakers want to know the odor threshold, the lowest concentration of a compound that the human nose can detect. The odor threshold for dimethyl sulfide (DMS) in trained wine tasters is about 25 micrograms per liter of wine ( μ g / L ). The untrained noses of consumers may be less sensitive, however. How sensitive are the untrained noses of students? You want to estimate the mean DMS odor threshold among all students and you would be satisfied to estimate the mean to within ± 0.1 with 90 % confidence. The standard deviation of the odor threshold for untrained noses is known to be σ = 8 micrograms per liter of wine. How large an SRS of untrained students do you need?

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

[tex]\bar X[/tex] represent the sample mean for the sample

[tex]\mu[/tex] population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)

The margin of error is given by this formula:

[tex] ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (2)

And on this case we have that ME =0.1 and we are interested in order to find the value of n, if we solve n from equation (2) we got:

[tex]n=(\frac{z_{\alpha/2} \sigma}{ME})^2[/tex] (3)

The critical value for 90% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.05;0;1)", and we got [tex]z_{\alpha/2}=1.640[/tex], replacing into formula (b) we got:

[tex]n=(\frac{1.64(8)}{0.1})^2 =17213.44 \approx 17214[/tex]

So the answer for this case would be n=17214 rounded up to the nearest integer