Sulfur compounds cause "off-odors" in wine, so winemaker 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. The untrained noses of consumers may be less sensitive, however. Here are the DMS odor thresholds for 10 untrained students:

30,30,42,35,22,33,31,29,19,23

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 99% confidence. The standard deviation of the odor threshold for untrained noses is known to be standard deviation = 7 micrograms per liter of wine. How large an SRS of untrained students do you need?

To estimate the mean DMS odor threshold among all students with a specified level of confidence and a given margin of error, you can use the formula for sample size calculation:

n = (Z * σ / E)^2

Where:
n = required sample size
Z = z-score corresponding to the desired confidence level (in this case, 99%, which corresponds to a z-score of 2.576)
σ = standard deviation of the population (7 micrograms per liter of wine)
E = desired margin of error (0.1)

Plugging in the values:

n = (2.576 * 7 / 0.1)^2
n = (18.0272 / 0.1)^2
n = 180.272^2
n ≈ 32,527

So, you would need a sample size of approximately 32,527 untrained students in order to estimate the mean DMS odor threshold among all students to within +- 0.1 with 99% confidence.

To determine how large of a simple random sample (SRS) of untrained students is needed to estimate the mean DMS odor threshold with the desired confidence level and precision, we can use the formula for sample size calculation:

n = (Z * σ / E)^2

where:
- n is the required sample size
- Z is the Z-score corresponding to the desired confidence level
- σ is the known standard deviation of the odor threshold
- E is the desired margin of error (precision)

In this case, the desired margin of error is ±0.1, and the desired confidence level is 99%. Thus, we need to find the appropriate Z-score for a 99% confidence level.

Using a Z-table or a statistical calculator, the Z-score for a 99% confidence level is approximately 2.576.

Plugging in the values:

n = (2.576 * 7 / 0.1)^2

Simplifying:

n = (18.032 / 0.1)^2
n = 180.32^2

Calculating:

n ≈ 32,517

Therefore, you would need a sample size of approximately 32,517 untrained students to estimate the mean DMS odor threshold within ±0.1 with 99% confidence.

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