Question

4 16.27 this wine stinks. sulfur compounds cause “off odors” in wine, and 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 (μg/l). the untrained noses of consumers may be less sensitive, however. here are the dms odor thresholds for 10 untrained students: wine2
30 30 42 35 22 33 31 29 19 23
a. assume that the standard deviation of the odor threshold for untrained noses is known to be σ = 7μg/l. briefly discuss the other two “simple conditions” and create a stemplot to verify that the distribution is roughly symmetric with no outliers.
b. give a 95% confidence interval for the mean dms odor threshold among all students. use the four - step process.

Explanation:

Step1: Calculate the sample mean

The sample data is \(x = [30,30,42,35,22,33,31,29,19,23]\).
The formula for the sample mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\), where \(n = 10\).
\(\sum_{i=1}^{10}x_{i}=30 + 30+42+35+22+33+31+29+19+23=294\)
\(\bar{x}=\frac{294}{10}=29.4\)

Step2: Determine the critical value

For a \(95\%\) confidence interval, the significance level \(\alpha=1 - 0.95 = 0.05\), and \(\alpha/2=0.025\).
The critical value \(z_{\alpha/2}\) for a normal distribution. Looking up in the standard normal table, \(z_{0.025}=1.96\)

Step3: Calculate the margin of error

The formula for the margin of error \(E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\), where \(\sigma = 7\) and \(n = 10\)
\(E=1.96\times\frac{7}{\sqrt{10}}\approx1.96\times2.214\approx4.34\)

Step4: Calculate the confidence interval

The confidence interval is \(\bar{x}-E<\mu<\bar{x} + E\)
Substitute \(\bar{x}=29.4\) and \(E = 4.34\)
\(29.4-4.34<\mu<29.4 + 4.34\)
\(25.06<\mu<33.74\)

Answer:

The \(95\%\) confidence interval for the mean DMS odor threshold among all students is \((25.06,33.74)\)