Question

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 (μg/l). The untrained noses of consumers may be less sensitive, how ever. Here are the DMS odor thresholds for 10 untrained f students:
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 = 7 μg/l. Briefly discuss the other two "simple conditions, " using 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.

Answer 1

Answer:

a) If we construct a stemplot for the data we have this:

Stem     Leaf

1       |     9

2      |     2 3 9

3      |     0 0 1 3 5

4      |     2

Notation: 1|9 means 19

b) $29.4-2.26\frac{6.75}{\sqrt{10}}=24.58$    

$29.4-2.26\frac{6.75}{\sqrt{10}}=34.22$    

So on this case the 95% confidence interval would be given by (24.58;34.22)    

Step-by-step explanation:

For this case we have the following data:

30 30 42 35 22 33 31 29 19 23

Part a

If we construct a stemplot for the data we have this:

Stem     Leaf

1       |     9

2      |     2 3 9

3      |     0 0 1 3 5

4      |     2

Notation: 1|9 means 19

As we can see the distribution is a little assymetrical to the right since we have not to much values on the right tail. But we can approximate roughly the distribution symmetric and with no outliers.

Part b

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".

$\bar X$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

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

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (1)

In order to calculate the mean and the sample deviation we can use the following formulas:  

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)  

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (3)  

The mean calculated for this case is $\bar X=29.4$

The sample deviation calculated $s=6.75$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=10-1=9$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,9)".And we see that $t_{\alpha/2}=2.26$

Now we have everything in order to replace into formula (1):

$29.4-2.26\frac{6.75}{\sqrt{10}}=24.58$    

$29.4-2.26\frac{6.75}{\sqrt{10}}=34.22$    

So on this case the 95% confidence interval would be given by (24.58;34.22)