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

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(6.75)/(√(10))=24.58`

`29.4-2.26(6.75)/(√(10))=34.22`

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

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)(s)/(√(n))` (1)

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

`\bar X= \sum_(i=1)^n (x_i)/(n)` (2)

`s=\sqrt{(\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(6.75)/(√(10))=24.58`

`29.4-2.26(6.75)/(√(10))=34.22`

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