Chapter 4, Problem 61CR

Acrylamide (a possible cancer-causing substance) forms in high-carbohydrate foods cooked at high temperatures and acrylamide levels can vary widely even within the same type of food. An article appearing in the journal Food Chemistry (March 2014, pages 204–211) included the following acrylamide content (in nanograms/gram) for five brands of biscuits:

1. a. Calculate the mean acrylamide level and the five deviations from the mean.
2. b. Verify that, except for the effect of rounding, the sum of the deviations from the mean is equal to 0 for this data set. (If you rounded the sample mean or the deviations, your sum may not be exactly zero, but it should be close to zero if you have calculated the deviations correctly.)
3. c. Calculate the variance and standard deviation for this data set.

To determine

Compute the values of mean acrylamide level and the five deviations from the mean.

Answer to Problem 61CR

The mean acrylamide level is 299 nanograms/gram.

The five deviations from the mean are 46, –7, 35,–23, and –51.

Explanation of Solution

Calculation:

The data represent the acrylamide content for five brands of biscuits. The summary statistics are given.

Mean, variance, and standard deviation:

Software procedure:

Step-by-step procedure to find the mean, variance, and standard deviation using the MINITAB software:

• Choose Stat > Basic Statistics > Display Descriptive Statistics.
• In Variables, enter the columns Content.
• Choose option Statistics, and select Mean, Variance and Standard deviation.
• Click OK.

Output using the MINITAB software is given below:

Thus, the mean acrylamide content is 299 nanograms/gram.

Here, the deviations from the mean are obtained by subtracting the mean value from each observation.

The five deviations from the mean are obtained as given below:

Observations	$\text{Deviation}=\text{Observation}-\text{Mean}$
345	$46\left(=345-299\right)$
292	$-7\left(=292-299\right)$
334	$35\left(=334-299\right)$
276	$-23\left(=276-299\right)$
248	$-51\left(=248-299\right)$

Therefore, the five deviations from the mean are 46, –7, 35, –23, and –51.

To determine

Verify that the sum of deviations from the mean is equal to 0.

Explanation of Solution

Properties of mean:

• It will be affected by the extreme values.
• The sum of the deviations of each value from the mean is zero.
• Every set of the interval and ratio level data have mean.

Based on the concept of measure of center, the sum of deviation from the mean will always be equal to zero.

In this context, the sum of the deviation is obtained as follows:

$\begin{array}{c}{\displaystyle \sum \text{Deviation}}={\displaystyle \sum \left(\text{Observation}-\text{mean}\right)}\\ =46+\left(-7\right)+35+\left(-23\right)+\left(-51\right)\\ =0\end{array}$

Hence, it has been proved that the sum of the deviations from the mean is equal to 0.

To determine

Calculate the variance and standard deviation of the acrylamide content.

Answer to Problem 61CR

The variance of the acrylamide content is 1,630 nanograms/gram.

The standard deviation of the acrylamide content is 40.4 nanograms/gram.

Explanation of Solution

From Part (a), the variance and standard deviation of the acrylamide content are 1,630 and 40.4.

Hence, the variance of the acrylamide content is 1,630 nanograms/gram, and the standard deviation of the acrylamide content is 40.4 nanograms/gram.