Chapter 4, Problem 55CR

a.

To determine

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

Answer to Problem 55CR

• The mean acrylamide level is 287.714.
• The seven deviations from the mean are 209.286, –94.714, 40.286, –132.714, 38.286, –42.714, and –17.714.

Explanation of Solution

Calculation:

The data represent the acrylamide content of French fries purchased at seven different locations.

Mean:

Mean is defined as the sum of observed values divided by the number of observations.

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 andStandard deviation.
• Click OK.

The output obtained using the MINITAB software is given below:

• Thus, the mean acrylamide content is 287.714.
• Here, the deviations from the mean are obtained by subtracting the mean value from each observation.
• The seven deviations from the mean are obtained as given below:

Observations	$\text{Deviation}=\text{Observation}-\text{Mean}$
497	$209.286\left(=497-287.714\right)$
193	$-94.714\left(=193-287.714\right)$
328	$40.286\left(=328-287.714\right)$
155	$-132.714\left(=155-287.714\right)$
326	$38.286\left(=326-287.714\right)$
245	$-42.714\left(=245-287.714\right)$
270	$-17.714\left(=270-287.714\right)$

• Therefore, the five deviations from the mean are 209.286, –94.714, 40.286, –132.714, 38.286, ­42.714, and –17.714.

b.

To determine

Verify that the sum of deviations from the mean is equal to 0, except for the effect of rounding.

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 the ratio level data have mean.

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

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

$\begin{array}{c}{\displaystyle \sum \text{Deviation}}={\displaystyle \sum \left(\text{Observation}-\text{mean}\right)}\\ =\left[\begin{array}{l}209.286+\left(-94.714\right)+40.286+\left(-132.714\right)+\\ \left(38.286\right)+\left(-42.714\right)+\left(-17.714\right)\end{array}\right]\\ =0.002\end{array}$

Here, it can be observed that the sum of the deviations from the mean is approximately equal to zero, since the sample mean and deviations are rounded.

Thus, except for the effect of rounding, the sum of deviations from the mean is equal to zero.

c.

To determine

Calculate the variance and standard deviation of the acrylamide content.

Answer to Problem 55CR

The variance of the acrylamide content is 12,601.9.

The standard deviation of the acrylamide content is 112.3.

Explanation of Solution

From Part (a), the variance and standard deviation of the acrylamide content are 12,601.9 and 112.3, respectively.

Hence, the variance of the acrylamide content is 12,601.9 units and the standard deviation of the acrylamide content is 112.3 units.