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Chapter 3: Problem 21

Acrylamide, a possible cancer-causing substance, forms in high-carbohydrate foods cooked at high temperatures. Acrylamide levels can vary widely even within the same type of food. An article appearing in the journal Food Chemistry (March 2014, 204-211) included the following acrylamide content (in nanograms/gram) for five brands of bisquits: $$ \begin{array}{lllll} 345 & 292 & 334 & 276 & 248 \end{array} $$ a. Calculate the mean acrylamide level. For each data value, calculate the deviation from the mean. b. Verify that, except for the effect of rounding, the sum of the five 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 still be close to zero.) c. Use the deviations from Part (a) to calculate the variance and standard deviation for this data set.

Short Answer

Expert verified
The mean acrylamide level is 299 ng/g, and the deviations are 46, -7, 35, -23, and -51, with a sum of 0. The variance is 1630 (ng/g)虏, and the standard deviation is approximately 40.37 ng/g.

Step by step solution

01

Part a. Calculate the mean#acrylamide level

To calculate the mean, first sum up the data points and then divide the total by the number of data points (in this case, 5). Mean formula: \[ \bar{x} = \frac{\sum x}{n} \] The data points are: \[345, 292, 334, 276, 248\] So, we can calculate the mean: \[\bar{x} = \frac{345+292+334+276+248}{5} \]
02

Calculate the mean#acrylamide level

Calculating the mean, we get: \[\bar{x} = \frac{1495}{5} = 299\] Hence, the mean acrylamide level is \(299\,ng/g\).
03

Calculate the deviation from the mean

Now, we need to calculate the deviation from the mean for each data value, using the formula: \[ d_i = x_i - \bar{x} \] For each data value, calculate its deviation as follows: \[ d_1 = 345 - 299 = 46 \] \[ d_2 = 292 - 299 = -7 \] \[ d_3 = 334 - 299 = 35 \] \[ d_4 = 276 - 299 = -23 \] \[ d_5 = 248 - 299 = -51 \] So, the deviations are: \(46, -7, 35, -23, -51\)
04

Part b. Verify the sum of deviations from the mean is 0

Now, we'll verify if the sum of the deviations from the mean is equal to 0 or not: \[ \sum d_i = 46 + (-7) + 35 + (-23) + (-51) \]
05

Sum of deviations

Calculating the sum of deviations, we get: \[ \sum d_i = 0 \] Here, the sum of deviations from the mean is \(0\). Hence, the result has been verified.
06

Part c. Calculate the variance and standard deviation

Now, let's calculate the variance and standard deviation for this data set. The formula for variance is: \[s^2 = \frac{\sum d_i^2}{n-1}\] The formula for standard deviation is: \[s = \sqrt{s^2}\] Let's square the deviations and calculate the variance:
07

Variance calculation

\[s^2 = \frac{(46)^2 + (-7)^2 + (35)^2 + (-23)^2 + (-51)^2 }{5 - 1} \] \[s^2 = \frac{2116 + 49 + 1225 + 529 + 2601}{4}\] Calculating the variance, we get: \[s^2 = \frac{6520}{4} = 1630\] Now, we obtain the standard deviation from the variance:
08

Standard deviation calculation

\[s = \sqrt{1630} \] Calculating the standard deviation, we get: \[s \approx 40.37\] Hence, the variance of the data set is \(1630\,(ng/g)^2\), and the standard deviation is approximately \(40.37\,ng/g\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mean Calculation
In statistics, the mean represents the average value of a data set, providing a measure of the central tendency of the data. To calculate the mean, sum all the individual data points and divide by the number of points. In the context of acrylamide content in biscuits, this gives us a quick way to understand the 'typical' level of acrylamide across multiple brands.

For example, let's consider five brands of biscuits with acrylamide content as follows: 345, 292, 334, 276, and 248 nanograms per gram. To find the mean, you add up these values, resulting in a total of 1495 ng/g, and then divide this sum by the number of data points, which is 5 in our case. This calculation yields a mean of 299 ng/g. It's a simple yet powerful calculation that allows us to summarize an entire data set with a single number, providing a benchmark for comparison with individual data points.
Deviation from the Mean
After calculating the mean, the next step is understanding how individual data points deviate from it. This is where we calculate the deviation from the mean for each data value, which tells us how far a point is from the average. This deviation can be positive or negative, indicating whether the data point is above or below the mean, respectively.

In our acrylamide example, we subtract the mean (299 ng/g) from each brand's acrylamide content to find the deviations: 46, -7, 35, -23, and -51 ng/g. These deviations illustrate the spread of the data around the mean. But here's an intriguing statistical truth: when you sum up all the deviations (as long as you haven't rounded the individual deviations), they should always add up to zero. This is because the deviations above the mean effectively cancel out those below the mean. If you find that the sum of the deviations isn't zero, this could indicate an error in calculation.
Variance and Standard Deviation
To further analyze the dataset, we turn to the concepts of variance and standard deviation. These are measures of dispersion or how spread out the data points are in relation to the mean. Variance is calculated by squaring each deviation from the mean, summing those squares, and then dividing by the number of data points minus one. This gives us the average of the squared deviations and is expressed in the units of the data squared.

The standard deviation, on the other hand, is the square root of variance, bringing the measure of spread back into the original units of the data. It is a more intuitive measure of spread because it can be directly compared to the mean. For our biscuits acrylamide content, the variance comes out to be 1630 (ng/g)^2, and the standard deviation is approximately 40.37 ng/g, suggesting that there is a moderate spread of acrylamide levels among the different brands of biscuits.

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Most popular questions from this chapter

The article "Caffeinated Energy Drinks-A Growing Problem" (Drug and Alcohol Dependence [2009]: 1-10) gave the accompanying data (on the next page) on caffeine concentration (mg/ounce) for eight top-selling energy drinks. a. What is the mean caffeine concentration for this set of energy drinks? b. Coca-Cola has 2.9 mg/ounce of caffeine and Pepsi Cola has \(3.2 \mathrm{mg} /\) ounce of caffeine. Write a sentence explaining how the caffeine concentration of top-selling energy drinks compares to that of these colas. $$ \begin{array}{|lc|} \hline \text { Energy Drink } & \begin{array}{c} \text { Caffeine Concentration } \\ \text { (mg/ounce) } \end{array} \\ \hline \text { Red Bull } & 9.6 \\ \text { Monster } & 10.0 \\ \text { Rockstar } & 10.0 \\ \text { Full Throttle } & 9.0 \\ \text { No Fear } & 10.9 \\ \text { Amp } & 8.9 \\ \text { SoBe Adrenaline Rush } & 9.5 \\ \text { Tab Energy } & 9.1 \\ \hline \end{array} $$

The report titled "State of the News Media \(2013^{\text {" }}\) (Pew Research Center, May 7,2013 ) included the weekday circulation numbers for the top 20 newspapers in the country. Here are the data for the 6 months ending September 2012: $$ \begin{array}{rrrrr} 2,293798 & 1,713,833 & 1,613,866 & 641,369 & 535,875 \\ 529,999 & 522,868 & 462,228 & 432,455 & 412,669 \\ 411,960 & 410,130 & 392,989 & 325,814 & 313,003 \\ 311,504 & 300,277 & 296,427 & 293,139 & 285,088 \end{array} $$ a. Calculate and interpret the value of the median of this data set. b. Explain why the median is preferable to the mean for describing center for this data set. c. Explain why it would be unreasonable to generalize from this sample of 20 newspapers to the population of all daily newspapers in the United States.

In a study investigating the effect of car speed on accident severity, the vehicle speed at impact was recorded for 5000 fatal accidents. For these accidents, the mean speed was 42 mph and the standard deviation was 15 mph. A histogram revealed that the vehicle speed distribution was mound shaped and approximately symmetric. a. Approximately what percentage of the vehicle speeds were between 27 and \(57 \mathrm{mph} ?\) b. Approximately what percentage of the vehicle speeds exceeded \(57 \mathrm{mph} ?\)

Suppose that your younger sister is applying to college and has taken the SAT exam. She scored at the 83 rd percentile on the verbal section of the test and at the 94 th percentile on the math section. Because you have been studying statistics, she asks you for an interpretation of these values. What would you tell her?

Although bats are not known for their eyesight, they are able to locate prey (mainly insects) by emitting high-pitched sounds and listening for echoes. A paper appearing in Animal Behaviour ("The Echolocation of Flying Insects by Bats" [1960]: 141-154) gave the following distances (in centimeters) at which a bat first detected a nearby insect: $$ \begin{array}{lllllllllll} 62 & 23 & 27 & 56 & 52 & 34 & 42 & 40 & 68 & 45 & 83 \end{array} $$ a. Calculate and interpret the mean distance at which the bat first detects an insect. b. Calculate the sample variance and standard deviation for this data set. Interpret these values.
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