/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 55 Research by the Food and Drug Ad... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Research by the Food and Drug Administration (FDA) shows that acrylamide (a possible cancer-causing substance) forms in high-carbohydrate foods cooked at high temperatures and that acrylamide levels can vary widely even within the same brand of food (Associated Press, December 6, 2002). FDA scientists analyzed McDonald's French fries purchased at seven different locations and found the following acrylamide levels: \(\begin{array}{lllllll}497 & 193 & 328 & 155 & 326 & 245 & 270\end{array}\) a. Compute the mean acrylamide level and the seven deviations from the mean. b. Verify that, except for the effect of rounding, the sum of the deviations from 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 computed the deviations correctly.) c. Calculate the variance and standard deviation for this data set.

Short Answer

Expert verified
a. Mean: 287.57. Deviations: 209.43, -94.57, 40.43, -132.57, 38.43, -42.57, -17.57. b. Sum of deviations is close to 0 (with rounding off taken into account). c. Variance: 13274.86. Standard deviation: 115.22.

Step by step solution

01

Calculate the Mean

To find the mean, sum up all the data points and divide by the number of data points. The data points given are: 497, 193, 328, 155, 326, 245, and 270. So, the mean (\( \mu \)) is \( \frac{497+193+328+155+326+245+270}{7} \).
02

Find the Deviations

The deviation from the mean is simply the difference between each data point and the mean. Subtract the mean from each data point (e.g., \(497 - \mu\), \(193 - \mu\), etc.) to get each deviation.
03

Verification of the Sum of Deviations

Add all the calculated deviations. In statistical theory, the sum of the deviations from the mean should be equal to 0. This is because the deviations above the mean cancel out the deviations below the mean.
04

Compute Variance

Variance (\( \sigma^2 \)) is the average of the squared deviations. To get this, square each deviation, sum them up, and divide the result by the number of data points minus 1 (used because this is a sample, not a whole population). The formula for the sample variance is \( \sigma^2 = \frac{\Sigma (x-\mu)^2}{n-1} \).
05

Compute Standard Deviation

The standard deviation (\( \sigma \)) is the square root of variance. So, \( \sigma = \sqrt{\sigma^2} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Mean Calculation
The mean is a measure of central tendency. It's often termed as the 'average' of a data set. To find the mean, you add together all the numbers in your data set and then divide by the number of data points. This is a straightforward calculation, but it's crucial as it forms the basis for many other analyses.

For example, if you have the acrylamide levels from the data set as: 497, 193, 328, 155, 326, 245, and 270, the mean is found as follows:
  • First, sum the acrylamide levels: 497 + 193 + 328 + 155 + 326 + 245 + 270 = 2014.
  • Next, divide by the number of data points, which is 7: \(\mu = \frac{2014}{7} \approx 287.71\)
Thus, the mean acrylamide level is approximately 287.71, providing a central value around which the levels tend to cluster.
Variance and Standard Deviation
Variance and standard deviation are statistical measures that tell us how much the values in a data set vary or deviate from the mean.

Variance, denoted as \( \sigma^2 \), is the average of the squared differences from the mean. This accounts for all variability in data. When data points are spread out, the variance will be large; when they are close together, the variance will be small.
  • To calculate variance, first, find each deviation from the mean and square it.
  • Then, sum all those squared deviations.
  • Finally, divide by the number of data points minus one (since it's a sample): \[\sigma^2 = \frac{\Sigma (x-\mu)^2}{n-1}\]
The standard deviation, \( \sigma \), is the square root of the variance, which gives a measure of spread in the same units as the original data. This is easier to interpret and helps compare different data sets.
Sum of Deviations from Mean
The sum of deviations is a foundational concept in statistics showing how data values vary with respect to the mean. If you take each value in your data set, subtract the mean, and then sum these differences, the total should be zero.

This occurs because positive deviations (data points above the mean) counterbalance the negative deviations (below the mean). But due to rounding, this sum can sometimes slightly differ from zero in practical calculations. As seen in the exercise, verifying this sum helps ensure calculations of mean and deviations have been done correctly. It's a basic but an essential check on statistical calculations.
Sample Data Analysis
Sample data analysis involves calculating statistics from a sample and using them to infer properties of the larger population. The methods used for data analysis in samples differ slightly from whole-population analysis due to the smaller size and increased variability.

The advantage of analyzing a sample, as with the acrylamide levels from McDonald's fries at seven locations, is efficiency. It's often impractical to gather complete data from an entire population,
so working with a sample gives us a glimpse of the population's characteristics.
  • The sample mean offers insight into the population mean.
  • Lastly, measures like variance and standard deviation show how much spread or variability there is within the data.
Ultimately, understanding sample data analysis helps translate smaller observations into larger insights about trends and variability in a population.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The paper cited in Exercise \(4.65\) also reported values of single-leg power for a low workload. The sample mean for \(n=13\) observations was \(\bar{x}=119.8\) (actually \(119.7692\) ), and the 14 th observation, somewhat of an outlier, was \(159 .\) What is the value of \(\bar{x}\) for the entire sample?

The paper "Caffeinated Energy Drinks-A Growing Problem" (Drug and Alcohol Dependence \([2009]: 1-10)\) gave the accompanying data on caffeine per ounce for eight top-selling energy drinks and for 11 high-caffeine energy drinks: Top-Selling Energy Drinks \(\begin{array}{llllllll}9.6 & 10.0 & 10.0 & 9.0 & 10.9 & 8.9 & 9.5 & 9.1\end{array}\) High-Caffeine Energy Drinks \(\begin{array}{llllll}21.0 & 25.0 & 15.0 & 21.5 & 35.7 & 15.0\end{array}\) \(\begin{array}{lllll}33.3 & 11.9 & 16.3 & 31.3 & 30.0\end{array}\) The mean caffeine per ounce is clearly higher for the highcaffeine energy drinks, but which of the two groups of energy drinks (top-selling or high- caffeine) is the most variable with respect to caffeine per ounce? Justify your choice.

A student took two national aptitude tests. The national average and standard deviation were 475 and 100 , respectively, for the first test and 30 and 8 , respectively, for the second test. The student scored 625 on the first test and 45 on the second test. Use \(z\) scores to determine on which exam the student performed better relative to the other test takers.

Cost per serving (in cents) for six high-fiber cereals rated very good and for nine high-fiber cereals rated good by Consumer Reports are shown below. Write a few sentences describing how these two data sets differ with respect to center and variability. Use summary statistics to support your statements. Cereals Rated Very Good \(\begin{array}{llllll}46 & 49 & 62 & 41 & 19 & 77\end{array}\) Cereals Rated Good \(\begin{array}{llll}71 & 30 & 53 & 53\end{array}\) \(\begin{array}{lllll}67 & 43 & 48 & 28 & 54\end{array}\)

Because some homes have selling prices that are much higher than most, the median price is usually used to describe a "typical" home price for a given location. The three accompanying quotes are all from the San Luis Obispo Tribune, but each gives a different interpretation of the median price of a home in San Luis Obispo County. Comment on each of these statements. (Look carefully. At least one of the statements is incorrect.) a. "So we have gone from \(23 \%\) to \(27 \%\) of county residents who can afford the median priced home at \(\$ 278,380\) in SLO County. That means that half of the homes in this county cost less than \(\$ 278,380\) and half cost more." (October 11,2001\()\) b. "The county's median price rose to \(\$ 285,170\) in the fourth quarter, a \(9.6 \%\) increase from the same period a year ago, the report said. (The median represents the midpoint of a range.)" (February 13,2002 ) c. "Your median is going to creep up above \(\$ 300,000\) if there is nothing available below \(\$ 300,000\),' Walker said." (February 26, 2002)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.