/*! 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 66 The accompanying data on milk vo... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 66

The accompanying data on milk volume (in grams per day) were taken from the paper "Smoking During Pregnancy and Lactation and Its Effects on Breast Milk Volume" (American Journal of Clinical Nutrition [1991]: \(1011-1016)\) : \(\begin{array}{lllrrll}\text { Smoking } & 621 & 793 & 593 & 545 & 753 & 655 \\\ \text { mothers } & 895 & 767 & 714 & 598 & 693 & \\ \text { Nonsmoking } & 947 & 945 & 1086 & 1202 & 973 & 981 \\ \text { mothers } & 930 & 745 & 903 & 899 & 961 & \end{array}\) Compare and contrast the two samples.

Short Answer

Expert verified
As per the analysis, the mean milk volume tends to be higher with a more variable distribution for non-smoking mothers than for smoking mothers. It suggests that smoking during pregnancy and lactation could potentially have a detrimental effect on breast milk volume.

Step by step solution

01

List Out the Data

Firstly, list out the milk volumes for smoking and nonsmoking mothers. Smoking mothers milk volume (in g/day): \(621, 793, 593, 545, 753, 655, 895, 767, 714, 598, 693\) and nonsmoking mothers milk volume (in g/day): \(947, 945, 1086, 1202, 973, 981, 930, 745, 903, 899, 961\)
02

Calculate the Mean

Next, we calculate the mean (average) of the two samples using the formula: \(\text{mean} = \frac{\text{sum of all data}}{\text{number of data points}}\). To do this, add up all the values in each data set, and then divide by the number of data points in each set.
03

Calculate the Standard Deviation

Then find the standard deviation of each sample. The standard deviation is another measure of spread which shows the average distance from each data point to the mean. We can calculate it using the formula: \(\text{standard deviation} = \sqrt{\frac{\sum (x-\text{mean})^2}{\text{number of data points}-1}}\). It involves subtracting the mean from each data point, squaring the result, summing these squared results, dividing by the number of data points minus 1, and then taking the square root.
04

Inspecting the Data Distributions

For a more comprehensive comparison, try plotting a histogram for each dataset. Maps the milk volumes on the x-axis and the frequency of each volume on the y-axis for both datasets. This would provide a visual representation of both datasets, thereby assisting in their comparison and contrast.
05

Compare and Contrast the Results

Finally, after obtaining the above results, compare and contrast the measures of central tendency and variability (mean and standard deviation) for both groups. You can also describe the shape of the histograms and highlight any noticeable differences in the distributions of the two samples.

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 91影视!

Key Concepts

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

Mean Calculation
To begin understanding datasets, calculating the mean is crucial. The mean tells us the average of a dataset, providing a central value that represents the dataset as a whole. Calculating the mean involves adding up all the values in the dataset and then dividing by the number of values.
For instance, consider the two data sets for smoking and nonsmoking mothers from our exercise. The mean gives us a quick idea about the central tendency of milk volume in each group.
Knowing the mean helps us compare the average milk production between smoking and nonsmoking mothers. With this average, you can start to see which group generally produces more milk.
Standard Deviation
While the mean tells us about the average, the standard deviation gives insight into the variability of the data. It measures how spread out the numbers are around the mean.
To compute it, subtract the mean from each number, square the result, then average those squared differences. Finally, take the square root of that average.
This statistic helps us understand how consistent the milk production is within each group. A smaller standard deviation indicates that most data points are close to the mean, while a larger one means more variation within the sample. Understanding this concept helps us gauge the reliability and consistency of the data.
Data Distribution
Data distribution describes how values are spread across a dataset. Knowing the distribution helps in understanding the general pattern and any outliers.
For our datasets, after calculating both the mean and standard deviation, we can describe how the milk production data points are dispersed for each group.
  • A tightly clustered set of data points would suggest a normal distribution.
  • Conversely, data points with frequent high or low values can indicate skewed distributions.
Understanding the distribution is essential to fully interpret our data and make accurate comparisons between groups.
Histogram Plotting
Creating histograms is a practical way to visualize distribution. A histogram is a type of bar chart that shows the frequency of data points within specified intervals.
Plotting histograms for our datasets from smoking and nonsmoking mothers will clearly highlight the frequency of different milk volumes.
By looking at the histograms, you can easily see where most data points lie and identify any skewness or patterns in the datasets. This visual representation is helpful in comparing the overall dispersion and central tendency of the datasets.

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

Houses in California are expensive, especially on the Central Coast where the air is clear, the ocean is blue, and the scenery is stunning. The median home price in San Luis Obispo County reached a new high in July 2004 soaring to \(\$ 452,272\) from \(\$ 387,120\) in March 2004. (San Luis Obispo Tribune, April 28,2004\()\). The article included two quotes from people attempting to explain why the median price had increased. Richard Watkins, chairman of the Central Coast Regional Multiple Listing Services was quoted as saying "There have been some fairly expensive houses selling, which pulls the median up." Robert Kleinhenz, deputy chief economist for the California Association of Realtors explained the volatility of house prices by stating: "Fewer sales means a relatively small number of very high or very low home prices can more easily skew medians." Are either of these statements correct? For each statement that is incorrect, explain why it is incorrect and propose a new wording that would correct any errors in the statement.

In 1997 a woman sued a computer keyboard manufacturer, charging that her repetitive stress injuries were caused by the keyboard (Genessey v. Digital Equipment Corporation). The jury awarded about \(\$ 3.5\) million for pain and suffering, but the court then set aside that award as being unreasonable compensation. In making this determination, the court identified a "normative" group of 27 similar cases and specified a reasonable award as one within 2 standard deviations of the mean of the awards in the 27 cases. The 27 award amounts were (in thousands of dollars) \(\begin{array}{rrrrrrrr}37 & 60 & 75 & 115 & 135 & 140 & 149 & 150 \\ 238 & 290 & 340 & 410 & 600 & 750 & 750 & 750 \\\ 1050 & 1100 & 1139 & 1150 & 1200 & 1200 & 1250 & 1576 \\ 1700 & 1825 & 2000 & & & & & \end{array}\) What is the maximum possible amount that could be awarded under the "2-standard deviations rule"?

The Highway Loss Data Institute reported the following repair costs resulting from crash tests conducted in October 2002 . The given data are for a 5 -mph crash into a flat surface for both a sample of 10 moderately priced midsize cars and a sample of 14 inexpensive midsize cars. \(\begin{array}{lrrrrr}\text { Moderately Priced } & 296 & 0 & 1085 & 148 & 1065 \\ \text { Midsize Cars } & 0 & 0 & 341 & 184 & 370 \\ \text { Inexpensive } & 513 & 719 & 364 & 295 & 305 \\ \text { Midsize Cars } & 335 & 353 & 156 & 209 & 288 \\ & 0 & 0 & 397 & 243 & \end{array}\) a. Compute the standard deviation and the interquartile range for the repair cost of the moderately priced midsize cars. b. Compute the standard deviation and the interquartile range for the repair cost of the inexpensive midsize cars. c. Is there more variability in the repair cost for the moderately priced cars or for the inexpensive midsize cars? Justify your choice. d. Compute the mean repair cost for each of the two types of cars. e. Write a few sentences comparing repair cost for moderately priced and inexpensive midsize cars. Be sure to include information about both center and variability.

The standard deviation alone does not measure relative variation. For example, a standard deviation of \(\$ 1\) would be considered large if it is describing the variability from store to store in the price of an ice cube tray. On the other hand, a standard deviation of \(\$ 1\) would be considered small if it is describing store-to-store variability in the price of a particular brand of freezer. A quantity designed to give a relative measure of variahility is the \(\mathrm{co-}\) efficient of variation. Denoted by CV, the coefficient of variation expresses the standard deviation as a percentage of the mean. It is defined by the formula \(C V=100\left(\frac{s}{\bar{x}}\right)\). Consider two samples. Sample 1 gives the actual weight (in ounces) of the contents of cans of pet food labeled as having a net weight of 8 oz. Sample 2 gives the actual weight (in pounds) of the contents of bags of dry pet food labeled as having a net weight of \(50 \mathrm{lb}\). The weights for the two samples are: \(\begin{array}{lrrrrr}\text { Sample 1 } & 8.3 & 7.1 & 7.6 & 8.1 & 7.6 \\ & 8.3 & 8.2 & 7.7 & 7.7 & 7.5 \\ \text { Sample 2 } & 52.3 & 50.6 & 52.1 & 48.4 & 48.8 \\ & 47.0 & 50.4 & 50.3 & 48.7 & 48.2\end{array}\) a. For each of the given samples, calculate the mean and the standard deviation. b. Compute the coefficient of variation for each sample. Do the results surprise you? Why or why not?

O The paper "The Pedaling Technique of Elite Endurance Cyclists" (International Journal of Sport Biomechanics [1991]: 29-53) reported the following data on single-leg power at a high workload: $$ \begin{array}{llllllll} 244 & 191 & 160 & 187 & 180 & 176 & 174 & 205 \\ 183 & 211 & 180 & 194 & 200 & & & \end{array} $$ a. Calculate and interpret the sample mean and median. b. Suppose that the first observation had been 204 , not 244\. How would the mean and median change? c. Calculate a trimmed mean by eliminating the smallest and the largest sample observations. What is the corresponding trimming percentage? d. Suppose that the largest observation had been 204 rather than \(244 .\) How would the trimmed mean in Part (c) change? What if the largest value had been 284 ?
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Statistics

Read Explanation

Decision Maths

Read Explanation

Geometry

Read Explanation
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.