/*! 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 A deficiency of the trace elemen... [FREE SOLUTION] | 91Ó°ÊÓ

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A deficiency of the trace element selenium in the diet can negatively impact growth, immunity, muscle and neuromuscular function, and fertility. The introduction of selenium supplements to dairy cows is justified when pastures have low selenium levels. Authors of the paper "Effects of Short-Term Supplementation with Selenised Yeast on Milk Production and Composition of Lactating Cows" (Australian J. of Dairy Tech., 2004: 199-203) supplied the following data on milk selenium concentration \((\mathrm{mg} / \mathrm{L})\) for a sample of cows given a selenium supplement and a control sample given no supplement, both initially and after a 9-day period. $$ \begin{array}{rrrrr} \text { Obs } & \text { Init Se } & \text { Init } & & \text { Final } \\ 1 & 11.4 & \text { Cont } & \text { Final Se } & \text { Cont } \\ 2 & 9.6 & 8.7 & 104.0 & 8.8 \\ 3 & 10.1 & 9.7 & 96.4 & 8.8 \\ 4 & 8.5 & 10.8 & 89.0 & 10.1 \\ 5 & 10.3 & 10.9 & 88.0 & 9.6 \\ 6 & 10.6 & 10.6 & 103.8 & 8.6 \\ 7 & 11.8 & 10.1 & 147.3 & 10.4 \\ 8 & 9.8 & 12.3 & 97.1 & 12.4 \\ 9 & 10.9 & 8.8 & 172.6 & 9.3 \\ 10 & 10.3 & 10.4 & 146.3 & 9.5 \\ 11 & 10.2 & 10.9 & 99.0 & 8.4 \\ 12 & 11.4 & 10.4 & 122.3 & 8.7 \\ 13 & 9.2 & 11.6 & 103.0 & 12.5 \\ 14 & 10.6 & 10.9 & 117.8 & 9.1 \\ 15 & 10.8 & & 121.5 & \\ 16 & 8.2 & & 93.0 & \end{array} $$ a. Do the initial Se concentrations for the supplement and control samples appear to be similar? Use various techniques from this chapter to summarize the data and answer the question posed. b. Again use methods from this chapter to summarize the data and then describe how the final Se concentration values in the treatment group differ from those in the control group.

Short Answer

Expert verified
Initial Se concentrations are similar; treatment group shows increased and variable final Se levels.

Step by step solution

01

Organize Initial Se Concentrations for Supplement and Control Groups

Organize the initial selenium concentration data for both the supplement and control groups from the table. For the supplement group, the initial data are: 11.4, 9.6, 10.1, 8.5, 10.3, 10.6, 11.8, 9.8, 10.9, 10.3, 10.2, 11.4, 9.2, 10.6, 10.8, 8.2. For the control group, the initial data available are: 8.7, 9.7, 10.8, 10.9, 10.6, 10.1, 12.3, 8.8, 10.4, 10.9, 11.6, 10.9, 10.4, and 8.8.
02

Calculate Descriptive Statistics for Initial Se Concentrations

Calculate the mean, median, and standard deviation for both groups' initial selenium levels. For the supplement group: Mean = \(10.29\), Median = \(10.3\), SD ≈ \(1.07\). For the control group: Mean = \(10.38\), Median = \(10.5\), and SD ≈ \(1.05\). These statistics provide insights into the central tendency and dispersion of initial selenium levels in each group.
03

Compare Initial Se Concentrations

Compare the calculated descriptive statistics for initial Se levels in both groups. Both groups have similar means and medians, suggesting that initial selenium concentrations are relatively comparable between the supplement and control groups.
04

Organize and Analyze Final Se Concentrations for Treatment and Control Groups

Identify the final selenium concentration values for the treatment (supplement) group: 104.0, 96.4, 89.0, 88.0, 103.8, 147.3, 97.1, 172.6, 146.3, 99.0, 122.3, 103.0, 117.8, 121.5, 93.0. For the control group, the final concentrations are: 8.8, 8.8, 10.1, 9.6, 8.6, 10.4, 12.4, 9.3, 9.5, 8.4, 8.7, 12.5, and 9.1.
05

Calculate Descriptive Statistics for Final Se Concentrations

Calculate the mean and standard deviation for final selenium levels in both groups. For the treatment group: Mean ≈ \(114.8\), SD ≈ \(27.47\). For the control group: Mean ≈ \(9.74\), SD ≈ \(1.27\). These statistics illustrate the differences in final selenium concentrations between the two groups.
06

Describe Differences in Final Se Concentrations

Analyze the differences in final selenium concentrations between the supplement and control groups. The treatment group shows a significantly higher mean and greater variability in selenium levels compared to the control group, indicating a substantial effect of selenium supplementation on increasing selenium concentration in milk.

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Key Concepts

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

Descriptive Statistics
When dealing with sets of data, it's important to use descriptive statistics to understand the patterns and key characteristics. In essence, descriptive statistics helps to summarize the data in a meaningful way. Some of the most common measures include:
  • Mean: This is the average value, offering a central point of the data set.
  • Median: The middle value when data is ordered, giving a sense of the central tendency.
  • Standard Deviation: This measures dispersion or how much the values deviate from the mean.
In the original exercise, both the supplement and control groups' initial selenium levels were analyzed using these statistics. For instance, both groups displayed similar mean and median values (\( 10.29\) and \( 10.3 \) for the supplement group vs \( 10.38 \) and \( 10.5 \) for the control group), indicating that there was not a significant initial difference. Understanding these basic yet essential metrics helps in deciding whether further steps, like testing, are needed.
Comparative Analysis
Comparative analysis is crucial when you want to evaluate differences between groups or conditions. This kind of analysis was used in the exercise to compare the initial and final selenium concentrations between the cows given supplements and those that were not.

It's essential to first calculate the mean and standard deviation to observe any major differences. In the exercise, the supplement group's final mean selenium concentration was \( 114.8 \) mg/L compared to only \( 9.74 \) mg/L for the control group. Note also the significantly larger standard deviation in the supplement group (\( 27.47 \)) versus the control group (\( 1.27 \)), highlighting greater variability.

In essence, comparative analysis serves to draw efficient and informative discernments about interventions, like a supplement's efficacy, focusing not only on average results but also on variability and distribution across data sets.
Selenium Supplementation
Selenium is a trace element crucial for various bodily functions, including immunity and reproductive health. Selenium supplementation in animals, such as dairy cows, can be necessary, especially in regions where natural forage lacks this essential nutrient.

The study in the exercise explores the effects of selenium supplementation on milk production in cows. Initially, both the supplementation and control groups had similar selenium levels, indicating no bias or initial advantage. However, post-supplementation, there was a significant increase in the selenium concentration in milk of cows in the treatment group, exemplifying that selenium supplements can effectively introduce the nutrient where it is deficient.

The implication is clear: in areas where pastures are naturally low in selenium, providing supplements can enhance the nutritional value of milk, thereby benefiting both the livestock and end consumers. This highlights the importance of targeted nutritional interventions in improving agricultural outputs and animal health.

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Most popular questions from this chapter

Every score in the following batch of exam scores is in the \(60 \mathrm{~s}, 70 \mathrm{~s}, 80 \mathrm{~s}\), or \(90 \mathrm{~s}\). A stem-and-leaf display with only the four stems \(6,7,8\), and 9 would not give a very detailed description of the distribution of scores. In such situations, it is desirable to use repeated stems. Here we could repeat the stem 6 twice, using \(6 \mathrm{~L}\) for scores in the low 60 s (leaves \(0,1,2,3\), and 4 ) and \(6 \mathrm{H}\) for scores in the high 60 s (leaves \(5,6,7,8\), and 9 ). Similarly, the other stems can be repeated twice to obtain a display consisting of eight rows. Construct such a display for the given scores. What feature of the data is highlighted by this display? \(\begin{array}{lllllllllllll}74 & 89 & 80 & 93 & 64 & 67 & 72 & 70 & 66 & 85 & 89 & 81 & 81 \\ 71 & 74 & 82 & 85 & 63 & 72 & 81 & 81 & 95 & 84 & 81 & 80 & 70 \\ 69 & 66 & 60 & 83 & 85 & 98 & 84 & 68 & 90 & 82 & 69 & 72 & 87\end{array}\)

Many universities and colleges have instituted supplemental instruction (SI) programs, in which a student facilitator meets regularly with a small group of students enrolled in the course to promote discussion of course material and enhance subject mastery. Suppose that students in a large statistics course (what else?) are randomly divided into a control group that will not participate in SI and a treatment group that will participate. At the end of the term, each student's total score in the course is determined. a. Are the scores from the SI group a sample from an existing population? If so, what is it? If not, what is the relevant conceptual population? b. What do you think is the advantage of randomly dividing the students into the two groups rather than letting each student choose which group to join? c. Why didn't the investigators put all students in the treatment group? Note: The article "Supplemental Instruction: An Effective Component of Student Affairs Programming" (J. of College Student Devel., 1997: 577-586) discusses the analysis of data from several SI programs.

The accompanying data set consists of observations on shear strength (lb) of ultrasonic spot welds made on a certain type of alclad sheet. Construct a relative frequency histogram based on ten equal-width classes with boundaries \(4000,4200, \ldots\). [The histogram will agree with the one in "Comparison of Properties of Joints Prepared by Ultrasonic Welding and Other Means" (J. of Aircraft, 1983: 552-556).] Comment on its features. $$ \begin{array}{lllllll} 5434 & 4948 & 4521 & 4570 & 4990 & 5702 & 5241 \\ 5112 & 5015 & 4659 & 4806 & 4637 & 5670 & 4381 \\ 4820 & 5043 & 4886 & 4599 & 5288 & 5299 & 4848 \\ 5378 & 5260 & 5055 & 5828 & 5218 & 4859 & 4780 \\ 5027 & 5008 & 4609 & 4772 & 5133 & 5095 & 4618 \\ 4848 & 5089 & 5518 & 5333 & 5164 & 5342 & 5069 \\ 4755 & 4925 & 5001 & 4803 & 4951 & 5679 & 5256 \\ 5207 & 5621 & 4918 & 5138 & 4786 & 4500 & 5461 \\ 5049 & 4974 & 4592 & 4173 & 5296 & 4965 & 5170 \\ 4740 & 5173 & 4568 & 5653 & 5078 & 4900 & 4968 \\ 5248 & 5245 & 4723 & 5275 & 5419 & 5205 & 4452 \\ 5227 & 5555 & 5388 & 5498 & 4681 & 5076 & 4774 \\ 4931 & 4493 & 5309 & 5582 & 4308 & 4823 & 4417 \\ 5364 & 5640 & 5069 & 5188 & 5764 & 5273 & 5042 \\ 5189 & 4986 & & & & & \end{array} $$

The first four deviations from the mean in a sample of \(n=5\) reaction times were \(.3, .9,1.0\), and \(1.3\). What is the fifth deviation from the mean? Give a sample for which these are the five deviations from the mean.

Consider a sample \(x_{1}, x_{2}, \ldots, x_{n}\) and suppose that the values of \(\bar{x}, s^{2}\), and \(s\) have been calculated. a. Let \(y_{i}=x_{i}-\bar{x}\) for \(i=1, \ldots, n\). How do the values of \(s^{2}\) and \(s\) for the \(y_{i}\) s compare to the corresponding values for the \(x_{i} s\) ? Explain. b. Let \(z_{i}=\left(x_{i}-\bar{x}\right) / s\) for \(i=1, \ldots, n\). What are the values of the sample variance and sample standard deviation for the \(z_{i} s\) ?

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