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Chapter 1: Q5E (page 12)

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.]

Short Answer

Expert verified

a. No, All the students taking a large statistics course who participate in an SI program.

b. The randomization eliminates any biases and ensures that the students in the SI group are as similar as possible to the students in the control group.

c. The comparison of the SI scores cannot be done if all the students are put in the treatment group.

Step by step solution

01

Given information

It is given that the students in a large statistics course are divided into the control group that will not participate in the supplemental instruction (SI)program and a treatment group that will participate in SI.

02

Check whether the scores from the SI group a sample from an existing population

a.

The scores from the SI group are not a sample from an existing population because the population is divided into control and treatment group.

The relevant conceptual population is,

All the students taking a large statistics course who participate in an SI program.

03

State the advantage of dividing the students into two groups.

b.

The advantage of randomly dividing the students into two groups rather than letting each student choose which group to join is that the randomization eliminates any biases and ensures that the students in the SI group are as similar as possible to the students in the control group.

04

Explain why all the students cannot be in the treatment group.

c.

All students cannot be put in the treatment group because there will be no criteria for assessing the effectiveness of the SI.

Therefore, the comparison of the SI scores can not be done if all the students are put in the treatment group.

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Most popular questions from this chapter

A transformation of data values by means of some mathematical function, such as\(\sqrt x \)or\(1/x\), can often yield a set of numbers that has “nicer” statistical properties than the original data. In particular, it may be possible to find a function for which the histogram of transformed values is more symmetric (or, even better, more like a bell-shaped curve) than the original data. As an example, the article “Time Lapse Cinematographic Analysis of Beryllium–Lung FibroblastInteractions” (Environ. Research,1983: 34–43) reported the results of experiments designed to study the behavior of certain individual cells that had been exposed to beryllium. An important characteristic of such an individual cell is its interdivision time (IDT). IDTs were determined for a large number of cells, both in exposed (treatment) and unexposed(control) conditions. The authors of the articleused a logarithmic transformation, that is, transformed value=log(original value). Consider the following representative IDT data:

IDT log10(IDT) IDT log10(IDT) IDT log10(IDT)

28.1 1.45 60.1 1.78 21.0 1.32

31.2 1.49 23.7 1.37 22.3 1.35

13.7 1.14 18.6 1.27 15.5 1.19

46.0 1.66 21.4 1.33 36.3 1.56

25.8 1.41 26.6 1.42 19.1 1.28

16.8 1.23 26.2 1.42 38.4 1.58

34.8 1.54 32.0 1.51 72.8 1.86

62.3 1.79 43.5 1.64 48.9 1.69

28.0 1.45 17.4 1.24 21.4 1.33

17.9 1.25 38.8 1.59 20.7 1.32

19.5 1.29 30.6 1.49 57.3 1.76

21.1 1.32 55.6 1.75 40.9 1.61

31.9 1.50 25.5 1.41

28.9 1.46 52.1 1.72

Use class intervals 10-<20, 20-<30,… to construct a histogram of the original data. Use intervals1.1-<1.2, 1.2-<1.3,… to do the same for the transformed data. What is the effect of the transformation?

Let \({\rm{X}}\) denote the amount of space occupied by an article placed in a \({\rm{1 - f}}{{\rm{t}}^{\rm{3}}}\) packing container. The pdf of \({\rm{X}}\) is

\({\rm{f(x) = }}\left\{ {\begin{array}{*{20}{c}}{{\rm{90}}{{\rm{x}}^{\rm{8}}}{\rm{(1 - x)}}}&{{\rm{0 < x < 1}}}\\{\rm{0}}&{{\rm{ otherwise }}}\end{array}} \right.\)

a. Graph the pdf. Then obtain the cdf of \({\rm{X}}\) and graph it. b. What is \({\rm{P(X}} \le {\rm{.5)}}\) (i.e., \({\rm{F(}}{\rm{.5)}}\))? c. Using the cdf from (a), what is \({\rm{P(}}{\rm{.25 < X}} \le {\rm{.5)}}\)? What is \({\rm{P(}}{\rm{.25}} \le {\rm{X}} \le {\rm{.5)}}\)? d. What is the \({\rm{75th}}\) percentile of the distribution? e. Compute \({\rm{E(X)}}\) and \({{\rm{\sigma }}_{\rm{X}}}\). f. What is the probability that \({\rm{X}}\) is more than \({\rm{1}}\) standard deviation from its mean value?

The cumulative frequency and cumulative relativefrequency for a particular class interval are the sum offrequencies and relativefrequencies, respectively, forthat interval and all intervals lying below it. If, forexample, there are four intervals with frequencies 9,16, 13, and 12, then the cumulative frequencies are 9,25, 38, and 50, and the cumulative relative frequenciesare .18, .50, .76, and 1.00. Compute the cumulativefrequencies and cumulative relative frequencies for thedata of Exercise 24.

a. Give three different examples of concrete populations and three different examples of hypothetical populations.

b. For one each of your concrete and your hypothetical populations, give an example of a probability question and an example of an inferential statistics question.

Let \({\bar x_n}\) and \(s_n^2\) denote the sample mean and variance for the sample \({x_1},{x_2},...,{x_n}\) and let \({\bar x_{n + 1}}\) and \(s_{n + 1}^2\) denote these quantities when an additional observation \({x_{n + 1}}\) is added to the sample.

a. Show how\({\bar x_{n + 1}}\)can be computed from\({\bar x_n}\)and\({x_{n + 1}}\).

b. Show that

\(ns_{n + 1}^2 = \left( {n - 1} \right)s_n^2 + \frac{n}{{n + 1}}{\left( {{x_{n + 1}} - {{\bar x}_n}} \right)^2}\)

so that\(s_{n + 1}^2\)can be computed from\({x_{n + 1}}\),\({\bar x_n}\), and\(s_n^2\).

c. Suppose that a sample of 15 strands of drapery yarn has resulted in a sample mean thread elongation of 12.58 mm and a sample standard deviation of .512 mm. A 16th strand results in an elongation value of 11.8. What are the values of the sample mean and sample standard deviation for all 16 elongation observations?
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