Tutored Problem Solving vs. Pure Worked Examples and …web.cs.wpi.edu/~nth/pubs_and_grants/papers/2009/Tech Report 1.pdfanalysis is a discussion on how best to foster student self-explaining

Tutored Problem Solving vs. Pure Worked Examples and Appendix

"The paper on page 1‐6 was submitted to Cognitive Science 2009. Space limits means that we could include the appendix to allow others to replicate this result. Therefore, there are two attached documents at the end. The first, starting on page 8 is the seven‐item pretest (repeated as the posttest). The second attachment beginning on page 11 is a thirty‐four page document that shows all the material within the experimental conditions. We hope by making these materials available, others will be able to more easily replicate and understand our results.

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Tutored Problem Solving vs. “Pure” Worked Examples

Ryung S. Kim ([email protected]) Department of Mathematical Sciences, Worcester Polytechnic Institute, 100 Institute Road Worcester, MA 01609 USA

Rob Weitz ([email protected]) Department of Computing and Decision Sciences, Seton Hall University, South Orange, NJ 07079

Neil T. Heffernan ([email protected]) Department of Computer Science, Worcester Polytechnic Institute

Nathan Krach ([email protected]) Department of Computer Science, Worcester Polytechnic Institute

Abstract At present a handful of comparisons have been made of different variants of worked examples and tutored problem solving in several different domains. There is some evidence to report a benefit of adding worked examples (WE) to current tutored problem solving (TPS) environments. Our research investigated whether a “pure” WE condition could compete with an intelligent tutoring condition. By pure we mean the WE condition does not include tutoring, a self-explanation component, or fading. We report on two experiments. We found that WE was superior to TPS on conceptual knowledge, and also used less time (i.e. it was more efficient). On the other hand, TPS was superior at producing more learning on procedural knowledge, but took more time. These results provide evidence that “pure” WE can be as efficient as TPS. We speculate that with self-disciplined students, pure WE is effective, but more research is needed before generalizing this result to other contexts.

Keywords: tutored problem solving, worked examples.

Introduction This study compares student learning using two approaches: tutored problem solving (TPS) and worked examples (WE). There is a long history of research in each area. Figuring out the relative strengths and weaknesses of worked examples and computer-based tutored solving is of interest to the Cognitive Science community as indicated by the awarding of the Cognition and Student Learning Prize at last year’s conference to researchers in this area. our work adds to the body of research in three ways. First we compare “pure” TPS with “pure” WE conditions. Students in the TPS condition received TPS remediation, while students in the WE condition received solely WE remediation (as opposed to TPS remediation, which appears to be the case in previous studies). Neither condition included a self-explanation component. In other words, the WE condition was a “passive instructional event” (Koedinger and Aleven 2007). Second we examined the effect of prior knowledge in the subject area as a mediating factor. Third we investigated how well a student’s preference for a particular form of instruction predicted which approach

was actually superior for that student in terms of learning outcomes.

The area of instructions was college-level introductory statistics. Problem solutions generally required multiple steps. The domain is naturally suited to both procedural and conceptual problems. We conducted two experiments; the first focused on the application of the binomial and Normal probability distributions and the second dealt with confidence intervals.

Simple Problem Solving vs. Worked Examples A number of studies have shown the benefits of learning from WE. Ward and Sweller (1990) and Sweller and Cooper (1985) compared simple problem solving with an alternating WE with problem solving condition. Building on this work, further research explored the advantages of adding a self-explanation component to WE Atkinson, Derry, Renkl, Wortham (2000) provide a comprehensive review of the WE (vs. simple problem solving) literature with a focus on how best to design WE. Included in their analysis is a discussion on how best to foster student self-explaining. One of their overarching conclusions (p. 197) is that “students who self-explain tend to outperform student who do not.”

Renkl, Atkinson and Maier (2000) and Renkl, Atkinson, Maier, & Staley (2002) explored the effectiveness of fading (successively removing worked-out solution steps) WE vs. traditional WE. Atkinson, Renkl and Merrill (2003) combined fading with prompts “designed to encourage learners to identify the underlying principle illustrated in each worked-out solution step.” They reported improved far transfer over WE with fading alone.

Intelligent Tutoring vs. Worked Examples Koedinger and Aleven (2007) review the literature regarding adding worked examples to cognitive tutors.

McLaren, Lim and Koedinger (2008a, 2008b) compare a cognitive tutor with a WE in the domain of chemistry (stoichiometry). The WE condition included an interactive self-explanation component. (The explanations are checked for correctness.) They found that students in

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the WE conditions did not learn significantly more than students in the TPS condition, however the WE condition was more efficient,

Schwonke, Wittwer, Alevan, Salden, Krieg and Renkel (2007) describe two studies, both comparing a cognitive tutor with faded worked examples (in the field of high school geometry). Both conditions included an interactive self-explanation element. The results of the first study showed no difference in conceptual or transfer learning though the WE group took less time. In the second study the example condition yielded better conceptual understanding (and again, reduced time). There was no difference in students’ transfer knowledge.

Salden. Aleven, Renkl, and Schwonke (2008) built on the above work, this time adding an adaptive fading WE condition to the cognitive tutor and fixed-fading WE conditions . (Adaptive here means that the rate of fading is based on student’s level of understanding.) The two experiments (lab and classroom) they conducted indicated an advantage to the adaptively faded condition.

Finally, Schwonke, Renkl, Krieg, Wittwer, Aleven & Salden (in press) undertook two experiments comparing cognitive tutor and faded WE conditions; both conditions included self-explaining activities. The first experiment yielded no difference in learning and less time for the WE condition. The second experiment indicated an advantage to WE for conceptual learning, no difference regarding procedural learning and, again, a time advantage for WE.

In each of the above cases, the WE example condition included self-explanation requirements and the WE condition provided tutoring support when the student was unable to solve the isomorphic problem. In the experiments described below we instead use a “pure” worked example condition that does not include any intelligent tutoring, self explanations or fading of prompts. This condition is meant to represent a “cleaner” test of the WE condition compared to TPS alone. Certainly, from a practical perspective if learning under worked examples can be comparable or better than tutored problem solving, then we can save time, money and effort on building tutors.

The Experiments As noted previously, our study involved college students taking an introductory statistics course. Statistics is a good domain for this research as it includes both procedural and conceptual components. The problems we categorized as conceptual measure what Garfield (2002) has called as the third level of statistical reasoning, or transitional reasoning. They measures student's ability "to correctly identify one or two dimensions of a statistical process without fully integrating these dimensions, such as, that a larger sample size leads to a narrower confidence interval, that a smaller standard error leads to a narrower confidence interval."

We performed two experiments: one for probability distributions (Binomial and Normal) and one for confidence intervals. The methodology undertaken for each experiment is described below.

Student Characteristics Participating students were enrolled in an introductory statistics course at Worcester Polytechnic Institute (WPI), a private university specializing in engineering and the sciences. Ninety-five students participated in each experiment. The tutorials and associated assessments were conducted as part of the course’s regular statistics lab sessions and as such were integrated elements of the course. Students in this study comprise freshmen (17%), sophomores (61%), juniors (15%), and seniors (7%). Student majors comprise Engineering (65%), Math/ Physics/Chemistry (7%), and Social Science/Computer Science/ Biology (27%).

Experiment 1 In the first experiment, we compared the effect TPS and WE on learning of Binomial and Normal probability distribution. The problems were all procedural in nature, and are typical of problems given in introductory statistics courses. The subject matter was taught on days preceding the experiment. There were no assignments or tests on these topics due before the experiment.

Each student was randomly assigned to one of the conditions listed in Table 1. Each student experienced both tutorial types. Each tutorial (TPS or WE) was composed of two, two-part isomorphic problems.

Table1: No. Students in Each Condition of Experiment 1

First tutorial (Method/Topic)

Second Tutorial

Students numbers

TPS/Binomial WE/Normal 20 TPS/Normal WE/Binomial 30 WE/Binomial TPS/Normal 30 WE/Normal TPS/Binomial 16

The ASSISTment System Our experiment was conducted via the ASSISTment.org intelligent tutoring system built by a team lead by Heffernan and Koedinger. It’s an intelligent tutoring system similar to the CTAT(Koedinger et al. 2004) used in some of the previously mentioned studies (McLaren, Lim & Koedinger, 2008a). It is similar in that the system provides the student with tutoring on the individual steps of a problem, generally breaking a problem down into 3-4 steps. For each step, a student would be asked to provide an answer, and would get feedback on their answer until they got it correct. In this study ASSISTments was used for the TPS condition and the WE condition. In order to

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help others understand, and possibly replicate our work, we have archived all of our materials (Heffernan 2009). Our system differs from the CTAT structure in several ways including that there is only one solution path and the intermediate solution goals are highlighted. Tutored Problem Solving Condition In this study the system was modified to force students to work through the TPS for the first problem of each pair. This “forced TPS” approach ensures that each student experiences tutoring. After completion of the first problem of the pair, the student is presented with an isomorphic problem and is asked by the system to provide the answer. If the student gets this second question correct, the student is done with the problem. If the student gets the answer incorrect or indicates that s/he needs help solving the problem, the system provides TPS support (and records that the student was unable to solve the problem).

“Pure” Worked Example Condition Student is presented with the first problem (same as the first problem under the TPS condition) and a worked solution to that problem. The student is then presented with an isomorphic problem (same second problem as in the TPS condition), which the student is expected to solve. The student has access to the first WE while trying to solve the second. If the student gets this second question correct, the student is done with the problem. If the student gets the answer incorrect or indicates that s/he needs help solving the problem, the system provides the worked solution for the problem for review by the student (and records that the student was unable to solve the problem).

Table 2: A Comparison of Intelligent Tutoring and Worked Examples

Tutored Problem Solving (TPS)

Worked Examples (WE)

First Problem

Student studies with forced TPS

Student studies WE.

Second Problem

Student is given opportunity to answer the question. If student answer is incorrect, the problem is marked incorrect and, TPS is provided. WE is provided.

Due to relatively little workload for these tutorials,

the students were allowed to work though both tutorials at their own pace. Less than 5% of the students failed to finish the tutorials on time. Students were allowed to move to the second tutorial once they complete first tutorial.

Statistical Models Used

We use two statistical models to analyze our data. We use the marginal regression model using Generalize Estimating Equations, or GEE, (Liang and Zeger 1986) and the Generalized Linear Mixed Model, or GLMM, using maximum likelihood estimation (Bates and Sarkar 2007). Emberston & Reise (2000) advise that cognitive psychologist use modern statistical methods that have more statistical power. To help readers who are not familiar with these statistical methods, but interested in our cognitive science findings, we include details of the statistical models in Appendix 1. Result of Experiment 1 In this experiment students worked on two problems, each with two parts. The design for the experiment is a pair-matched randomized design with two conditions (TPS and WE), each with two problems. We define learning in each case (TPS and WE) if the student gets the second isomorphic problem correct. From the number of students with discordant performances between two tutorials (i.e. off-diagonal numbers in table 3), it is clear more students did better under TPS. For example, there were 10 students who got both questions correct after TPS but no problems correct after WE.

From the marginal regression model (M1; See Appendix 1), the probability of a student solving the problem after a WE (pooled over two topics) is estimated as 53% (log odds 0.137) and that after a TPS is 63% (log odds 0.526). This difference between two tutorials was significant (p=0.047).

Table 3: Number of question answered correctly by

condition

Number of questions answered correctly after WE

0 1 2

Number of questions answered correctly after TPS

0 14 6 2 1 11 5 11 2 10 9 27

Experiment 2 The second experiment utilizes questions from the domain of one-sample confidence interval of the mean (with continuous observations). There were two types of problems: procedural and conceptual in nature. As in experiment 1, the general concepts of the topic were taught during the days preceding the trial, there were no assignments or tests on this topic due before the trial, and on the day of experiment there was no additional teaching from the instructor prior to the tutorials. The experiment consisted of three parts: pre-test, tutorial, and post-test. The pre-test and post-test were identical, and comprised of three conceptual problems and four procedural

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problems. In the experiment, we used a completely randomized design: approximately half the students took the TPS version of the tutorial and the other half took a WE version. The students were given 20 minutes to go through pre-test without any feedback, 40 minutes for one of two types of tutorial, and 20 minutes for the post test (Table 4). In order to control time, students were not allowed to move to next step until designated time passed. The design of the tutorials is equivalent to that of experiment 1. That is to say, the problems are presented in pairs using the same approach as experiment 1, the contents of the two tutorials was as equivalent as possible. The tutorials in experiment 2 were comprised of three problems. The first two problems were procedural and the last one (composed of four sub-problems) was conceptual.

Table 4: Outline of Experiment 2

One Sample Confidence Interval for the Mean Several Days Prior to Lab Session • Lecture on the topic During Lab Session 1. Pre-Test (20 min; students’ initial knowledge)

• 20 minutes • four procedural and three conceptual.

2. Condition (TPS or WE) • 40 minutes • 3 pairs of Problems: 2 procedural, one

conceptual (3 parts) 3. Post-Test (20 min; students’ knowledge after trial)

• Same problems as Pre-Test

Results of Experiment 2 Item-wise Learning Pooled Over the Two Conditions Student learning was clearly shown in all items. Item easiness, or the probability for the seven problems to be solved correctly in two tests (pre-test/post-test) were 22%/56%, 11%/21%, 35%/71%, 15%/46%, 75%/87%, 42%/61%, and 33%/69%, respectively (These numbers are estimated use M2c; See Appendix 1). Itemized Learning Pooled Over the Two Conditions Table 5 shows WE improves learning significantly for both problem types (p=0.031 and 0.020), and TPS improves learning significantly for procedural problems (p < 0.0001) but only moderately in conceptual problems (p=0.740). The table shows the size of learning, in probability estimated by our model (Appendix 1; M2a). For example, on average, TPS helps 23% more students to answer the procedural problems correctly (column four, row three) and only 1% for the conceptual problems (column 4, row 1).

Table 5: Log odds (and probabilities in parenthesis) to solve problem

PRE POST Diff p-value

conceptual:TPS -0.296 (43%*)

-0.231 (44%)

0.064 (1%) 0.740

conceptual:WE -0.355 (41%)

0.066 (52%)

0.421 (11%) 0.031

procedural:TPS -1.102 (25%)

-0.088 (48%)

1.014 (23%) <0.0001

procedural:WE -0.780 (31%)

-0.240 (44%)

0.540 ((13%) 0.020

Interaction of Tutor Type and Problem Type We note

the following trends in Table 4: in procedural problems, learning from TPS is 10% (Log Odds Ratio = = 1.014-0.540) greater than that from the WE. In conceptual problems, on the contrary, learning from WE is 10% (Log Odds Ratio = - = 0.421-0.064) greater than that of TPS. Neither of these two main differences was significant. However the interaction in learning between tutor types and problem types was significant ( = 0.831; p = 0.0347; See M2a in Appendix 1). In other words, the experiment shows significant evidence of learning benefit from changing tutorial types according to problem types. In practical terms, WE was more effective for conceptual problems, while TPS was more effective for procedural problems.

Comparison of Rate of Learning In addition to comparing the amount of learning, we took into account time students spend on tutorial. While we tried to control for the number of problems done, and not for time, for practical reasons of running a classroom we set a 40 minute time window to complete the problems. We thought 40 minutes represents a reasonable amount of time to complete the problems. Unfortunately, of the 95 student in the experiment, 16 did not complete the problems (13 in TPS condition and 3 in WE condition). Hence, student that did not finish the tutorial were recorded at 40 minutes. This represents uneven censoring of the data. We address this issue at the end of this section. On average, students spent 31 minutes (s=10.4) on TPS and 22 minutes (s=10.0) on WE. We estimated the learning rates per minute in two tutorial types. The rate in log odds ratio per minute by each condition is in Table 6 (M2b in Appendix 1). WE is significantly more efficient in conceptual learning than TPS (p=0.008). On the contrary, two tutorial types have similar efficiencies in procedural learning. The difference in rates are clear in Figure 1, which shows how the probability to solve problem changes as students spend more time on tutorials. The low learning rate of TPS for conceptual problems is notable in the figure (flat line). To see if uneven censoring caused bias, we repeated our analyses comparing only students who completed the tutorials, and the results remained unchanged.

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Table 6: Learning Rate (log odds ratio per minute)

TPS WE Diff Pvalue Conceptual 0.0002 0.0256 -0.025 0.008 Procedural 0.0358 0.0288 0.007 0.550

Figure 1: Change of probabilities to solve a problem type as student spends time on each tutorial type.

Interaction of Tutorial Effect and Student Preference We investigated whether there were interactions between tutorial type and student preference. Students were overwhelmingly positive about the laboratory experience. We asked students to indicate which method of instruction they preferred after the first experiment. Student responses were about evenly split. There was a trend indicating that students performed better on experiment two when given their preferred method of instruction but these results were not statistically significant. Interaction of Tutorial Effect and Student Prior Knowledge We also investigated the interaction between prior knowledge and tutor type. We measured student prior knowledge from previous quizzes in class. The topics of quizzes were not related to confidence interval: they were theoretical probability distributions and statistical design. The trend indicated that students with high prior knowledge learned more with WE, and those with low knowledge learned more with TPS. However, these results were not significant.

Conclusions and Future Research We report on two experiments and compared both

student procedural and conceptual knowledge on tutored and “pure” WE conditions. We found that the pure WE was superior to TPS on conceptual knowledge, and also used less time (i.e. more efficient). On the other hand, TPS was superior at producing more learning on procedural knowledge but took more time. On procedural

questions the efficiency of TPS and WE were not reliably different from each other. Given that creating worked examples is so much more cost effective than creating intelligent tutoring system, these results are of practical importance. However, we hesitate to generalize these conclusions to other environments where the self-discipline to use WE might be lower. In fact in a separate paper submitted to this conference we report (Razzaq et al. submitted) that in an inner city middle school, albeit a slightly different implementation of a worked example condition, we found that TPS was more effective than the worked example condition. This suggests to us that worked example can be effective for students that have the self-discipline needed to actively engage with the material. WPI undergrads might have the self-discipline needed, but in the case of inner middle school classroom, the extra interactivity of TPS might become more important. More work is required to further explore why and under which conditions this kind of straightforward WE condition performs so well.

Appendix 1 – Statistical Approaches Two popular regression models exist to analyze data with correlated responses. They are the marginal regression model using GEE, and GLMM using maximum likelihood estimation. The merits of inference based on GEE are that it is valid as long as only the marginal mean is correctly specified, and that it allows unstructured covariance with balanced data. The merits of GLMM are the ability to model multi-level effects and to predict individual performances. GEE is recommended when the main scientific question is on population effects and when design is balanced with moderate number of items. GLMM is recommended when there are multiple levels of effects, when it is necessary to estimate individual student performance or item difficulties, or when the correlation structure are different across experimental conditions (Fitzmaurice, 2004). In the first experiment, which was under a pair-matched randomized design with repeated binary responses, we used GEE with unstructured association to model , the ith student’s probability of solving jth problem correctly, to be determined by the tutorial type (k=1,2). That is,

log 1 M1

for 1, … , 95, 1,2. For the second experiment, we used the following three models. First, we use model M2a where the probability is determined by problem type (m=1,2), tutorial type (k=1,2), and the test (s=1,2). Two interesting parameters in this model are the learning difference between two tutorials in procedural problems,

, , , , , and that in conceptual problems, , , , , .

0 10 20 30 40

0.0

0.2

0.4

0.6

0.8

Minutes on Tutorial

The

prob

abilit

y of

co

rrec

tly s

olvi

ng a

pro

blem

TPS conceptualWE conceptualTPS proceduralWE procedural

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log 1 , , M2a

We used model M2b to take into account the time each student spent on tutorial (ti). In this model, , , is the rate of learning per minute for each problem type (m=1,2) , and each tutorial type (k=1,2),.

log 1 , , M2b

Finally, we used a multilevel GLMM (M2c) to estimate item easiness in the second experiment. In this model, we have Gaussian random effects corresponding to student knowledge on pre-test, , and on post-test,

, and problem difficulty, . This is an extension of Rasch IRT model (Doran et al. 2007) to measure student performance on each test. We defined item easiness, , , averaging easiness over two tutorial types.

log 1 , M2c

References Atkinson, R. K., Derry, S. J., Renkl, A., & Wortham, D.

(2000). Learning From Examples: Instructional Principles From The Worked Examples Research. Review of Educational Research, 70(2), 181–214.

Atkinson, R.K., Renkl, A. & Merrill, M.M. (2003). Transitioning From Studying Examples to Solving Problems: Effects of Self-Explanation Prompts and Fading Worked-Out Steps. Journal of Educational Psychology, 95(4), 774–783.

Bates D. & Sarkar D. (2007). lme4: Linear Mixed-Effects Models Using S4 Classes. R package version 0.9975-12, URL http://CRAN.R-project.org/.

Doran H., Bates D., Bliese P., Dowling M. (2007). Estimating the Multilevel Rasch Model: With the lme4 Package, Journal of Statistical Software. 20 (2). http://www.jstatsoft.org/v20/i02

Embretson, S. & Reise, S (2002) Item Response Theory for Psychologists. LEA: Mahaw, NJ.

Fitzmaurice G. M., Laird N. M., Ware J. H., (2004). Applied Longitudinal Analysis, Wiley-IEEE.

Heffernan, N. (2009) WPI Department of Computer Science Technical Report Number 2009-01. http://www.cs.wpi.edu/Research/techreports.html

Garfield, J. (2002) The Challenge of Developing Statistical Reasoning, Journal of Statistics Education Volume 10, Number 3 (2002)

Koedinger, K. R., Aleven, V., Heffernan. T., McLaren, B. & Hockenberry, M. (2004). Opening the door to non-programmers: Authoring intelligent tutor behavior by demonstration. In Lester et al (Eds.) Proceedings of 7th Annual Intelligent Tutoring Systems Conference, Springer pp.162-173.

McLaren, B.M., Lim, S., & Koedinger, K.R. (2008a). When is Assistance Helpful to Learning? Results in Combining Worked Examples and Intelligent Tutoring. In B. Woolf, E. Aimeur, R. Nkambou, S. Lajoie (Eds), Proceedings of the 9th International Conference on Intelligent Tutoring Systems, Lecture Notes in Computer Science, 5091 (pp. 677-680). Berlin: Springer.

McLaren, B.M., Lim, S., & Koedinger, K.R. (2008b). When and How Often Should Worked Examples be Given to Students? New Results and a Summary of the Current State of Research. In B. C. Love, K. McRae, & V. M. Sloutsky (Eds.), Proceedings of the 30th Annual Conference of the Cognitive Science Society (pp. 2176-2181). Austin, TX: Cognitive Science Society

Razzaq, L., Heffernan, N. Shrestha, P., Wei, X., Maharjan, A. & Heffernan, C. (in submission) Are Worked Examples an Effective Feedback Mechanism During Problem Solving? Submitted to Cognitive Science 2009.

Renkl, A., Atkinson, R. K., & Maier, U. H. (2000). From studying examples to solving problems: Fading worked-out solution steps helps learning. In L. Gleitman & A. K. Joshi (Eds.), Proceeding of the 22nd

Annual Conference of the Cognitive Science Society (pp. 393–398). Mahwah, NJ: Erlbaum.

Renkl, A., Atkinson, R. K., Maier, U. H., & Staley, R. (2002). From Example Study To Problem Solving: Smooth Transitions Help Learning. Journal of Experimental Education, 70, 293–315.

Salden, R., Aleven, V., Renkl, A., & Schwonke, R. (2008). Worked Examples And Tutored Problem Solving: Redundant Or Synergistic Forms Of Support? In C. Schunn (Ed.) Proceedings of the 30th Annual Meeting of the Cognitive Science Society, CogSci 2008 (pp. 659-664). New York, NY: Lawrence Earlbaum

Schwonke, R, Wittwer, J., Alevan, V., Salden, R., Krieg, C., & Renkel, A. (2007). Can Tutored Problem Solving Benefit From Faded Worked-Out Examples? In S. Vosniadou, D. Kayser & A. Protopapas (Eds.), Proceedings of EuroCogSci 07. The European Cognitive Science Conference 2007 (pp. 59 – 64). New York, NY: Erlbaum.

Schwonke, R., Renkl, A., Krieg, C., Wittwer, J., Aleven, V., & Salden, R. (in press). The Worked-Example Effect: Is It Just An Artefact Of Lousy Control Conditions? Computers in Human Behavior.

Sweller, J. & Cooper, G. A. (1985). The Use of Worked Examples As a Substitute For Problem Solving In Learning Algebra. Cognition and Instruction, 2, 59–89.

Ward, M. and Sweller, J. (1990). Structuring Effective Worked Examples. Cognition and Instruction, 7(1), 1- 39.

Problem Set "3. Problem set after exercise" id:[5823]

1) Assistment #29672 "29672 - Please answer the..."Please answer the following seven problems. We would like to know the average amount of time students spend studying for their classes eachweek. A random sample of 20 students are surveyed. They report spending an average of 7.6hours studying, with s = 1.2 hours. Based on these data, find a 95% confidence interval for theaverage amount of time a student studies for classes each week. You may assume thatdistribution of the time students spend is approximately normal. Please round to two decimalplaces.Multiple choice:

(7.07, 8.13) hours(6.91, 8.29) hours(7.16, 8.04) hours(5.09, 10.11) hours(6.83, 8.37) hours(7.14, 8.06) hours

(7.04, 8.16) hours

(7.47, 7.73) hours

2) Assistment #29674 "29674 - Problem 2 of 7 &n..."Problem 2 of 7 A particular psychological test is used to measure academic motivation. The average test score forall university students nationwide is 115. A university in the south of the country estimates themean test score for students in its economic faculty by testing a random sample of n students andconstructing a confidence interval based on their scores. Which of the following statements aboutthe confidence interval are true? Check all that apply.Check all that apply:

The resulting interval will contain 115.The 95 percent confidence interval for n = 100 will generally be more narrow than the 95 percentconfidence interval for n = 50.For n = 100, the 95 percent confidence interval will be wider than the 90 percent confidence interval.

3) Assistment #29675 "29675 - Problem 3 of 7 &n..."Problem 3 of 7 We would like to know: "What is the average price of a house in Franklin County, OH? " Wetook a random sample of 101 houses, and found the average to be $120,128 and the standarddeviation to be $50,324. Provide an answer using the 99% confidence interval. Please round tothe nearest ten dollars.Multiple choice:

(108290, 131970) dollars

Assistment - Printing Content http://www.assistment.org/build/print/sequence/5823?mode=custom&op_...

1 of 3 1/20/2009 10:22 AM

(106980, 133280) dollars

(118820, 121440) dollars(-12020, 252270) dollars(96700, 143560) dollars(108000, 132360) dollars(110100,130160) dollars(95100, 145160) dollars

4) Assistment #29676 "29676 - Problem 4 of 7 &n..."Problem 4 of 7 A study was planned to examine the length of a certain species of fish on Gull Lake. The initialplan was to take a random sample of 25 fish from this lake using a special net, and examine ourresults. Numerical summaries on lengths of the fish measured in this study are given. What is yourmost informative estimate of the average fish length for the entire lake, based on information fromthis sample? It is reasonable to assume that the length to be normally distributed. Please round totwo decimal places. mean 25.018median 25.295standard deviation 4.1831range 20.73min 12.67max 33.4N 25Multiple choice:

A 95% confidence interval of (23.59, 26.45) inchesA 95% confidence interval of (23.38, 26.66) inchesA 95% confidence interval of (24.67, 25.36) inchesA 95% confidence interval of (16.38, 33.65) inchesA 95% confidence interval of (23.81, 26.23) inches

A 95% confidence interval of (23.29, 26.74) inches

A 95% confidence interval of (23.91, 26.13) inches25.30 inches25.02 inches

5) Assistment #29679 "29679 - Problem 5 of 7 &n..."Problem 5 of 7 Suppose there is a population of test scores on a large, standardized exam for which the mean andstandard deviation are unknown. Two different random samples of 50 data values are taken fromthe population. One sample has a larger sample standard deviation (SD) than the other. Each ofthe samples is used to construct a 95% confidence interval. How do you think these twoconfidence intervals would compare?


2 of 3 1/20/2009 10:22 AM

Multiple choice:

The two samples would produce identical values for the lower and upper bounds of the twoconfidence intervals.The confidence interval based on the sample with the smaller standard deviation would be wider.

The confidence interval based on the sample with the larger standard deviation would be wider.

The two confidence intervals would have the same width because they are both 95% intervals.

6) Assistment #29680 "29680 - Problem 6 of 7 &n..."Problem 6 of 7

The average number of fruit candies in a large bag is estimated. The 95% confidence interval is(40,48). After calculating this confidence interval, you discover an additional 10 bags of thesecandies. You add that data to the original set of counts and recalculate the confidence interval.Which of the following confidence intervals is most likely to result?Multiple choice:

41 to 4638 to 4838 to 50

42 to 46

40 to 50

7) Assistment #30812 "30812 - Problem 7 of 7 &n..."Problem 7 of 7 At a certain plant, batteries are being produced with a life expectancy that has a populationstandard deviation of 5.76 months. Suppose the mean life expectancy in a simple random sampleof 64 batteries is 12.35 months. Find a 90% confidence interval estimate of life expectancy for allthe batteries produced at this plant. Please round to two decimal places.Multiple choice:

(10.67, 14.02) months(11.15, 13.55) months

(11.17, 13.53) months

(11.43, 13.27) months(12.2, 12.5) months(2.9, 21.8) months(10.70, 14.00) months(10.50, 14.20) months


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Problem Set "Practice Confidence Interval " id:[5834]

1) Assistment #30253 "30253 - An auto manufactu..."a) An auto manufacturer makes a type of car that is known to have an average of 31.5 mpg on thehighway and standard deviation 6.6 mpg. They want to put a smaller engine in this type of car,and they want to see if this increases the average mpg. They select a random sample of 100 carswith the new engine, and those cars average 32.8 mpg. You may assume the same (population)standard deviation in mpg as that for the old engine.

They want to estimate the average mpg for cars with the new smaller engine, and therefore havedecided to compute the 95% confidence interval. Carry out the procedure and specify the intervalbelow. Please give your answer to one decimal place.

Let's break this problem down and solve it step by step.

Our best point estimate for the population mean goes in the center of the confidence interval.What’s your best point estimate (or best one-number guess) for the populations mean?Algebra:

32.8 Hints:The best point estimate for the population mean is the sample mean.The center of the confidence interval for the population mean is the sample mean. For thisproblem, the sample mean is 32.8 mpg. Please type in 32.8.

b) We’ve established that the center of the confidence interval is at the sample mean of 32.8 mpg. We now need to calculate the interval on either side of the sample mean that comprises the 95%confidence interval. This interval is comprised of two components: a critical value of the z or tstatistic and the standard error. Put another way, the general expression for the confidenceinterval is given by:

sample mean ± (critical value of z or t) * (standard error).

For this problem, should we use the z or t statistic?

Multiple choice: t statistic z statistic

Hints:The population standard deviation is known.In this case, we use the z statistic as the population standard deviation is known.

c) Again, the general expression for the confidence interval is given by:


We’ve established that we should use the z statistic for this problem. What value for z should weuse?Algebra:

1.96 Hints:The z value is defined by the area the statistic cuts off in the tail of the Normal distribution.For a 95% confidence interval, we’re interested in the z value that cuts off a total area of 0.05in the tails of the distribution – that is, an area of 0.025 in each tail.z0.025 = 1.96

d) Again, the general expression for the confidence interval is given by:



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Which of the following is the correct mathematical expression for the standard error for thisproblem?

Multiple choice: (a) (b) (c) (d)

Hints:The standard error is the standard deviation of individual observations divided by the squareroot of the sample size.If we know the population standard deviation (σ), we use that value. If not, we use the samplestandard deviation (s) in its place.We’re given the population standard deviation: 6.6 mpg. The standard error is given by

e) So, returning to the original question,

An auto manufacturer makes a type of car that is known to have an average of 31.5 mpg on thehighway and standard deviation 6.6 mpg. They want to put a smaller engine in this type of car,and they want to see if this increases the average mpg. They select a random sample of 100 carswith the new engine, and those cars average 32.8 mpg. They want to estimate the average mpg forcars with the new smaller engine, and therefore have decided to compute the 95% confidenceinterval. Carry out the procedure and specify the interval below. Provide your answer to onedecimal place.

Multiple choice: (30.1, 32.9) mpg (31.7, 33.9) mpg (31.0, 34.6) mpg (31.4, 34.2) mpg (19.9,

45.7) mpg (31.6, 34.0) mpg (29.7, 33.3) mpg 30.4, 32.6) mpg Hints:


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2) Assistment #30254 "30254 - Now that you have..."Now that you have worked through the example, here is a similar problem! The fire endurance of a wall is the time required for a fire of a given intensity on one side of thewall to breach the wall, to raise the temperature at any single point on the opposite side of thewall to one pre-specified level or to raise the overall temperature on the opposite side of the wallto another pre-specified level. In 16 tests of a new type of wall, the mean fire endurance is 126minutes. It is known that the standard deviation for this type of wall is 16 minutes. The data showno evidence of non-normality. The building code requires 99% confidence that the man fireendurance of walls exceeds 120 minutes. Provide the 99% confidence interval to the nearestminute.

(Press 'Break this problem into steps' only if you do not know how to solve this problem.)Multiple choice:

(114, 138) minutes (117, 135) minutes (116, 136) minutes (119, 133) minutes (85, 177) minutes (123, 129) minutes (118, 134) minutes (113, 139) minutes

Scaffold:Our best point estimate for the population mean goes in the center of the confidence interval.What’s your best point estimate (or best one-number guess) for the populations mean?Algebra:

126 Hints:The best point estimate for the population mean is the sample mean.What value does the problem indicate: the sample proportion or sample mean?The center of the confidence interval for the population mean is the sample mean. For thisproblem, the sample mean is 126 minutes. Please type in 126.

Scaffold:We’ve established that the center of the confidence interval is at the sample mean of 126minutes. We now need to calculate the interval on either side of the sample mean thatcomprises the 95% confidence interval. This interval is comprised of two components: acritical value of the z or t statistic and the standard error. Put another way, the generalexpression for the confidence interval is given by:


For this problem, should we use the z or t statistic?Multiple choice:

z statistic t statistic Hints:The population standard deviation is known.In this case, we use the z statistic as the population standard deviation is known.

Scaffold:Again, the general expression for the confidence interval is given by:


We’ve established that we should use the z statistic for this problem. What value for z shouldwe use?Algebra:

2.58 Hints:The z value is defined by the area the statistic cuts off in the tail of the Normal distribution.For a 99% confidence interval, we’re interested in the z value that cuts off a total area of0.01 in the tails of the distribution – that is, an area of 0.005 in each tail.z0.025 = 2.58


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Hints:The standard error is the standard deviation of individual observations divided by thesquare root of the sample size.If we know the population standard deviation (σ), we use that value. If not, we use thesample standard deviation (s) in its place.

We’re given the population standard deviation: 16 minues. The standard error is given by

Scaffold:So, returning to the original question,

The fire endurance of a wall is the time required for a fire of a given intensity on one side of thewall to breach the wall, to raise the temperature at any single point on the opposite side of thewall to one pre-specified level or to raise the overall temperature on the opposite side of the wallto another pre-specified level. In 16 tests of a new type of wall, the mean fire endurance is 126minutes. It is known that the standard deviation for this type of wall is 16 minutes. The data showno evidence of non-normality. The building code requires 99% confidence that the man fireendurance of walls exceeds 120 minutes. Provide the 99% confidence interval to the nearestminute.

Multiple choice: (119, 133) minutes (114, 138) minutes (118, 134) minutes (117, 135) minutes

(85, 177) minutes (123, 129) minutes (116, 136) minutes (113, 139) minutes Hints:


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3) Assistment #30858 "30858 - Before beginning ..."a) Before beginning a study investigating the ability of the drug heparin to preventbronchoconstriction, baseline values of pulmonary function were measured for a sample of 12individuals with a history of exercise-induced asthma. The mean value of forced vital capacity(FVC) for the sample is = 4.49 liters and the standard deviation is s = 0.83 liters.

(a) Compute a two-sided 95% confidence interval for µ, the true population mean FVC. Provideyour answer to two decimal places.

Let's break this problem down and solve it step by step.

Our best point estimate for the population mean goes in the center of the confidence interval.What’s your best point estimate (or best one-number guess) for the populations mean?Algebra:

4.49 Hints:The best point estimate for the population mean is the sample mean.The center of the confidence interval for the population mean is the sample mean. For thisproblem, the sample mean is 4.49 liters. Please type in 4.49

b) We’ve established that the center of the confidence interval is at the sample mean of 4.49liters. We now need to calculate the interval on either side of the sample mean that comprises the95% confidence interval. This interval is comprised of two components: a critical value of the z ort statistic and the standard error. Put another way, the general expression for the confidenceinterval for the mean is given by:



z statistic t statistic Hints:The population standard deviation is not known.In this case we use the t statistic as the population standard deviation is not known (and we’reusing the sample standard deviation as an estimate in its place).

c) Again, the general expression for the confidence interval is given by:


We’ve established that we should use the t statistic for this problem. What value for t should weuse?Algebra:

2.2010 Hints:The t value is defined by two numbers, the number of degrees of freedom and the area thestatistic cuts off in the tail of its distribution.The number of degrees of freedom in this case is given by (n – 1).


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For a 95% confidence interval, we’re interested in the t value that cuts off a total area of .05 inthe tails of the distribution – that is, an area of .025 in each tail.tn-1, α/2 = t11, 0.025 = 2.2010

d) Again, the general expression for the confidence interval is given by:




Hints:The standard error is the standard deviation of individual observations divided by the squareroot of the sample size.If we know the population standard deviation (σ), we use that value. If not, we use the samplestandard deviation (s) in its place.

The answer is given by:

e) So, returning to the original question,

Before beginning a study investigating the ability of the drug heparin to preventbronchoconstriction, baseline values of pulmonary function were measured for a sample of 12individuals with a history of exercise-induced asthma. The mean value of forced vital capacity(FVC) for the sample is =4.49 liters and the standard deviation is s = 0.83 liters.

(a) Compute a two-sided 95% confidence interval for µ, the true population mean FVC.Multiple choice:

(3.96, 5.02) liters (4.02, 4.96) liters (2.66, 6.32) liters (4.10, 4.88) liters (4.34,4.64) liters (3.87, 5.11) liters (3.74, 5.24) liters (4.06, 4.92) liters

Hints:


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f) Before beginning a study investigating the ability of the drug heparin to preventbronchoconstriction, baseline values of pulmonary function were measured for a sample of 12individuals with a history of exercise-induced asthma. The mean value of forced vital capacity(FVC) for the sample is = 4.49 liters and the standard deviation is s = 0.83 liters. (b) In order for the confidence interval computed in (a) to really have 95% confidence level, whatassumption(s) is/are needed?

i) The sample size should be large enough for us to use the central limit theorem. ii) The sample used is a simple random sample. iii) The distribution of FVC is normal or approximately normal. iv) ii and iiiMultiple choice:

i) ii) iii) iv) Hints:We always need a random sample in order to compute an accurate confidence level.Given that the sample size is small (n < 30) the distribution of the sample means will not benormal if the distribution of the underlying population (in the case, the distribution of FVC) isnot at least approximately normal.The sample used must be a random sample -- this is a requirement for any sample to be used tocompute an accurate confidence level. For small sample sizes (n< 30), the underlyingdistribution, in this case the distribution of FVC, must be normal or approximately normal. (Forlarge sample sizes, we don’t have this requirement as the central limit theorem assures us thatthe distribution of the sample mean is normal.) Therefore the answer is (iv).

4) Assistment #30250 "30250 - Now that you have..."a) Now that you have worked through the example, here is a similar problem! Researchers measured the seated systolic blood pressure of 27 health white males. The resultingpublication reports =114.9 (mmHg) and s = 9.3 (mmHg).

(a) Compute a two-sided 99% confidence interval for the mean blood pressure in the populationform which the subjects were recruited. Provide your answer to one decimal place. (Press 'Break this problem into steps' only if you do not know how to solve this problem.)Multiple choice:

(109.9, 119.9) mmHg (111.2, 118.6) mmHg (111.4, 118.4) mmHg (110.3, 119.5)mmHg (89.1, 140.7) mmHg (111.9, 117.9) mmHg (113.9, 115.9) mmHg (111.8,118.0) mmHg

Scaffold:Our best point estimate for the population mean goes in the center of the confidence interval.What’s your best point estimate (or best one-number guess) for the populations mean?Algebra:

114.9 Hints:The best point estimate for the population mean is the sample mean.What value does the problem indicate: the sample proportion or sample mean?The center of the confidence interval for the population mean is the sample mean. For thisproblem, the sample mean is 114.9 mmHg. Please type in 114.9.


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Scaffold:We’ve established that the center of the confidence interval is at the sample mean of 114.9mmHg. We now need to calculate the interval on either side of the sample mean thatcomprises the 99% confidence interval. This interval is comprised of two components: acritical value of the z or t statistic and the standard error. Put another way, the generalexpression for the confidence interval for the mean is given by:



t statistic z statistic Hints:The population standard deviation is not known. In this case we use the t statistic as the population standard deviation is not known (andwe’re using the sample standard deviation as an estimate in its place).



We’ve established that we should use the t statistic for this problem. What value for t shouldwe use? Please round to three decimal places.Algebra:

2.7787 Hints:The t value is defined by two numbers, the number of degrees of freedom and the area thestatistic cuts off in the tail of its distribution.The number of degrees of freedom in this case is given by (n – 1).For a 99% confidence interval, we’re interested in the t value that cuts off a total area of.01 in the tails of the distribution – that is, an area of .005 in each tail.tn-1, α/2 = t26, 0.005 = 2.7787





Hints:The standard error is the standard deviation of individual observations divided by the


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square root of the sample size.If we know the population standard deviation (σ), we use that value. If not, we use thesample standard deviation (s) in its place.We don’t know the population standard deviation, so we use the sample standard

deviation: 0.83 liters. The standard error is given by

Scaffold:So, returning to the original question,

Researchers measured the seated systolic blood pressure of 27 health white males. The resultingpublication reports =114.9 (mmHg) and s = 9.3 (mmHg).

(a) Compute a two-sided 99% confidence interval for the mean blood pressure in the populationform which the subjects were recruited. Provide your answer to one decimal place.Multiple choice:


Hints:

b) Researchers measured the seated systolic blood pressure of 27 health white males. The resultingpublication reports =114.9 (mmHg) and s = 9.3 (mmHg).

(b)What conditions for the population and the study design are required for the validity of theprocedure you used in (a)?

i) The distribution of blood pressure measurements should be normal or approximately normal. ii) The sample size should be large enough for us to use the central limit theorem. iii) The sample used should be a simple random sample. iv) i and iii.Multiple choice:

i) ii) iii) iv) Hints:We always need a random sample in order to compute an accurate confidence level.Given that the sample size is small (n < 30) the distribution of the sample means will not benormal if the distribution of the underlying population (in the case, the distribution of bloodpressure measurements) is not at least approximately normal.The sample used must be a random sample -- this is a requirement for any sample to be used tocompute an accurate confidence level. For small sample sizes (n< 30), the underlying


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distribution, in this case the distribution of FVC, must be normal or approximately normal. (Forlarge sample sizes, we don’t have this requirement as the central limit theorem assures us thatthe distribution of the sample mean is normal.) Therefore the answer is (iv).

5)Duplicate assistment: Assistment #30858 "30858 - Before beginning ..." was not displayed.6)Duplicate assistment: Assistment #30250 "30250 - Now that you have..." was not displayed.7)Duplicate assistment: Assistment #30253 "30253 - An auto manufactu..." was not displayed.8)Duplicate assistment: Assistment #30254 "30254 - Now that you have..." was not displayed.9) Assistment #30257 "30257 - 5A"a) Suppose that a random sample of 64 state college students is asked to measure the length oftheir right foot in centimeters. The sample mean foot length is 23.4 centimeters. It is known fromrepeated data collections over time that the population standard deviation of right foot lengths forstate college students is 5.52 centimeters. The 95% confidence interval is calculated from thisdata. a) Imagine the sample had included 256 students instead of 64 students (and all else is the same).Another 95% confidence interval is calculated for the revised sample size of 256. How will thewidth of the two confidence intervals compare? i) The confidence interval when n = 256 will be twice as wide as the confidence interval when n =64.ii) The confidence interval when n = 256 will be four times as wide as the confidence intervalwhen n = 64.iii) The confidence interval when n = 256 will be half the width of the confidence interval when n= 64.iv) The confidence interval when n = 256 will be ¼ the width of the confidence interval when n =64.v) The two confidence intervals will be the same size. Let's break this problem down and solve it step by step. Which of the following is the correct mathematical expression for the confidence interval for thisproblem?


Hints:The general expression for the confidence interval for the mean is given by: sample mean ±(critical value of z or t) * (standard error).Here the population standard deviation is known, so we should use the z statistic.

The correct answer is .

b) We’ve established that the confidence interval for this problem is given by the expression.Which of the following is the expression for the width of the confidence interval?


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Multiple choice: (a) (b) (c)

Hints:

c)


Hints:

The value of is determined solely by the confidence level (e.g., 95%, 99%, etc.)

The value of stays the same as the sample size changes.

d)


Hints:Note that n is in the denominator of the expression for the width of the confidence interval.


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As n increases, the width of the confidence interval decreases.

e)

Suppose that a random sample of 64 state college students is asked to measure the length of theirright foot in centimeters. The sample mean foot length is 23.4 centimeters. It is known fromrepeated data collections over time that the population standard deviation of right foot lengths forstate college students is 5.52 centimeters. The 95% confidence interval is calculated from this data.

a) Imagine the sample had included 256 students instead of 64 students (and all else is the same).Another 95% confidence interval is calculated for the revised sample size of 256. How will thewidth of the two confidence intervals compare?

i) The confidence interval when n = 256 will be twice as wide as the confidence interval when n =64.ii) The confidence interval when n = 256 will be four times as wide as the confidence interval whenn = 64.iii) The confidence interval when n = 256 will be half the width of the confidence interval when n =64.iv) The confidence interval when n = 256 will be ¼ the width of the confidence interval when n =64.v) The two confidence intervals will be the same size.

Multiple choice: i) ii) iii) iv) v)

Hints:

f) (b) Consider the original data given in part (a):

Suppose that a random sample of 64 state college students is asked to measure the length of theirright foot in centimeters. The sample mean foot length is 23.4 centimeters. It is known fromrepeated data collections over time that the population standard deviation of right foot lengths forstate college students is 5.52 centimeters. The 95% confidence interval is calculated from thisdata.

b) Imagine the population standard deviation (σ) is 2.76 centimeters instead of 5.52 centimeters


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(and all else is the same). How will the width of the two confidence intervals compare?

i) The confidence interval when σ = 2.76 will be twice as wide as the confidence interval when σ= 5.52.ii) The confidence interval when σ = 2.76 will be four times as wide as the confidence intervalwhen σ = 5.52.iii) The confidence interval when σ = 2.76 will be half the width of the confidence interval whenσ = 5.52.iv) The confidence interval when σ = 2.76 will be ¼ the width of the confidence interval when σ= 5.52.v) The two confidence intervals will be the same size. Let's break this problem down and solve it step by step.



The correct answer is

g)


Hints:


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The confidence interval in this case, as noted above, is given by the expression

The width of a confidence interval is 2 * (critical value of z or t) * (standard error)

The width of the confidence interval is given by

h)


Hints:


The value of stays the same as the sample standard deviation changes.

i) You concluded that the width of the CI is and as σ decreases the quantity

remains unchanged. Now, as the standard deviation (σ) decreases, what happens to the quantity

?

a) It increases.b) It decreases.c) It stays the same.Multiple choice:

(a) (b) (c) Hints:Note that σ is in the numerator of the expression for the width of the confidence interval.As σ decreases, the width of the confidence interval decreases.

j) You concluded that the width of the confidence interval is , and as σ decreases,

the quantity remains unchanged, and the quantity decreases therefore narrowing the

width of the confidence interval. Returning to the original question:


b) Imagine the standard deviation (σ) is 2.76 centimeters instead of 5.52 centimeters (and all elseis the same). How will the width of the two confidence intervals compare?

i) The confidence interval when σ = 2.76 will be twice as wide as the confidence interval when σ


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= 5.52.ii) The confidence interval when σ = 2.76 will be four times as wide as the confidence intervalwhen σ = 5.52.iii) The confidence interval when σ = 2.76 will be half the width of the confidence interval whenσ = 5.52.iv) The confidence interval when σ = 2.76 will be ¼ the width of the confidence interval when σ= 5.52.v) The two confidence intervals will be the same size.Multiple choice:

i) ii) iii) iv) v) Hints:

The width of the confidence interval is

The width of the confidence interval decreases as the standard deviation decreases, because as

the standard deviation (σ) decreases the quantity decreases. We can look at this

intuitively as well, and say that as the standard deviation decreases we are more likely to get anaccurate estimate for µ, and therefore the width of the confidence interval decreases, all otherthings being equal.

k) c) Consider the original data given in part (a):


Imagine a 99% confidence interval is calculated as well. (The data remain the same.) How will thewidth of the two confidence intervals compare?

i) The confidence interval when the confidence level is 99% will be wider than the confidenceinterval when the confidence level is 95%ii) The confidence interval when the confidence level is 99% will be narrower than theconfidence interval when the confidence level is 95%iii) The confidence interval when the confidence level is 99% will be the same width as theconfidence interval when the confidence level is 95% Let's break this problem down and solve it step by step. Which of the following is the correct mathematical expression for the confidence interval for thisproblem?


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The correct answer is (d)

l)


Hints:

The confidence interval in this case, as noted above, is given by the expression .



m) You concluded that the width of the confidence interval is , so let’s take a look

at the two components of this expression, and , in turn. As the confidence level

increases, what happens to the quantity ?

a) It increases.b) It decreases.c) It stays the same.


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Hints:The standard deviation, σ, is unaffected by the confidence level.The sample size, n, is unaffected by the confidence level.

The quantity stays the same as the confidence level is increased, as neither the standard

deviation nor the sample size is affected by the choice of confidence level.

n) You concluded that the width of the CI is , and as the confidence level increases,

the quantity remains unchanged. Now, as the confidence level increases, what happens to

the quantity ?


(a) (b) (c) Hints:


The quantity represents the z value that cuts off an area of in each tail of the z

distribution. As the confidence level increases, the area cut off in the tails of the z distributiondecreases.

As the confidence level increases, the value of increases.

o) You concluded that the width of the confidence interval is , and as the

confidence level increases, the quantity remains unchanged, and the quantity increases.

Returning to the original question:



i) The confidence interval when the confidence level is 99% will be wider than the confidenceinterval when the confidence level is 95%ii) The confidence interval when the confidence level is 99% will be narrower than theconfidence interval when the confidence level is 95%iii) The confidence interval when the confidence level is 99% will be the same width as theconfidence interval when the confidence level is 95%Multiple choice:

i) ii) iii) Hints:

The expression for the confidence interval in this case is .


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The width of the confidence interval is .

As the confidence level increases from 95% to 99%, increases from 1.96 to 2.58 andtherefore the width of the confidence interval increases. We can look at this intuitively as well,and say that the wider the interval, the more confident we can be about that interval containingµ.

10) Assistment #30258 "30258 - Now that you have..."a) Now that you have worked through the example, here is a similar problem! A researcher at a biotechnology company is testing an artificial pancreas on laboratory rats. Shegives 40 diabetic rats, which have had this pancreas implanted, an initial dose of glucose insolution and then measures their blood-sugar levels (serum/plasma glucose) after 1 hour. Theaverage blood-sugar level was 199 mg/100 ml. It is known that the (population) standarddeviation blood-sugar levels for diabetic rats one hour after an initial dose of glucose in solution is55.5 mg/100ml. A 99% confidence interval was calculated from this data.

a) If the sample had included 360 rats (and all else had turned out the same), would theconfidence interval be narrower or wider than the original interval? By what factor would it benarrower or wider? How will the width of the two confidence intervals compare?

i) The confidence interval when n = 360 will be 1/3 the width of the confidence interval when n =40.ii) The confidence interval when n = 360 will be 1/9 the width of the confidence interval when n= 40.iii) The confidence interval when n = 360 will be three times as wide as the confidence intervalwhen n = 40.iv) The confidence interval when n = 360 will be nine times as wide as the confidence intervalwhen n = 40.v) The two confidence intervals will be the same size. (Press 'Break this problem into steps' only if you do not know how to solve this problem.)Multiple choice:

i) ii) iii) iv) v) Scaffold:

Which of the following is the correct mathematical expression for the confidence interval forthis problem?




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The correct answer is (d) .

Scaffold:We’ve established that the confidence interval for this problem is given by the expression .Which of the following is the expression for the width of the confidence interval?


Hints:The confidence interval in this case, as noted above, is given by the expression

.


The width of the confidence interval is given by .

Scaffold:

You concluded that the width of the confidence interval is , so let’s take a look

at the two components of this expression, and , in turn. As the sample size increases

(from 40 to 360), what happens to the quantity ?


It increases. It decreases. It stays the same. Hints:The value of z α/2 is determined solely by the confidence level (e.g., 95%, 99%, etc.)The value of z α/2 stays the same as the sample size changes.

Scaffold:

You concluded that the width of the confidence interval is and as n increases

the quantity remains unchanged. Now, as n increases (from 40 to 360), what happens to

the quantity (and therefore the width of the confidence interval)?

a) It increases.b) It decreases.c) It stays the same.

Multiple choice: It increases. It decreases. It stays the same.

Hints:Note that n is in the denominator of the expression for the width of the confidence interval.


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As n increases, the width of the confidence interval decreases.

Scaffold:

You concluded that the width of the confidence interval is , and as n increases, the

quantity remains unchanged, and the quantity decreases therefore narrowing the width

of the confidence interval. Returning to the original question:

A researcher at a biotechnology company is testing an artificial pancreas on laboratory rats. Shegives 40 diabetic rats, which have had this pancreas implanted, an initial dose of glucose insolution and then measures their blood-sugar levels (serum/plasma glucose) after 1 hour. Theaverage blood-sugar level was 199 mg/100 ml. It is known that the (population) standarddeviation blood-sugar levels for diabetic rats one hour after an initial dose of glucose in solution is55.5 mg/100ml. A 99% confidence interval was calculated from this data.


i) The confidence interval when n = 360 will be 1/3 the width of the confidence interval when n= 40.ii) The confidence interval when n = 360 will be 1/9 the width of the confidence interval when n= 40.iii) The confidence interval when n = 360 will be three times as wide as the confidence intervalwhen n = 40.iv) The confidence interval when n = 360 will be nine times as wide as the confidence intervalwhen n = 40.v) The two confidence intervals will be the same size.Multiple choice:

i) ii) iii) iv) v) Hints:

b) (b) Consider the original data given in part (a):

A researcher at a biotechnology company is testing an artificial pancreas on laboratory rats. She gives 40diabetic rats, which have had this pancreas implanted, an initial dose of glucose in solution and thenmeasures their blood-sugar levels (serum/plasma glucose) after 1 hour. The average blood-sugar level was199 mg/100 ml. It is known that the (population) standard deviation blood-sugar levels for diabetic rats onehour after an initial dose of glucose in solution is 55.5 mg/100ml. A 99% confidence interval was calculatedfrom this data.

b) Imagine the standard deviation (σ) had turned out to be 44.4 ml/100ml instead of 55.5 ml/100ml (and allelse is the same). How will the width of the two confidence intervals compare?


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(Press 'Break this problem into steps' only if you do not know how to solve this problem.)


Scaffold:Which of the following is the correct mathematical expression for the confidence interval forthis problem?

Multiple choice: a) b) c) d)



Scaffold:We’ve established that the confidence interval for this problem is given by the expression

. Which of the following is the expression for the width of the confidence

interval?




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.The width of a confidence interval is 2 * (critical value of z or t) * (standard error)


Scaffold:


at the two components of this expression, and , in turn. As the standard deviation

decreases (from 5.52 to 2.76) what happens to the quantity ?


(a) (b) (c) Hints:The value of z α/2 is determined solely by the confidence level (e.g., 95%, 99%, etc.)The value of z α/2 stays the same as the sample standard deviation changes.

Scaffold:

You concluded that the width of the and as σ decreases the quantity

remains unchanged. Now, as the standard deviation (σ) decreases, what happens to the

quantity ?


(a) (b) (c) Hints:Note that σ is in the numerator of the expression for the width of the confidence interval.As σ decreases, the width of the confidence interval decreases.

Scaffold:

You concluded that the width of the confidence interval is , and as σ decreases, the quantity

remains unchanged, and the quantity decreases therefore narrowing the width of the confidence

interval. Returning to the original question:

A researcher at a biotechnology company is testing an artificial pancreas on laboratory rats. She gives 40diabetic rats, which have had this pancreas implanted, an initial dose of glucose in solution and thenmeasures their blood-sugar levels (serum/plasma glucose) after 1 hour. The average blood-sugar level was199 mg/100 ml. It is known that the (population) standard deviation blood-sugar levels for diabetic rats onehour after an initial dose of glucose in solution is 55.5 mg/100ml. A 99% confidence interval was calculatedfrom this data.

b) Imagine the standard deviation (σ) had turned out to be 44.4 ml/100ml instead of 55.5 ml/100ml (and allelse is the same). How will the width of the two confidence intervals compare?


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Hints:


The width of the confidence interval decreases as the standard deviation decreases, because as the

standard deviation (σ) decreases the quantity decreases. We can look at this intuitively as well,

and say that as the standard deviation decreases we are more likely to get an accurate estimate for µ,and therefore the width of the confidence interval decreases, all other things being equal.As the standard deviation of 44.4 is 4/5 that of 55.5, we would expect the new confidence interval to be4/5 of the original.

c) c) Consider the original data given in part (a):



i) The confidence interval when the confidence level is 95% will be wider than the confidenceinterval when the confidence level is 99%ii) The confidence interval when the confidence level is 95% will be narrower than theconfidence interval when the confidence level is 99%iii) The confidence interval when the confidence level is 95% will be the same width as theconfidence interval when the confidence level is 99%


i) ii) iii) Scaffold:

Which of the following is the correct mathematical expression for the confidence interval forthis problem?


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Scaffold:We’ve established that the confidence interval for this problem is given by the expression

. Which of the following is the expression for the width of the confidence

interval?



.



Scaffold:


at the two components of this expression, and , in turn. As the confidence level

decreases, what happens to the quantity ?

a) It increases.b) It decreases.


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c) It stays the same.Multiple choice:

(a) (b) (c) Hints:The standard deviation, σ, is unaffected by the confidence level.The sample size, n, is unaffected by the confidence level.

The quantity stays the same as the confidence level is increased, as neither the

standard deviation nor the sample size is affected by the choice of confidence level.

Scaffold:

You concluded that the width of the , and as the confidence level increases, the

quantity remains unchanged. Now, as the confidence level decreases, what happens to

the quantity ?


(a) (b) (c) Hints:The value of z α/2 is determined solely by the confidence level (e.g., 95%, 99%, etc.)The quantity z α/2 represents the z value that cuts off an area of in each tail of the zdistribution. As the confidence level decreases, the area cut off in the tails of the zdistribution increases.As the confidence level decreases, the value of z α/2 decreases.

Scaffold:

You concluded that the width of the confidence interval is , and as the

confidence level decreases, the quantity remains unchanged, and the quantity

decreases. Returning to the original question:

A researcher at a biotechnology company is testing an artificial pancreas on laboratory rats.She gives 40 diabetic rats, which have had this pancreas implanted, an initial dose of glucosein solution and then measures their blood-sugar levels (serum/plasma glucose) after 1 hour.The average blood-sugar level was 199 mg/100 ml. It is known that the (population) standarddeviation blood-sugar levels for diabetic rats one hour after an initial dose of glucose insolution is 55.5 mg/100ml. A 99% confidence interval was calculated from this data.

Imagine a 95% confidence interval is calculated as well. (The data remain the same.) Howwill the width of the two confidence intervals compare?

i) The confidence interval when the confidence level is 95% will be wider than theconfidence interval when the confidence level is 99%ii) The confidence interval when the confidence level is 95% will be narrower than theconfidence interval when the confidence level is 99%iii) The confidence interval when the confidence level is 95% will be the same width as theconfidence interval when the confidence level is 99%Multiple choice:

i) ii) iii) Hints:

The expression for the confidence interval in this case is .


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As the confidence level decreases from 99% to 95%, decreases from 2.58 to 1.96 andtherefore the width of the confidence interval decreases. We can look at this intuitively aswell, and say that the narrower the interval, the less confident we can be about thatinterval containing µ.

11) Assistment #30218 "30218 - The following two..."a) The following two problems are given with a visible solution. Study them carefully as you will be asked to solve a very similarproblem. You may come back and read this solution at that time. Before beginning a study investigating the ability of the drug heparin to prevent bronchoconstriction, baseline values of pulmonaryfunction were measured for a sample of 12 individuals with a history of exercise-induced asthma. The mean value of forced vitalcapacity (FVC) for the sample is = 4.49 liters and the standard deviation is s = 0.83 liters.

(a) Compute a two-sided 95% confidence interval for µ, the true population mean FVC. Solution:

Multiple choice: Okay, I have finished reading.

b) (b) In order for the confidence interval computed in (a) to really have 95% confidence level,what assumption(s) is/are needed?


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i) The sample size should be large enough for us to use the central limit theorem. ii) The sample used is a simple random sample. iii) The distribution of FVC is normal or approximately normal. iv) ii) and iii)

Solution:

In order for to have a T-distribution, we must have a simple random sample and thedistribution of FVC must be (at least approximately) normal. We do not need sample size to belarge since we are not using central limit theorem. Therefore, (iv) is the correct answer.Multiple choice:

Okay, I have finished reading.

c) Now that you have worked through the example, here is a similar problem!Researchers measured the seated systolic blood pressure of 27 health white males. The resultingpublication reports = 114.9 (mmHg) and s = 9.3 (mmHg).

(a) Compute a two-sided 99% confidence interval for the mean blood pressure in the populationfrom which the subjects were recruited. (Press 'Break this problem into steps' only if you do not know how to solve this problem.)Multiple choice:


Scaffold:Sorry, that answer is incorrect. The following is the solution. (a) Compute a two-sided 99% confidence interval for the mean blood pressure in the population from which the subjects wererecruited.

Solution:


27 of 36 1/20/2009 10:17 AM


d) Researchers measured the seated systolic blood pressure of 27 health white males. The resultingpublication reports = 114.9 (mmHg) and s = 9.3 (mmHg).

(b)What conditions for the population and the study design are required for the validity of theprocedure you used in (a)?

i) The distribution of blood pressure measurements should be normal or approximately normal. ii) The sample size should be large enough for us to use the central limit theorem. iii) The sample used is a simple random sample. iv) i) and iii) (Press 'Break this problem into steps' only if you do not know how to solve this problem.)Multiple choice:

i) ii) iii) iv) Scaffold:

Sorry, that answer is incorrect. The following is the solution.

(b)What conditions for the population and the study design are required for the validity of theprocedure you used in (a)? Solution:


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In order for to have the t-distribution, we must have a simple random sample and thedistribution of blood pressure must be (at least approximately) normal. We do not need thesample size to be large since we are not using central limit theorem. Therefore, (iv) is thecorrect answer.Multiple choice:


12) Assistment #30222 "30222 - The following pro..."a) The following problem is given with a visible solution. Study it carefully as you will be asked to solve a very similar problem. You macome back and read this solution at that time.

An auto manufacturer makes a type of car that is known to have an average of 31.5 mpg on the highway and standard deviation 6.6 mpgThey want to put a smaller engine in this type of car, and they want to see if this increases the average mpg. They select a random sample100 cars with the new engine, and those cars average 32.8 mpg. You may assume the same (population) standard deviation in mpg as thathe old engine. They want to estimate the average mpg for cars with the new smaller engine, and therefore have decided to compute the 95% confidenceinterval. Carry out the procedure and specify the interval below. Solution:

Multiple choice:



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b) Now that you have worked through the example, here is a similar problem!

The fire endurance of a wall is the time required for a fire of a given intensity on one side of thewall to breach the wall, to raise the temperature at any single point on the opposite side of thewall to one pre-specified level or to raise the overall temperature on the opposite side of the wallto another pre-specified level. In 16 tests of a new type of wall, the mean fire endurance is 126minutes. It is known that the standard deviation for this type of wall is 16 minutes. The data showno evidence of non-normality. The building code requires 99% confidence that the mean fireendurance of walls exceeds 120 minutes. Provide 99% confidence interval. (Press 'Break this problem into steps' only if you do not know how to solve this problem.)Multiple choice:

(114, 138) minutes (118, 134) minutes (117, 135) minutes (119, 133) minutes (85, 177) minutes (123, 129) minutes (113, 139) minutes (116, 136) minutes

Scaffold:Sorry, that answer is incorrect. The following is the solution.

Multiple choice:


13)Duplicate assistment: Assistment #30222 "30222 - The following pro..." was not displayed.14)Duplicate assistment: Assistment #30218 "30218 - The following two..." was not displayed.15) Assistment #30226 "30226 - The following thr..."


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a) The following three problems are given with a visible solution. Study them carefully as you will be asked to solve a very similar setof problems. You may come back and read these solutions at that time.

Suppose that a random sample of 64 state college students is asked to measure the length of their right foot in centimeters. The samplemean foot length is 23.4 centimeters. It is known from repeated data collections over time that the population standard deviation of rightfoot lengths for state college students is 5.52 centimeters. The 95% confidence interval is calculated from this data.

a) Imagine the sample had included 256 students instead of 64 students (and all else is the same). Another 95% confidence interval iscalculated for the revised sample size of 256. How will the width of the two confidence intervals compare?

i) The confidence interval when n = 256 will be twice as wide as the confidence interval when n = 64.ii) The confidence interval when n = 256 will be four times as wide as the confidence interval when n = 64.iii) The confidence interval when n = 256 will be half the width of the confidence interval when n = 64.iv) The confidence interval when n = 256 will be ¼ the width of the confidence interval when n = 64.v) The two confidence intervals will be the same size.

Solution:


b) (b) Consider the original data given in part (a):

Suppose that a random sample of 64 state college students is asked to measure the length of their right foot in centimeters. The samplemean foot length is 23.4 centimeters. It is known from repeated data collections over time that the population standard deviation ofright foot lengths for state college students is 5.52 centimeters. The 95% confidence interval is calculated from this data.

b) Imagine the population standard deviation (σ) is 2.76 centimeters instead of 5.52 centimeters (and all else is the same). How will thewidth of the two confidence intervals compare?

i) The confidence interval when σ = 2.76 will be twice as wide as the confidence interval when σ = 5.52.ii) The confidence interval when σ = 2.76 will be four times as wide as the confidence interval when σ = 5.52.iii) The confidence interval when σ = 2.76 will be half the width of the confidence interval when σ = 5.52.iv) The confidence interval when σ = 2.76 will be ¼ the width of the confidence interval when σ = 5.52.v) The two confidence intervals will be the same size.


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Solution:

Multiple choice:


c) c) Consider the original data given in part (a):

Suppose that a random sample of 64 state college students is asked to measure the length of their right foot in centimeters. The samplemean foot length is 23.4 centimeters. It is known from repeated data collections over time that the population standard deviation of rightfoot lengths for state college students is 5.52 centimeters. The 95% confidence interval is calculated from this data.

Imagine a 99% confidence interval is calculated as well. (The data remain the same.) How will the width of the two confidence intervalscompare?

i) The confidence interval when the confidence level is 99% will be wider than the confidence interval when the confidence level is 95%ii) The confidence interval when the confidence level is 99% will be narrower than the confidence interval when the confidence level is95%iii) The confidence interval when the confidence level is 99% will be the same width as the confidence interval when the confidence leveis 95% Solution:


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d) Now that you have worked through the example, here is a similar problem! Now that you have worked through the example, here is a similar problem!



i) The confidence interval when n = 360 will be 1/3 the width of the confidence interval when n= 40.ii) The confidence interval when n = 360 will be 1/9 the width of the confidence interval when n= 40.iii) The confidence interval when n = 360 will be three times as wide as the confidence intervalwhen n = 40.iv) The confidence interval when n = 360 will be nine times as wide as the confidence intervalwhen n = 40.v) The two confidence intervals will be the same size.


i) ii) iii) iv) v) Scaffold:Sorry, that answer is incorrect. The following is the solution.


33 of 36 1/20/2009 10:17 AM

Multiple choice:


e) (b) Consider the original data given in part (a):


b) Imagine the standard deviation (σ) had turned out to be 44.4 ml/100ml instead of 55.5ml/100ml (and all else is the same). How will the width of the two confidence intervals compare?

i) The confidence interval when σ = 44.4 will be 4/5 the width of the confidence interval when σ= 55.5.ii) The confidence interval when σ = 44.4 will be 5/4 the width of the confidence interval when σ= 55.5.iii) The confidence interval when σ = 44.4 will be the width of the confidence interval when σ =55.5.iv) The confidence interval when σ = 44.4 will be the width of the confidence interval when σ =55.5.v) The two confidence intervals will be the same size. (Press 'Break this problem into steps' only if you do not know how to solve this problem.)Multiple choice:

i) ii) iii) iv) v) Scaffold:Sorry, that answer is incorrect. The following is the solution.


34 of 36 1/20/2009 10:17 AM


f) c) Consider the original data given in part (a):



i) The confidence interval when the confidence level is 95% will be wider than the confidenceinterval when the confidence level is 99%ii) The confidence interval when the confidence level is 95% will be narrower than theconfidence interval when the confidence level is 99%iii) The confidence interval when the confidence level is 95% will be the same width as theconfidence interval when the confidence level is 99% (Press 'Break this problem into steps' only if you do not know how to solve this problem.)Multiple choice:

i) ii) iii) Scaffold:Sorry, that answer is incorrect. The following is the solution.


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36 of 36 1/20/2009 10:17 AM

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