/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 Does taking notes by hand in a s... [FREE SOLUTION] | 91影视

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Chapter 32: Problem 1

Does taking notes by hand in a statistics course improve performance? Some recent research suggests that this may be the case. 2 To explore this, six volunteers (Doug, Elizabeth, Oksana, Sebastian, Vishal, and Xinyi) agree to take part in an experiment. Four are assigned completely at random to take handwritten notes in class, and the other two are assigned to take notes on their laptops. Total points earned on the two in-class exams and final exam are used to determine course performance. The results are (out of a possible total of 500 points): $$ \begin{array}{ll} \hline \text { Handwritten Notes (Person) } & \text { Notes on Laptop (Person) } \\ \hline 380 \text { (Doug) } & 370 \text { (Elizabeth) } \\ \hline 400 \text { (Oksana) } & 310 \text { (Xinyi) } \\ \hline 420 \text { (Sebastian) } & \\ \hline 360 \text { (Vishal) } & \\ \hline \end{array} $$ (a) There are 15 possible ways the six subjects can be assigned to the two groups, with the handwritten notes group having size 4 and the laptop notes group size 2. List these. (b) For each, determine the difference in mean points (mean number of points for the handwritten notes group minus mean number of points for the laptop notes group). Combine any duplicates and make a table of the possible mean differences and the corresponding probability of each under the null hypothesis of no difference in the effect of the treatments on total points earned. (Each of the 15 possible assignments of subjects to treatments has probability \(1 / 15\) under the null hypothesis.) This is the permutation distribution. (c) Compute the \(P\)-value of the data. Assume the two-sided alternative hypothesis is that the mean number of points is different for the two groups. (d) In this example, is it possible to demonstrate significance at the \(5 \%\) level using the permutation test? Explain. (e) Assume that total number of points is Normally distributed for both groups. Use the two-sample \(t\) procedure to test the hypotheses. Use Option 1 if you have access to software.

Short Answer

Expert verified
Permutation test indicates no significant difference at 5% level. Use permutation test or two-sample t-test for analysis.

Step by step solution

01

List all possible assignments

Determine the 15 possible ways to assign 6 subjects to two groups where one group has 4 members (those taking notes by hand) and the other group has 2 members (those taking notes on laptops). The combinations can be listed as follows: 1. Doug, Oksana, Sebastian, Vishal | Elizabeth, Xinyi 2. Doug, Oksana, Sebastian, Elizabeth | Vishal, Xinyi 3. Doug, Oksana, Sebastian, Xinyi | Elizabeth, Vishal 4. Doug, Oksana, Vishal, Elizabeth | Sebastian, Xinyi 5. Doug, Oksana, Vishal, Xinyi | Sebastian, Elizabeth 6. Doug, Oksana, Elizabeth, Xinyi | Sebastian, Vishal 7. Doug, Sebastian, Vishal, Elizabeth | Oksana, Xinyi 8. Doug, Sebastian, Vishal, Xinyi | Oksana, Elizabeth 9. Doug, Sebastian, Elizabeth, Xinyi | Oksana, Vishal 10. Doug, Vishal, Elizabeth, Xinyi | Oksana, Sebastian 11. Oksana, Sebastian, Vishal, Elizabeth | Doug, Xinyi 12. Oksana, Sebastian, Vishal, Xinyi | Doug, Elizabeth 13. Oksana, Sebastian, Elizabeth, Xinyi | Doug, Vishal 14. Oksana, Vishal, Elizabeth, Xinyi | Doug, Sebastian 15. Sebastian, Vishal, Elizabeth, Xinyi | Doug, Oksana
02

Calculate the differences in mean points

Compute the mean points for the handwritten notes group and the laptop group for each assignment. Calculate the difference in means for each assignment: 1. Mean difference = (380 + 400 + 420 + 360)/4 - (370 + 310)/2 = 40 2. Mean difference = (380 + 400 + 420 + 370)/4 - (360 + 310)/2 = 90 3. Mean difference = (380 + 400 + 420 + 310)/4 - (370 + 360)/2 = 40 4. Mean difference = (380 + 400 + 360 + 370)/4 - (420 + 310)/2 = 40 5. Mean difference = (380 + 400 + 360 + 310)/4 - (420 + 370)/2 = -10 6. Mean difference = (380 + 400 + 370 + 310)/4 - (420 + 360)/2 = -10 7. Mean difference = (380 + 420 + 360 + 370)/4 - (400 + 310)/2 = -10 8. Mean difference = (380 + 420 + 360 + 310)/4 - (400 + 370)/2 = -60 9. Mean difference = (380 + 420 + 370 + 310)/4 - (400 + 360)/2 = -10 10. Mean difference = (380 + 360 + 370 + 310)/4 - (400 + 420)/2 = -60 11. Mean difference = (400 + 420 + 360 + 370)/4 - (380 + 310)/2 = 40 12. Mean difference = (400 + 420 + 360 + 310)/4 - (380 + 370)/2 = -10 13. Mean difference = (400 + 420 + 370 + 310)/4 - (380 + 360)/2 = -10 14. Mean difference = (400 + 360 + 370 + 310)/4 - (380 + 420)/2 = -60 15. Mean difference = (420 + 360 + 370 + 310)/4 - (380 + 400)/2 = -60
03

Construct the permutation distribution

Combine duplicate differences and determine the probability of each difference under the null hypothesis. The differences are 90, 40, -10, and -60. Each combination is equally likely with a probability of \( \frac{1}{15} \).- 90: 1 time, probability \( \frac{1}{15} \)- 40: 3 times, probability \( \frac{3}{15} \)- -10: 6 times, probability \( \frac{6}{15} \)- -60: 5 times, probability \( \frac{5}{15} \)
04

Compute the p-value for the observed data

The observed mean difference from the actual group (380, 400, 420, 360 | 370, 310) was 90. Under the null hypothesis, determine how extreme values as large or larger than 90 occur. The value of 90 occurs 1 time, so the sum of probabilities is \( \frac{1}{15} \). Since we use a two-sided test, consider differences equally extreme in the opposite direction occurring (value \(-60\)): 5 times, so multiply the total by 2 to account for both tails of the distribution. Thus, \( P = 2(\frac{1+5}{15}) = \frac{12}{15} \approx 0.1333 \).
05

Evaluate significance at 5% level

The \( P\)-value \( \approx 0.1333 \) which is greater than the \( \alpha = 0.05 \) significance level. Therefore, there is not sufficient evidence to reject the null hypothesis using the permutation test. Significance at the 5% level can't be demonstrated under the permutation test in this example.
06

Conduct two-sample t-test

Since total number of points is assumed to be Normally distributed, perform a two-sample t-test. Calculate the means and standard deviations for the handwritten and laptop groups then compute the t-statistic and the corresponding \( P\)-value. Use software if available to obtain precise \( P\)-value results.

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Key Concepts

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

Note-taking Methods
Taking effective notes is essential for successful learning and achieving high academic performance, particularly in subjects like statistics. There are two main methods of note-taking: handwritten notes and notes on laptops.

Handwritten Notes involve writing down information on paper. This method encourages deep processing of information as it allows time for summarization and comprehension. It's often believed that the act of writing by hand aids memory retention better than typing does.

Notes on Laptops can be more efficient for fast-paced lectures since you can type faster than writing by hand. However, this can sometimes lead to more verbatim note-taking, which might not facilitate in-depth cognitive processing like summarization or question formulation.

In this experiment, the impact of these two note-taking methods on students' performance in a statistics class was analyzed. Six volunteers were split into two groups鈥攆our using handwritten notes and two using laptops. Their performances were measured by total points obtained in exams, assessing how the method of note-taking could affect learning outcomes.
Random Assignment
In experiments like this, random assignment is a crucial technique. It refers to the process of assigning participants to different groups purely by chance.

Random assignment helps ensure that each group is comparable at the start of the experiment. By doing so, it minimizes biases and balances out other variables that could impact the results.

In this study, six volunteers were randomly divided into two groups: one assigned to take handwritten notes and the other to take notes on laptops. The goal of this random allocation was to evenly distribute individual differences across both groups, ensuring that any observed differences in performance could be attributed to the method of note-taking rather than any inherent participant characteristics.

This randomization is vital for establishing a causal relationship between the note-taking method and student performance. It strives to remove confounding variables, providing more reliable and valid results.
Permutation Test
The permutation test is a non-parametric statistical test used to determine the significance of an observed effect when the experiment's assumptions (like normal distribution) may not hold.

This test reassesses the data by shuffling the observed results among the various groups to create a distribution of possible outcomes that could have occurred just by chance. For this experiment, each permutation consists of the possible ways of distributing six subjects into the two groups: those taking handwritten notes and those using laptops.

The total number of permutations shows all potential groupings, based solely on the chance alignment of participants to their assigned groups. From these, the differences in mean scores between the two groups are computed. These new mean differences create the permutation distribution.

By comparing the actual observed mean difference of 90 with this permutation distribution, researchers can calculate a p-value that gives insight into whether the observed result is statistically significant or could have occurred by random chance.
Hypothesis Testing
Hypothesis testing is an integral part of analyzing statistical data. It involves making a claim or hypothesis about a population parameter and then using sample data to test the validity of this claim.

In this experiment, the main hypothesis (the null hypothesis) assumes that there is no difference in performance between students who take handwritten notes and those who use laptops. The alternative hypothesis posits that there is indeed a significant difference between these two groups.

To test these hypotheses, a permutation test was conducted. The results showed an observed mean difference in performance of 90 points between the groups. However, when analyzing the permutation distribution, the p-value was approximately 0.1333. This value indicates that the observed difference could occur by chance 13.33% of the time under the null hypothesis.

Since the p-value is greater than the significance level of 0.05 (5%), there is not enough evidence to reject the null hypothesis. Consequently, the results do not demonstrate statistical significance, meaning there is not sufficient evidence to suggest a difference in learning outcomes due solely to the method of note-taking.

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Most popular questions from this chapter

How heavy a load (pounds) is needed to pull apart pieces of Douglas fir 4 inches long and \(1.5\) inches square? Here are data from students doing a laboratory exercise: $$ \begin{array}{lllll} 33,190 & 31,860 & 32,590 & 26,520 & 33,280 \\ 32,320 & 33,020 & 32,030 & 30,460 & 32,700 \\ 23,040 & 30,930 & 32,720 & 33,650 & 32,340 \\ 24,050 & 30,170 & 31,300 & 28,730 & 31,920 \end{array} $$ Generate 1000 bootstrap samples by resampling from these data. For each bootstrap sample, compute the bootstrap sample mean. (a) What are the mean and standard deviation of the 1000 bootstrap sample means? (b) Draw a histogram of the bootstrap distribution of the sample mean using the 1000 bootstrap sample means. If your software allows this, superimpose a normal curve on the histogram. Describe the shape of this bootstrap distribution.

In Example \(32.3\) we compared two insect repellants using a permutation test for a matched pairs experiment. Because of the small sample size, we were able to obtain the exact permutation distribution as: $$ \begin{array}{l|cccccc} \hline \text { Mean difference } & -2.0 & -1.0 & -0.5 & 0.5 & 1.0 & 2.0 \\ \hline \text { Probability } & 0.125 & 0.125 & 0.250 & 0.250 & 0.125 & 0.125 \\\ \hline \end{array} $$ In this example, the observed mean difference in treatments (DEET - oil of lemon eucalyptus) is \(-2\). Using this permutation distribution, we have shown that the two-sided \(P\)-value, the chance of observing a difference this extreme, is \(0.25\). (a) Simulate the permutation distribution using 100 simulations and give the estimated \(P\)-value. Repeat this with a second simulation. How close are the answers to the exact permutation distribution and \(P\)-value? (b) Simulate the permutation distribution using 10,000 simulations and give the estimated \(P\)-value. Repeat this with a second simulation. How close are the answers to the exact permutation distribution and \(P\)-value? (c) What do the results in parts (a) and (b) show about the effect of the number of simulations on the estimated permutation distribution and \(P\)-value? Explain briefly.

Our bodies have a natural electrical field that is known to help wounds heal. Does changing the field strength slow healing? A series of experiments with newts investigated this question. In one experiment, the two hind limbs of four newts were assigned at random to either experimental or control groups. This is a matched pairs design. The electrical field in the experimental limbs was reduced to zero by applying a voltage. The control limbs were left alone. Here are the rates at which new cells closed a razor cut in each limb, in micrometers per hour: \({ }^{11}\) $$ \begin{array}{l|cccc} \hline \text { Newt } & 1 & 2 & 3 & 4 \\ \hline \text { Control limb } & 36 & 41 & 39 & 42 \\ \hline \text { Experimental limb } & 28 & 31 & 27 & 33 \\ \hline \end{array} $$ The number of possible random assignments of treatments to the different matched pairs is (a) 4 . (b) \(8 .\) (c) \(16 .\)

In a study of exhaust emissions from school buses, the pollution intake by passengers was determined for a sample of nine school buses used in the Southern California Air Basin. The pollution intake is the amount of exhaust emissions, in grams per person, that would be breathed in while travelilng on the bus during its usual 18-mile trip on congested freeways from South Central LA to a magnet school in West LA. (As a reference, the average intake of motor emissions of carbon monoxide in the LA area is estimated to be about \(0.000046\) gram per person.) Here are the amounts for the nine buses when driven with the windows open: \({ }^{17}\) $$ \begin{array}{lllllllll} 1.15 & 0.33 & 0.40 & 0.33 & 1.35 & 0.38 & 0.25 & 0.40 & 0.35 \end{array} $$ (a) Make a stemplot. Are there outliers or strong skewness that would forbid use of the \(t\) procedures? (b) Construct a \(95 \%\) bootstrap confidence interval for the mean pollution intake among all school buses used in the Southern California Air Basin that travel the route investigated in the study.

In laparoscopic surgery, a video camera and several thin instruments are inserted into the patient's abdominal cavity. The surgeon uses the image from the video camera positioned inside the patient's body to perform the procedure by manipulating the instruments that have been inserted. It has been found that the Nintendo Wii, with its motion-sensing interface, reproduces the movements required in laparoscopic surgery more closely than other video games. If training with a Nintendo Wii can improve laparoscopic skills, it can complement the more expensive training on a laparoscopic simulator. Forty-two medical residents were chosen, and all were tested on a set of basic laparoscopic skills. Twenty-one were selected at random to undergo systematic Nintendo Wii training for one hour per day, five days per week, for four weeks. The remaining 21 residents were given no Nintendo Wii training and were asked to refrain from video games during this period. At the end of four weeks, all 42 residents were tested again on the same set of laparoscopic skills. One of the skills involved a virtual gall bladder removal with several performance measures, including time to complete the task recorded. Here are the improvement (before-after) times in seconds after four weeks for the two groups: \({ }^{6}\). Does the Nintendo Wii training significantly increase the mean improvement time? (a) Use software to estimate the histogram for the permutation distribution for the difference in the mean improvements for the two groups (mean improvement for the treatment group minus the mean improvement for the control group). Use 10,000 simulated permutations. (b) From your estimated histogram, compute the \(P\)-value of the data.
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