/*! 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 84 The paper "Risk Behavior, Decisi... [FREE SOLUTION] | 91Ó°ÊÓ

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The paper "Risk Behavior, Decision Making, and Music Genre in Adolescent Males" (Marshall University, May 2009 ) examined the effect of type of music playing and performance on a risky, decision-making task. a. Participants in the study responded to a questionnaire that was used to assign a risk behavior score. Risk behavior scores (read from a graph that appeared in the paper) for 15 participants follow. Use these data to construct a normal probability plot (the normal scores for a sample of size 15 appear in the previous exercise). \(\begin{array}{llllllll}102 & 105 & 113 & 120 & 125 & 127 & 134 & 135\end{array}\) \(\begin{array}{lllllll}139 & 141 & 144 & 145 & 149 & 150 & 160\end{array}\) b. Participants also completed a positive and negative affect scale (PANAS) designed to measure emotional response to music. PANAS values (read from a graph that appeared in the paper) for 15 participants follow. Use these data to construct a normal probability plot (the normal scores for a sample of size 15 appear in the previous exercise). \(\begin{array}{llllllll}36 & 40 & 45 & 47 & 48 & 49 & 50 & 52\end{array}\) \(\begin{array}{lllllll}53 & 54 & 56 & 59 & 61 & 62 & 70\end{array}\) c. The author of the paper states that he believes that it is reasonable to consider both risk behavior scores and PANAS scores to be approximately normally distributed. Do the normal probability plots from Parts (a) and (b) support this conclusion? Explain.

Short Answer

Expert verified
Without constructing actual plots, a definite short answer can't be provided. However, after construction as guided in the steps, if the data points approximately follow a straight line in both plots, it supports the author's conclusion that risk behavior scores and PANAS values are approximately normally distributed.

Step by step solution

01

Construct Normal Probability Plot for Risk Behaviour Scores

First, arrange the risk behaviour scores in ascending order. Then, calculate cumulative probabilities for each value using the formula: \(\frac{rank}{n+1}\). Use a Z-table to find the Z-score associated with each cumulative probability. Plot each risk behaviour score against its respective Z-score.
02

Construct Normal Probability Plot for PANAS values

The same calculation process from Step 1 applies here. The PANAS values need to be sorted in ascending order, cumulative probabilities calculated, Z-scores found from the Z-table and plotted against the respective PANAS values.
03

Evaluate Normality Based on Plots

In each plot, if points closely follow a straight line, it suggests that the data are normally distributed. Deviations from the straight line indicate departures from normality. This structural analysis allows the validation or refutation of the author's statement regarding normal distribution.

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Key Concepts

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

Probability Plot
A probability plot is a graphical technique used to determine if a dataset follows a particular distribution, usually the normal distribution. The essence of creating a probability plot lies in comparing the distribution of your sample data to a theoretical distribution.
- **Procedure**: The data is plotted against a known distribution's cumulative distribution function (CDF). - **Linear Relationship**: If the plotted points closely follow a straight line, it indicates the data likely follows the specified distribution.
To construct a probability plot, follow these steps:
  • Sort your data in ascending order.
  • Assign ranks to the sorted data, which helps in determining their cumulative probability.
  • Calculate cumulative probabilities as the rank divided by the total number of data points plus one.
  • Find corresponding Z-scores from a standard normal distribution table for each cumulative probability.
This process is the same for any data set, such as risk behavior scores or PANAS values. A visual inspection of the resulting plot will guide you to confirm or challenge assumptions about data distribution.
Normal Distribution
The normal distribution, often referred to as the bell curve, is the most crucial probability distribution in statistics. It is characterized by its symmetrical shape centered around the mean, with data tapering off as it moves away from the mean.
**Key Properties of Normal Distribution**:
  • Symmetrical around the mean.
  • Mean, median, and mode are all equal.
  • The area under the curve represents the total probability and equals 1.
  • As you move away from the mean, the probability decreases exponentially.
In using normal distribution for analysis, it's essential to ensure data points follow this pattern.
When analyzing experimental data, like risk behavior scores or PANAS values, researchers use probability plots to verify normality. Recognizing normal distribution in data can guide further statistical tests and hypotheses.
Z-Scores
Z-scores are a statistical measurement that expresses how far a data point is from the mean of the dataset. It's crucial in standardizing different datasets, enabling comparisons across various units or scales.
**Calculating Z-Scores**:
  • Find the mean and standard deviation of the data set.
  • For each data point, subtract the mean and divide by the standard deviation:
\[Z = \frac{(X - \mu)}{\sigma}\]Here, \(X\) is the data point, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
Z-scores are instrumental in creating probability plots as they translate the data into a form that can be easily compared to a standard normal distribution. By converting cumulative probabilities into Z-scores, we can plot them against the data values to assess normality.
Risk Behavior
Risk behavior refers to actions or activities that expose individuals to potential harm or loss. Understanding risk behavior involves quantifying it in a measurable form to enable statistical analysis.
**Applications**:
  • Public health studies often examine risk behaviors to develop preventive strategies.
  • The evaluation of risk behavior scores helps in identifying patterns and predicting outcomes.
In the context of the given exercise, risk behavior scores must be analyzed to understand if they follow a normal distribution.
Analyzing risk behavior through statistical methods allows researchers to draw conclusions regarding decision-making processes, such as whether a group of individuals, like adolescent males listening to different music genres, takes more risks. By utilizing probability plots and Z-scores, the researcher can discern the spread and nature of risk-taking behavior among participants.

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Most popular questions from this chapter

Symptom validity tests (SVTs) are sometimes used to confirm diagnosis of psychiatric disorders. The paper "Developing a Symptom Validity Test for PostTraumatic Stress Disorder: Application of the Binomial Distribution" (journal of Anxiety Disorders [2008 ]: \(1297-1302\) ) investigated the use of SVTs in the diagnosis of post-traumatic stress disorder. One SVT proposed is a 60 -item test (called the MENT test), where each item has only a correct or incorrect response. The MENT test is designed so that responses to the individual questions can be considered independent of one another. For this reason, the authors of the paper believe that the score on the MENT test can be viewed as a binomial random variable with \(n=60 .\) The MENT test is designed to help in distinguishing fictitious claims of post-traumatic stress disorder. The items on the MENT test are written so that the correct response to an item should be relatively obvious, even to people suffering from stress disorders. Researchers have found that a patient with a fictitious claim of stress disorder will try to "fake" the test, and that the probability of a correct response to an item for these patients is \(.7\) (compared to \(.96\) for other patients). The authors used a normal approximation to the binomial distribution with \(n=60\) and \(p=.70\) to compute various probabilities of interest, where \(x=\) number of correct responses on the MENT test for a patient who is trying to fake the test. a. Verify that it is appropriate to use a normal approximation to the binomial distribution in this situation. b. Approximate the following probabilities: i. \(\quad P(x=42)\) ii. \(\quad P(x<42)\) iii. \(P(x \leq 42)\) c. Explain why the probabilities computed in Part (b) are not all equal. d. The authors computed the exact binomial probability of a score of 42 or less for someone who is not faking the test. Using \(p=.96\), they found \(p(x \leq 42)=.000000000013\) Explain why the authors computed this probability using the binomial formula rather than using a normal approximation. e. The authors propose that someone who scores 42 or less on the MENT exam is faking the test. Explain why this is reasonable, using some of the probabilities from Parts (b) and (d) as justification.

A gasoline tank for a certain car is designed to hold 15 gallons of gas. Suppose that the variable \(x=\) actual capacity of a randomly selected tank has a distribution that is well approximated by a normal curve with mean \(15.0\) gallons and standard deviation \(0.1\) gallon. a. What is the probability that a randomly selected tank will hold at most \(14.8\) gallons? b. What is the probability that a randomly selected tank will hold between \(14.7\) and \(15.1\) gallons? c. If two such tanks are independently selected, what is the probability that both hold at most 15 gallons?

Let \(x\) be the number of courses for which a randomly selected student at a certain university is registered. The probability distribution of \(x\) appears in the following table: $$ \begin{array}{cccccccc} x & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ p(x) & .02 & .03 & .09 & .25 & .40 & .16 & .05 \end{array} $$ a. What is \(P(x=4)\) ? b. What is \(P(x \leq 4)\) ? c. What is the probability that the selected student is taking at most five courses? d. What is the probability that the selected student is taking at least five courses? more than five courses? e. Calculate \(P(3 \leq x \leq 6)\) and \(P(3

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A contractor is required by a county planning department to submit anywhere from one to five forms (depending on the nature of the project) in applying for a building permit. Let \(y\) be the number of forms required of the next applicant. The probability that \(y\) forms are required is known to be proportional to \(y\); that is, \(p(y)=\) \(k y\) for \(y=1, \ldots, 5\) a. What is the value of \(k ?\) (Hint: \(\sum p(y)=1 .\).) b. What is the probability that at most three forms are required? c. What is the probability that between two and four forms (inclusive) are required? d. Could \(p(y)=y^{2} / 50\) for \(y=1,2,3,4,5\) be the probability distribution of \(y\) ? Explain.

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