/*! 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 53 An experiment was conducted to i... [FREE SOLUTION] | 91Ó°ÊÓ

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An experiment was conducted to investigate whether a graphologist (a handwriting analyst) could distinguish a normal person's handwriting from that of a psychotic. A well-known expert was given 10 files, each containing handwriting samples from a normal person and from a person diagnosed as psychotic, and asked to identify the psychotic's handwriting. The graphologist made correct identifications in 6 of the 10 trials (data taken from Statistics in the Real World, by R. J. Larsen and D. F. Stroup [New York: Macmillan, 1976\(]\) ). Does this evidence indicate that the graphologist has an ability to distinguish the handwriting of psychotics? (Hint: What is the probability of correctly guessing 6 or more times out of 10 ? Your answer should depend on whether this probability is relatively small or relatively large.)

Short Answer

Expert verified
The probability calculated can provide a statistical indication of whether the graphologist is making the correct identifications based on an ability to distinguish handwriting or if the identifications are simply happening by chance. The smaller the probability, the stronger the evidence that the graphologist has an ability to distinguish handwriting.

Step by step solution

01

Understand the Problem Context

We need to find out whether the graphologist was able to identify the handwriting correctly based on skill or by mere chance. In any one trial, the graphologist has two handwritings and he must choose one. So the probability of choosing a correct handwriting by guessing is 0.5 or 1/2.
02

Determine the Distribution

This is a binomial problem because there are two possible outcomes in each trial (correct or incorrect), the trials are independent, and the probability of success is the same for each trial. In this case, the number of trials \(n\) is 10, and the probability of success \(p\) is 0.5.
03

Calculate the Probability

We need to calculate the probability of getting 6 or more correct identifications out of 10 by chance. Applying the formula for the binomial probability, we can calculate the probabilities for 6, 7, 8, 9 and 10 successes and add them together.
04

Interpret the Results

If the calculated probability is quite small, it means that getting 6 or more correct identifications by chance is unlikely, and hence, it can be argued that the graphologist has an ability to distinguish handwriting. If the probability is large, it means that the results could have occurred by chance.

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Key Concepts

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

Understanding Probability
Probability is a fundamental concept in both everyday decision making and formal statistical analysis. It is simply a measure of how likely it is for a particular event to occur. This likelihood is expressed as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty. For instance, when flipping a fair coin, the probability of getting heads or tails is 0.5, also known as a 50% chance. In the graphologist's scenario, the probability of correctly identifying a handwriting by pure guess is also 0.5 per trial since there are only two equally likely outcomes - correct or incorrect identification.

There are different rules associated with probability including the addition rule for mutually exclusive events and the multiplication rule for independent events. Understanding how to apply these rules helps in solving more complex probability problems, such as determining the likelihood of a graphologist correctly identifying handwritings over multiple trials. In our example, the coin toss analogy is crucial as it reflects the binary nature of the outcomes, which leads us to the concept of binomial distribution.
Basics of Statistics
Statistics is the science concerned with the collection, analysis, interpretation, and presentation of data. It enables us to make sense of numerical data and allows us to make decisions based on statistical evidence. In our handwriting analysis problem, statistical methods help us determine whether the graphologist's performance is significantly better than chance.

The field of statistics is divided into descriptive and inferential statistics. Descriptive statistics summarize and describe the features of a data set while inferential statistics use samples to make generalizations about a population. In the given scenario, if we view the 10 files as a sample, we can use inferential statistics, such as hypothesis testing, to infer whether the graphologist has a real ability to distinguish the psychotics' handwriting from the sample provided. To draw conclusions, we need to calculate the probability of the observed outcome under the assumption of the null hypothesis, which typically states that there is no effect or difference.
Explaining Binomial Distribution
The binomial distribution is a discrete probability distribution that applies to scenarios where there are exactly two mutually exclusive outcomes in each trial, often referred to as 'success' and 'failure'. In the context of our exercise, a 'success' is the graphologist correctly identifying the psychotic's handwriting while a 'failure' is an incorrect identification.

A binomial distribution is defined by two parameters: the number of trials (n) and the probability of success in a single trial (p). The binomial formula gives the probability of having exactly k successes in n independent trials. The formula is expressed as:
\[ P(X = k) = \binom{n}{k} p^k (1-p)^{(n-k)} \]
where \( \binom{n}{k} \) is the binomial coefficient, calculated as \( \frac{n!}{k! (n-k)!} \), which represents the number of ways to choose k successes from n trials.

In our case, the appropriate calculation would be the probability of having 6 or more successes out of 10 trials, with each trial having a success probability of 0.5. This requires computing the cumulative probability of getting 6, 7, 8, 9, and 10 correct identifications and adding those probabilities together. Through this calculation, we can assess if the graphologist's abilities are statistically significant or if they could be attributed to random chance.

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Most popular questions from this chapter

Sophie is a dog that loves to play catch. Unfortunately, she isn't very good, and the probability that she catches a ball is only .1. Let \(x\) be the number of tosses required until Sophie catches a ball. a. Does \(x\) have a binomial or a geometric distribution? b. What is the probability that it will take exactly two tosses for Sophie to catch a ball? c. What is the probability that more than three tosses will be required?

Let \(x\) denote the duration of a randomly selected pregnancy (the time elapsed between conception and birth). Accepted values for the mean value and standard deviation of \(x\) are 266 days and 16 days, respectively. Suppose that the probability distribution of \(x\) is (approximately) normal. a. What is the probability that the duration of pregnancy is between 250 and 300 days? b. What is the probability that the duration of pregnancy is at most 240 days? c. What is the probability that the duration of pregnancy is within 16 days of the mean duration? d. A "Dear Abby" column dated January 20,1973, contained a letter from a woman who stated that the duration of her pregnancy was exactly 310 days. (She wrote that the last visit with her husband, who was in the navy, occurred 310 days before birth.) What is the probability that the duration of pregnancy is at least 310 days? Does this probability make you a bit skeptical of the claim? e. Some insurance companies will pay the medical expenses associated with childbirth only if the insurance has been in effect for more than 9 months ( 275 days). This restriction is designed to ensure that the insurance company pays benefits for only those pregnancies for which conception occurred during coverage. Suppose that conception occurred 2 weeks after coverage began. What is the probability that the insurance company will refuse to pay benefits because of the 275 -day insurance requirement?

A machine that produces ball bearings has initially been set so that the true average diameter of the bearings it produces is \(0.500\) in. A bearing is acceptable if its diameter is within \(0.004\) in. of this target value. Suppose, however, that the setting has changed during the course of production, so that the distribution of the diameters produced is well approximated by a normal distribution with mean \(0.499\) in. and standard deviation \(0.002\) in. What percentage of the bearings produced will not be acceptable?

The article on polygraph testing of FBI agents referenced in Exercise \(7.51\) indicated that the probability of a false-positive (a trustworthy person who nonetheless fails the test) is \(.15 .\) Let \(x\) be the number of trustworthy \(\mathrm{FBI}\) agents tested until someone fails the test. a. What is the probability distribution of \(x ?\) b. What is the probability that the first false-positive will occur when the third person is tested? c. What is the probability that fewer than four are tested before the first false-positive occurs? d. What is the probability that more than three agents are tested before the first false-positive occurs?

Let \(z\) denote a variable that has a standard normal distribution. Determine the value \(z^{*}\) to satisfy the following conditions: a. \(P\left(zz^{*}\right)=.02\) e. \(P\left(z>z^{*}\right)=.01\) f. \(P\left(z>z^{*}\right.\) or \(\left.z<-z^{*}\right)=.20\)

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