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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

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?

A local television station sells \(15-\mathrm{sec}, 30-\mathrm{sec}\), and 60 -sec advertising spots. Let \(x\) denote the length of a randomly selected commercial appearing on this station, and suppose that the probability distribution of \(x\) is given by the following table: $$ \begin{array}{lrrr} x & 15 & 30 & 60 \\ p(x) & .1 & .3 & .6 \end{array} $$ a. Find the average length for commercials appearing on this station. b. If a 15 -sec spot sells for $$\$ 500$$, a 30 -sec spot for $$\$ 800$$, and a 60 -sec spot for $$\$ 1000$$, find the average amount paid for commercials appearing on this station. (Hint: Consider a new variable, \(y=\) cost, and then find the probability distribution and mean value of \(y .\) )

Suppose that for a given computer salesperson, the probability distribution of \(x=\) the number of systems sold in one month is given by the following table: $$ \begin{array}{lrrrrrrrr} x & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ p(x) & .05 & .10 & .12 & .30 & .30 & .11 & .01 & .01 \end{array} $$ a. Find the mean value of \(x\) (the mean number of systems sold). b. Find the variance and standard deviation of \(x\). How would you interpret these values? c. What is the probability that the number of systems sold is within 1 standard deviation of its mean value? d. What is the probability that the number of systems sold is more than 2 standard deviations from the mean?

The lifetime of a certain brand of battery is normally distributed with a mean value of \(6 \mathrm{hr}\) and a standard deviation of \(0.8 \mathrm{hr}\) when it is used in a particular cassette player. Suppose that two new batteries are independently selected and put into the player. The player ceases to function as soon as one of the batteries fails. a. What is the probability that the player functions for at least \(4 \mathrm{hr}\) ? b. What is the probability that the cassette player works for at most \(7 \mathrm{hr}\) ? c. Find a number \(z^{*}\) such that only \(5 \%\) of all cassette players will function without battery replacement for more \(\operatorname{than} z^{*} \mathrm{hr}\)

The paper "Temperature and the Northern Distributions of Wintering Birds" (Ecology [1991]: 2274-2285) gave the following body masses (in grams) for 50 different bird species: $$ \begin{array}{rrrrrrrr} 7.7 & 10.1 & 21.6 & 8.6 & 12.0 & 11.4 & 16.6 & 9.4 \\ 11.5 & 9.0 & 8.2 & 20.2 & 48.5 & 21.6 & 26.1 & 6.2 \\ 19.1 & 21.0 & 28.1 & 10.6 & 31.6 & 6.7 & 5.0 & 68.8 \\ 23.9 & 19.8 & 20.1 & 6.0 & 99.6 & 19.8 & 16.5 & 9.0 \\ 448.0 & 21.3 & 17.4 & 36.9 & 34.0 & 41.0 & 15.9 & 12.5 \\ 10.2 & 31.0 & 21.5 & 11.9 & 32.5 & 9.8 & 93.9 & 10.9 \\ 19.6 & 14.5 & & & & & & \end{array} $$ a. Construct a stem-and-leaf display in which \(448.0\) is listed separately beside the display as an outlier on the high side, the stem of an observation is the tens digit, the leaf is the ones digit, and the tenths digit is suppressed (e.g., \(21.5\) has stem 2 and leaf 1 ). What do you perceive as the most prominent feature of the display? b. Draw a histogram based on class intervals 5 to \(<10,10\) to \(<15,15\) to \(<20,20\) to \(<25,25\) to \(<30,30\) to \(<40,40\) to \(<50,50\) to \(<100\), and 100 to \(<500\). Is a transformation of the data desirable? Explain. c. Use a calculator or statistical computer package to calculate logarithms of these observations, and construct a histogram. Is the log transformation successful in producing a more symmetric distribution? d. Consider transformed value \(=\frac{1}{\sqrt{\text { original value }}}\) and construct a histogram of the transformed data. Does it appear to resemble a normal curve?

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