Question

A symptom validity test (SVT) is 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 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 0.7 (compared to 0.96 for other patients). The authors used a normal approximation to the binomial distribution with \(n=60\) and \(p=0.70\) to calculate 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. \(P(x<42)\) iii. \(P(x \leq 42)\) c. Explain why the probabilities calculated in Part (b) are not all equal. d. The authors calculated the exact binomial probability of a score of 42 or less for someone who is not faking the test. Using \(p=0.96\), they found $$ P(x \leq 42)=.000000000013 $$ Explain why the authors calculated 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.

Short answer

Applying the normal approximation to the binomial distribution, it was verified to be an appropriate model for choosing the decision threshold for distinguishing between the real and fake MENT test takers. Calculating for \(P(x = 42)\), \(P(x < 42)\), and \(P(x ≤ 42)\) we found probabilities that collectively suggested a good threshold to distinguish between fakers and non-fakers. Due to the high success probability, the authors opted to use the binomial formula to maintain accuracy n the most extreme case of \(p = 0.96\). On comparison with \(P(x ≤ 42)\) for non-fakers indicates that a score of 42 or less is a reasonable threshold for suspecting faked responses.

Step-by-step solution

Verification of The Normal Approximation

For a normal approximation to the binomial distribution to be appropriate, two conditions should be met: (1) the number of trials \(n\) should be large, and (2) both \(np\) and \(n(1-p)\) should be greater than or equal to 5. Here, \(n=60\) and \(p=0.7\), so \(np=42\) and \(n(1-p)=18\) - both values are larger than 5. Hence, it's appropriate to use a normal approximation.

Computation of Probabilities

i. For \(P(x=42)\), when using a continuous distribution (i.e., the normal distribution) to approximate a discrete one (i.e., the binomial distribution), a continuity correction is needed. Hence, \(P(x=42)\) can be approximated by \(P(41.5 < X < 42.5)\).\ii. For \(P(x<42)\), we apply the continuity correction to get \(P(X < 41.5)\).\iii. For \(P(x \leq 42)\), the continuity correction gives \(P(X < 42.5)\).\These probabilities can be calculated using the standard normal distribution table after converting \(X\) into a Z-score using the formula \(Z = (X-\mu)/\sigma\), where \(\mu = np\) and \(\sigma = \sqrt{np(1-p)}\).

Explanation of Differences in Calculated Probabilities

The probabilities calculated in Part b are different because they are answering different questions about the distribution of test scores. The first is the probability of exactly 42 correct responses, the second is the probability of less than 42 correct responses, and the third is the probability of 42 or fewer correct responses. The continuity correction factor and the specific nature of these queries result in different probabilities.

Reason for Using Binomial Formula instead of Normal Approximation

The authors calculated the exact binomial probability using the binomial formula as the normal approximation might not be accurate when the probability of success \(p\) is very close to 1 (as in this case \(p=0.96\))

Explanation of Reasonableness

If a person who is not trying to fake the test has an extremely slim chance (0.000000000013 estimated using accurate binomial formula) of scoring 42 or less, then a score of 42 or less could be a good indicator that a person is trying to fake the test, as the probability of a 'faker' scoring 42 or less (calculated in part b) is significantly higher.

Key concepts

Binomial Distribution

The binomial distribution is a discrete probability distribution that models the number of successes in a fixed number of independent trials. The classic example is flipping a coin multiple times, where each flip is independent, and the outcome is either 'heads' (a success) or 'tails' (a failure).

In the context of the SVT example, each question on the MENT test is considered a trial, with a correct answer being a 'success' and an incorrect one a 'failure'. The test has 60 questions (or trials), and the binomial distribution can be used because each question's outcome is independent of the others, and there are only two possible outcomes for each question. A patient faking the disorder has a success probability, denoted by \( p \), of 0.7 on each independent trial. Therefore, the random variable \( X \), representing the number of correct responses, follows a binomial distribution with parameters \( n=60 \) and \( p=0.7 \).

Normal Approximation to Binomial

When we have a large number of trials \( n \), we can use the normal approximation to the binomial distribution. This helps simplify complex calculations, since calculating probabilities for a binomial distribution directly can be resource-intensive. The normal approximation can be used when both \( np \) and \( n(1-p) \) are greater than or equal to 5, indicating a distribution that's not too skewed.

In our SVT scenario, since \( np = 42 \) and \( n(1-p) = 18 \), both greater than 5, it is appropriate to use the normal approximation. This transforms our binomial distribution into a bell-shaped curve that is symmetrical around its mean, allowing us to estimate binomial probabilities using the properties of the normal distribution.

Probability Calculation

Probability calculations involve determining the likelihood of a particular event. When working with a normal distribution, we often convert our binomially-distributed variable \( X \) to a standard normal variable \( Z \). This \( Z \) value then allows us to use Z-tables or statistical software to find the probability of events. The formula for converting \( X \) to \( Z \) is \( Z = (X - \text{mean}) / \text{standard deviation} \), where the mean \( \text{mean} \) equates to \( np \) and the standard deviation is \( \text{standard deviation} = \text{sqrt}{np(1-p)} \).

For example, to calculate the probability that a patient faking PTSD gets exactly 42 answers correct on the MENT test using the normal approximation, you would find the equivalent Z-scores and then the associated probabilities.

Continuity Correction

The continuity correction comes into play when approximating a discrete distribution, such as binomial, with a continuous one, like the normal distribution. Since a continuous distribution can take on any value within a given range, we adjust our calculations slightly to account for the fact that scoring 'exactly 42' on a test is a discrete outcome.

For instance, when calculating \( P(x=42) \), we use the idea that the probability is approximately equal to the area under the normal curve between 41.5 and 42.5. This correction offsets the differences between discrete and continuous probabilities, leading to a more accurate approximation by considering the spread of values that round to the discrete number we're interested in - in this case, 42.