Question

A true-false test consists of 50 questions. How many does a student have to get right to convince you that he is not merely guessing? Explain.

Short answer

The student needs at least 33 correct answers to prove they are not merely guessing.

Step-by-step solution

Understand the Problem

The exercise involves a true-false test with 50 questions. We need to determine how many questions a student needs to answer correctly to prove they are not simply guessing.

Define the Probability of Guessing

If a student is guessing, the probability of getting any single question correct is 0.5 (since each question has two possible answers: true or false).

Determine the Expected Value

The expected number of correct answers by guessing is calculated by multiplying the total number of questions (50) by the probability of getting any question correct (0.5):\[ E = 50 \times 0.5 = 25 \]

Calculate Standard Deviation

For a binomial distribution, standard deviation is given by the formula: \[ \sigma = \sqrt{n \cdot p \cdot (1-p)} \]where \( n = 50 \), \( p = 0.5 \). Calculate:\[ \sigma = \sqrt{50 \times 0.5 \times 0.5} = \sqrt{12.5} \approx 3.54 \]

Apply the Statistically Significant Criterion

A common criterion to deem something statistically significant is if it is more than two standard deviations away from the mean of the distribution for random guessing. Thus, calculate:\[ M = E + 2\sigma = 25 + 2(3.54) \approx 32.08 \]

Interpret the Result

To be convincingly above random guessing, the student should correctly answer more than 32 questions. Thus, they need at least 33 correct answers.

Key concepts

Understanding Binomial Distribution

The binomial distribution is a fundamental concept in probability and statistics. It is used when you have a fixed number of identical, independent trials, each with two possible outcomes. In the context of a true-false test, each question represents a trial, with possible outcomes being either true or false.
The probability of getting a certain number of correct answers by guessing can be modeled using the binomial distribution. The key parameters of this distribution are the number of trials, denoted by \( n \), and the probability of success on a single trial, denoted by \( p \). For our true-false test scenario:

• Number of trials \( n = 50 \) (since there are 50 questions)
• Probability of correct answer \( p = 0.5 \) (for a true-false question)

Each question is an independent event where the outcome does not affect others. This independence is crucial for applying the binomial distribution, allowing each trial to undergo the same probability of success across the board.

Calculating Expected Value

Expected value, often called the mean, is a crucial concept in understanding what to 'expect' on average from a random experiment. In our true-false test scenario, it measures the average number of correct answers we would expect if a student were guessing.
To calculate the expected value for a binomial distribution, you use the formula:\[ E = n \cdot p \]where \( E \) is the expected value, \( n \) is the number of trials, and \( p \) is the probability of success (guessing correctly).
For our exercise:

• \( E = 50 \times 0.5 = 25 \)

This means, if a student answers randomly, we expect them to get 25 questions correct out of 50. Expected value offers a benchmark against which we can assess actual outcomes.

Understanding Standard Deviation

Standard deviation in a probability distribution, like the binomial distribution, provides a measure of the spread or variability of the distribution. It tells us how much results typically deviate from the mean.
In the case of our true-false test, the standard deviation helps us quantify how much variation we can expect in the number of correct answers if a student is guessing.
For a binomial distribution, the formula to calculate standard deviation is:\[ \sigma = \sqrt{n \cdot p \cdot (1-p)}\]where \( \sigma \) is the standard deviation, \( n \) is the number of trials, and \( p \) is the probability of success.
In our example:

• \( \sigma = \sqrt{50 \times 0.5 \times 0.5} = \sqrt{12.5} \approx 3.54 \)

This value, approximately 3.54, indicates the typical deviation from the expected 25 correct answers when guessing. A standard deviation provides a yardstick for determining outcomes that significantly deviate from random guessing, as seen through the use of the statistically significant criterion.