There are 21,000 students in a MOOC. Each student tests the fairness of a coin (yes, the same coin; the instructor somehow gets Tyche¢s help in getting the coin to each student in turn). Specifically, each student tests:
Null: p is equal to 0.5 Alternative: p is not equal to 0.5
using the 5% cutoff.
Suppose that, unknown to the students, the coin is in fact fair.
Find the expected number of students whose test will conclude that the coin is unfair. [The is an integer)
ya i am working on it
try 1050 if I'm wrong say I do recalculations
Thank u - 1050 is the ryt ans
To find the expected number of students whose test will conclude that the coin is unfair, we need to calculate the probability of each student's test concluding that the coin is unfair and then sum up these probabilities for all students.
In this case, the null hypothesis is that the probability (p) of the coin landing on heads is equal to 0.5, and the alternative hypothesis is that the probability is not equal to 0.5.
Since the coin is actually fair, the probability of each student's test concluding that the coin is unfair is the probability of observing a result as extreme or more extreme than what was actually observed, given that the null hypothesis is true.
Let's use the binomial distribution to calculate this probability. The binomial distribution is appropriate because each student's test can be considered a Bernoulli trial (success = test concludes the coin is unfair, failure = test concludes the coin is fair), and the outcomes are independent.
The formula for the binomial distribution is:
P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)
Where:
- P(X = k) is the probability of having exactly k "successes" (students concluding the coin is unfair)
- n is the number of trials (number of students)
- k is the number of successes (students concluding the coin is unfair)
- p is the probability of success (probability of a student's test concluding the coin is unfair)
- C(n, k) is the binomial coefficient, which represents the number of ways to choose k successes from n trials.
In this case, p = 0.05 because we have a 5% cutoff significance level.
Let's calculate the expected number of students whose test will conclude that the coin is unfair:
Expected number = n * p
Expected number = 21000 * 0.05
Expected number = 1050
Therefore, the expected number of students whose test will conclude that the coin is unfair is 1050.