There are two boxes, each with several million tickets marked “1” or “0”. The two boxes have the same number of tickets, but in one of the boxes, 49% of the tickets are marked “1” and in the other box 50.5% of the tickets are marked “1”. Someone hands me one of the boxes but doesn’t tell me which box it is.
Consider the following hypotheses:
Null: p = 0.49 Alternative: p = 0.505
Here is my proposed test: I will draw a simple random sample of 10,000 tickets, and if 5,000 or more of them are marked “1” then I will choose the alternative; otherwise I will stay with the null.
PROBLEM 10 :
The significance level of my test is _____%. [Please be careful to enter your answer as a percent; that is, if your answer is 50% then please enter 50 in the blank; not 50%, nor 0.5, nor 1/2, etc.]
PROBLEM 11 :
The power of my test is _____%. [Please be careful to enter your answer as a percent; that is, if your answer is 50% then please enter 50 in the blank; not 50%, nor 0.5, nor 1/2, etc.]
To determine the significance level and power of the test, we need to calculate the respective probabilities.
Significance level (Problem 10):
The significance level, denoted by α, is the probability of rejecting the null hypothesis when it is actually true. In this case, the null hypothesis states that the proportion of tickets marked "1" is 0.49.
To find the significance level, we need to calculate the probability of observing 5,000 or more "1" tickets out of a random sample of 10,000 tickets, assuming that the true proportion is 0.49.
We can use the binomial distribution formula to calculate this probability:
P(X ≥ 5000) = Σ [(n choose x) * p^x * (1-p)^(n-x)]
where n is the number of trials (10,000), x is the number of successes (≥ 5000), and p is the assumed proportion (0.49).
However, since calculating this sum can be time-consuming, we can use a normal approximation to the binomial distribution when n is large (which is the case here). The mean of the binomial distribution is n * p, and the standard deviation is sqrt(n * p * (1-p)).
Using the normal approximation, we can calculate the z-score as (5000 - n * p) / sqrt(n * p * (1-p)). We then look up the corresponding z-score in the standard normal distribution table to find the corresponding probability.
For the null hypothesis with p = 0.49, the probability of observing 5,000 or more "1" tickets can be calculated as:
P(X ≥ 5000) = 1 - P(X < 5000) ≈ 1 - P(Z < (5000 - n * p) / sqrt(n * p * (1-p)))
Finding the z-score in the standard normal distribution table, we can determine the corresponding probability. Let's assume it is P(Z < z). Then the significance level is 100 * [1 - P(Z < z)]%.
Power (Problem 11):
The power of a test is the probability of correctly rejecting the null hypothesis when it is false. In this case, the alternative hypothesis states that the proportion of tickets marked "1" is 0.505.
To calculate the power, we need to calculate the probability of observing 5,000 or more "1" tickets out of a random sample of 10,000 tickets, assuming that the true proportion is 0.505.
Using the same approach as before, we calculate the probability according to the alternative hypothesis:
P(X ≥ 5000) = 1 - P(X < 5000) ≈ 1 - P(Z < (5000 - n * p) / sqrt(n * p * (1-p)))
where p is the assumed proportion of 0.505.
Finding the z-score in the standard normal distribution table, we can determine the corresponding probability. Let's assume it is P(Z < z). Then the power is 100 * [1 - P(Z < z)]%.
To find the values for both the significance level and power, you will need to calculate the z-scores and look up their corresponding probabilities in the standard normal distribution table.