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.
The significance level of my test is _____%.
The power of my test is _____%.
2.33
84.13
given that a sample of size was taken from population of 90 witha population standard deviation of 1.8,what ould be the margin of roe ata 95% level of confidence
To find the significance level and power of your test, we can use the concept of binomial distributions and probability.
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 is 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 tickets marked "1" in a random sample of 10,000, assuming that the true proportion is 0.49.
To do this, we can use the binomial distribution formula. The probability mass function (PMF) of the binomial distribution is given by:
P(X = k) = (n choose k) * p^k * (1-p)^(n-k)
Where:
- n is the number of trials (sample size)
- k is the number of tickets marked "1"
- p is the probability of success (proportion marked "1")
To find the probability of observing 5,000 or more tickets marked "1", we need to calculate the cumulative sum of probabilities from 5,000 to 10,000:
P(X >= 5,000) = P(X = 5,000) + P(X = 5,001) + ... + P(X = 10,000)
Using statistical software or a calculator, we can find this probability. Let's assume it is denoted by p-value.
The significance level α is then equal to 1 - p-value. For example, if the p-value is 0.025, then the significance level is 1 - 0.025 = 0.975 or 97.5%.
The power of a test is the probability of correctly rejecting the null hypothesis when it is false. In this case, the alternative hypothesis is that the proportion of tickets marked "1" is 0.505. To find the power, we need to calculate the probability of observing 5,000 or more tickets marked "1" in a random sample of 10,000, assuming that the true proportion is 0.505.
Just like before, we can use the binomial distribution formula to find this probability. Let's assume it is denoted by power.
The power of the test is then equal to power. For example, if the power is 0.8, then the power of the test is 80%.
Please note that the actual calculations require the use of statistical software or a calculator with a binomial distribution function. The values provided here are for illustrative purposes only.