The central limit theorem (CLT) states that for sufficiently large sample sizes, the distribution of sample means follows an approximately normal distribution with mean equal to the population mean and standard deviation equal to the population standard deviation divided by the square root of the sample size.
In this case, the sample mean estimate Theta hat is equal to 22. Since the random variable has a variance of 1, the population standard deviation is also 1.
The CLT tells us that the standard deviation of the sample mean is equal to the population standard deviation divided by the square root of the sample size.
So, the standard deviation of the sample mean is 1 / sqrt(100) = 1/10.
To find the 80% confidence interval, we need to find the z-score corresponding to the desired confidence level. For a one-tailed 80% confidence level, the z-score is approximately 1.28.
The margin of error is equal to the z-score multiplied by the standard deviation of the sample mean.
Margin of error = 1.28 * (1/10) = 0.128
To calculate the lower bound of the confidence interval, subtract the margin of error from the sample mean estimate:
Lower bound = 22 - 0.128 = 21.872 (approximately)
To calculate the upper bound of the confidence interval, add the margin of error to the sample mean estimate:
Upper bound = 22 + 0.128 = 22.128 (approximately)
Therefore, the 80% confidence interval is approximately [21.872, 22.128].
The sample mean estimate Theta hat of the mean of a random variable with variance 1, based on 100 samples, happened to be 22. The 80% confidence interval provided by the CLT is of the form [a,b] with a=? and b=?
1 answer
1 answer