Let X be a continuous random variable. We know that it takes values between 0 and 3, but we do not know its distribution or its mean and variance. We are interested in estimating the mean of X, which we denote by h. We will use 1.5 as a conservative value (upper bound) for the standard deviation of X. To estimate h, we take n i.i.d. samples X1,X2,…,Xn, which all have the same distribution as X, and compute the sample mean
H=1n∑i=1nXi.
Express your answers for this part in terms of h and n using standard notation.
E[H]=- unanswered
Given the available information, the smallest upper bound for var(H) is: - unanswered
Calculate the smallest possible positive value of n such that the standard deviation of H is guaranteed to be at most 0.01.
This minimum value of n is: - unanswered
We would like to be at least 99% sure that our estimate is within 0.05 of the true mean h. Using the Chebyshev inequality, calculate the minimum value of n that will achieve this.
This minimum value of n is: - unanswered
Assume that X is uniformly distributed on [0,3]. Using the Central Limit Theorem, identify the most appropriate expression for a 95% confidence interval for h.
- unanswered [H−1.96n−−√,H+1.96n−−√] [H−1.96⋅3−−−−−−√4n−−√,H+1.96⋅3−−−−−−√4n−−√] [H−1.963√4n−−√,H+1.963√4n−−√] [H−1.96⋅34n−−√,H+1.96⋅34n−−√
5. H-1.96*(3^0.5)/((4*n)^0.5),
H+1.96*(3^0.5)/((4*n)^0.5)
does someone have an answer for this question? i am stuck
please provide answer . anyone here to provide help
Help is much appreciated, please!
To answer these questions, we need to use some properties of random variables and statistical concepts. Let's go step by step:
1. E[H]:
E[H] represents the expected value or mean of the sample mean H. Since X is a continuous random variable, we can use the property that the expected value of the sample mean is equal to the true mean of the random variable X. Therefore, E[H] = h.
2. Smallest upper bound for var(H):
The variance of the sample mean H is given by var(H) = Var(X)/n, where Var(X) denotes the variance of the random variable X and n is the number of samples. We are given that the standard deviation of X is at most 1.5. Since the standard deviation is the square root of the variance, we can set an upper bound for the variance as (1.5)^2. Therefore, the smallest upper bound for var(H) is (1.5)^2/n.
3. Smallest possible value of n such that the standard deviation of H is guaranteed to be at most 0.01:
The standard deviation of H is the square root of the variance, so we want to find the smallest value of n such that the square root of (1.5)^2/n is at most 0.01. Solving this inequality will give us the minimum value of n.
4. Minimum value of n to achieve at least 99% confidence interval:
To achieve a certain level of confidence interval, we can use the Chebyshev inequality. The Chebyshev inequality states that for any random variable X with mean μ and standard deviation σ, the probability that X deviates from μ by more than k standard deviations is at most 1/k^2. In our case, we want the probability of X deviating from h by more than 0.05 (k = 0.05) to be at most 1/0.01 (since we want at least 99% confidence). Using the inequality, we can solve for the minimum value of n.
5. 95% confidence interval for h using the Central Limit Theorem:
The Central Limit Theorem states that as the sample size n goes to infinity, the distribution of the sample mean approaches a normal distribution regardless of the distribution of the individual samples. Assuming X is uniformly distributed on [0,3], we can use the Central Limit Theorem to establish a 95% confidence interval for h. The appropriate expression would be [H - 1.96 * 3/(4 * n)^0.5, H + 1.96 * 3/(4 * n)^0.5].
Remember, these answers are expressed in terms of h and n using standard notation.