5. H-1.96*(3^0.5)/((4*n)^0.5),
H+1.96*(3^0.5)/((4*n)^0.5)
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−−√
H+1.96*(3^0.5)/((4*n)^0.5)
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.