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.

1.1. h

1.2. 2.25/n

2. 22500

3. 90000

4.

H-1.96*(3^0.5)/((4*n)^0.5),
H+1.96*(3^0.5)/((4*n)^0.5)

To estimate the mean of X, denoted by h, using the given information, we can follow the following steps:

1. E[H] - To find the expected value of H, we take the expected value of each Xi and use linearity of expectation:
E[H] = E[1/n * (X1 + X2 + ... + Xn)]
= 1/n * (E[X1] + E[X2] + ... + E[Xn])
= 1/n * (h + h + ... + h) (since all the Xi have the same distribution and expected value of h)
= h

Therefore, E[H] = h.

2. Smallest upper bound for var(H) - To find the smallest upper bound for the variance of H, we use the fact that the samples are independent and use the variance properties:
var(H) = var(1/n * (X1 + X2 + ... + Xn))
= 1/n^2 * (var(X1) + var(X2) + ... + var(Xn))
≤ 1/n^2 * (1.5^2 + 1.5^2 + ... + 1.5^2) (since the upper bound for the standard deviation is 1.5)
= 1/n^2 * (n * 1.5^2)
= 1.5^2/n

Therefore, the smallest upper bound for var(H) is 1.5^2/n.

3. Smallest possible value of n for standard deviation of H ≤ 0.01 - To find the minimum value of n such that the standard deviation of H is guaranteed to be at most 0.01, we set up the following inequality:
sqrt(1.5^2/n) ≤ 0.01
1.5/n ≤ 0.0001
n ≥ 1.5/0.0001

Therefore, the minimum value of n is ceil(1.5/0.0001).

4. Minimum value of n for 99% confidence within 0.05 of h using Chebyshev inequality - To find the minimum value of n using the Chebyshev inequality, we use the following inequality:
P(|H - h| ≥ 0.05) ≤ (var(H))/(0.05)^2 (Chebyshev inequality)
P(|H - h| ≥ 0.05) ≤ (1.5^2/n)/(0.05)^2
P(|H - h| ≥ 0.05) ≤ (1.5^2)/(n * 0.05^2)

Since we want to be at least 99% sure, we set up the following equality:
1 - (1.5^2)/(n * 0.05^2) = 0.01
(1.5^2)/(n * 0.05^2) = 0.99
n = (1.5^2)/(0.99 * 0.05^2)

Therefore, the minimum value of n is ceil((1.5^2)/(0.99 * 0.05^2)).

5. 95% confidence interval for h using Central Limit Theorem - Since X is uniformly distributed on [0,3], we can use the Central Limit Theorem to approximate the distribution of H as approximately normal. The mean of H will still be h, and the standard deviation of H will be approximately 1.5/sqrt(n) according to the Central Limit Theorem.

Therefore, the 95% confidence interval for h can be expressed as h ± Z * (1.5/sqrt(n)), where Z is the critical value corresponding to a 95% confidence level (e.g., Z ≈ 1.96 for a large enough sample size).

Note: Please substitute the unanswered parts with the respective answers obtained from the explanations above.