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]=h - unanswered
h
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: 0,1! - 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:
1.1. h
1.2. 2.25/n
2. 22500
3. 90000
4. 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
H-1.96*(3^0.5)/((4*n)^0.5),
H+1.96*(3^0.5)/((4*n)^0.5)
what about others ?anyone to answer them ?
To find the expected value of H (E[H]), we can use the fact that E[H] is equal to the expected value of a single sample Xi. Since all the samples have the same distribution as X, we have:
E[H] = E[X1] = h,
where h is the mean of X (the variable we are trying to estimate).
To find the smallest upper bound for var(H), we can use the fact that the variance of H is equal to the variance of a single sample divided by n (the number of samples). Since each sample has an upper bound for its standard deviation of 1.5, the variance of a single sample is upper bounded by (1.5)^2 = 2.25. Therefore, the smallest upper bound for var(H) is 2.25/n.
To calculate the smallest possible positive value of n such that the standard deviation of H is guaranteed to be at most 0.01, we can use the fact that the standard deviation of H is equal to the square root of the variance of H. So we need:
sqrt(2.25/n) ≤ 0.01.
Solving for n, we have:
1/sqrt(n) ≤ 0.01/sqrt(2.25).
Squaring both sides and solving for n, we get:
1/n ≤ (0.01/sqrt(2.25))^2.
Taking the reciprocal, we have:
n ≥ 1/((0.01/sqrt(2.25))^2).
To be at least 99% sure that our estimate is within 0.05 of the true mean h, we can use the Chebyshev inequality. The Chebyshev inequality states that the probability that a random variable deviates from its mean by more than k standard deviations is at most 1/k^2.
In our case, we want the probability of deviation to be at most 1 - 0.99 = 0.01, and we want the deviation to be at most 0.05 standard deviations (k = 0.05). Using the inequality, we have:
1/k^2 ≥ 1 - 0.01.
Simplifying, we get:
1/0.05^2 ≥ 1 - 0.01.
Solving for n, we have:
1/n ≥ 1 - 0.01.
Taking the reciprocal, we get:
n ≥ 1/(1 - 0.01).