Let X be a continuous random variable. We know that it takes values between 0 and 6 , but we do not know its distribution or its mean and variance, although we know that its variance is at most 4 . We are interested in estimating the mean of X , which we denote by h . To estimate h , we take n i.i.d. samples X1,…,Xn , which all have the same distribution as X , and compute the sample mean
H=1/n*∑i=1nXi.
1. Express your answers for this part in terms of h and n using standard notation.
E[H]=
Given the available information, the smallest upper bound for Var(H) that we can assert/guarantee is:
2. Var(H)≤
Calculate the smallest possible value of n such that the standard deviation of H is guaranteed to be at most 0.01.
3. This minimum value of n is:
We would like to be at least 96% sure that our estimate is within 0.02 of the true mean h . Using the Chebyshev inequality, calculate the minimum value of n that will achieve this.
4. This minimum value of n is:
Suppose now that X is uniformly distributed on [h−3,h+3], for some unknown h . Using the Central Limit Theorem, identify the most appropriate expression for a 95% confidence interval for h . You may want to refer to the normal table.
1. h
2. 4/n
3. 40000
4. 250000
5. H+/- 1.96 * sqrt(3)/sqrt(n) i.e option (c)
4. 250,000
1. h
2. (var)/n , so its 4/n
3. 40000
4 - not sure on cheb's inequality
5. calculate the variance of x based on uniform dist (0,6), so it equals to 3, then apply CLT resulting in H+/- 1.96 * sqrt(3)/sqrt(n)
1. E[H] = h
2. Var(H) ≤ 4/n
3. To ensure the standard deviation of H is at most 0.01, we set an upper bound for Var(H):
0.01^2 ≤ 4/n
Simplifying, we get:
0.0001 ≤ 4/n
n ≥ 4/0.0001
n ≥ 40,000
Therefore, the smallest possible value of n is 40,000.
4. Using the Chebyshev inequality, we have:
P(|H - h| ≥ 0.02) ≤ (Var(H)/(0.02)^2)
Since Var(H) ≤ 4/n, we can substitute:
P(|H - h| ≥ 0.02) ≤ (4/n) / (0.02)^2
Since we want to be at least 96% sure, we set an upper bound of 0.04 (1 - 0.96) on the probability:
(4/n) / (0.02)^2 ≤ 0.04
Simplifying, we get:
n ≥ 4 / (0.04 * (0.02)^2)
n ≥ 250,000
Therefore, the minimum value of n to achieve at least 96% confidence level is 250,000.
5. When X is uniformly distributed on [h−3,h+3], we know that the mean is h and the standard deviation is 3/sqrt(3) = sqrt(3). Using the Central Limit Theorem, a 95% confidence interval for h would be:
[h - 1.96 * (sqrt(3)/sqrt(n)), h + 1.96 * (sqrt(3)/sqrt(n))]
Therefore, the most appropriate expression for a 95% confidence interval for h is:
[h - 1.96 * (sqrt(3)/sqrt(n)), h + 1.96 * (sqrt(3)/sqrt(n))]
1. To calculate the expected value of the sample mean H, we can use the fact that the expected value of a sum is the sum of the expected values and the expected value of a constant times a random variable is the constant times the expected value of that random variable.
Given that the sample mean is calculated as H = (1/n) * ∑(i=1 to n) Xi, where Xi are the individual samples, the expected value of H can be calculated as:
E[H] = E[(1/n) * ∑(i=1 to n) Xi]
Since each Xi has the same distribution as X, and the expected value of X is denoted as h, we can rewrite the expression:
E[H] = (1/n) * ∑(i=1 to n) E[Xi] = (1/n) * ∑(i=1 to n) h = (1/n) * n * h = h
Therefore, E[H] = h.
2. To find an upper bound for Var(H), we can use the fact that the variance of a sum of random variables is the sum of their variances, as long as the random variables are independent.
Given that the variances of each sample Xi are at most 4, we can write the variance of H as:
Var(H) = Var((1/n) * ∑(i=1 to n) Xi)
Since each Xi has the same distribution as X, and the variances of Xi are at most 4, we can rewrite the expression:
Var(H) = (1/n^2) * Var(∑(i=1 to n) Xi)
Since the Xi are assumed to be independent, we can use the property of variances to rewrite the expression:
Var(H) = (1/n^2) * ∑(i=1 to n) Var(Xi)
Since each Var(Xi) is at most 4, we have:
Var(H) ≤ (1/n^2) * n * 4 = 4/n
Therefore, Var(H) ≤ 4/n.
3. To guarantee that the standard deviation of H is at most 0.01, we can use the fact that the standard deviation is the square root of the variance. So, we want to find the smallest possible value of n such that:
sqrt(Var(H)) ≤ 0.01
Substituting the expression for Var(H) from above:
sqrt(4/n) ≤ 0.01
Squaring both sides:
4/n ≤ 0.0001
Solving for n:
n ≥ 4/0.0001
n ≥ 40000
Therefore, the minimum value of n is 40000.
4. To use the Chebyshev inequality to be at least 96% sure that our estimate is within 0.02 of the true mean h, we need to find the minimum value of n such that:
P(|H - h| ≥ 0.02) ≤ (1 - (1/0.96^2))
Using the Chebyshev inequality, we can write:
Var(H) / (0.02)^2 ≤ 1 / (0.96)^2
Substituting the expression for Var(H) from above:
4/n ≤ 1 / (0.96)^2
Simplifying:
n ≥ 4 / (0.96)^2
n ≥ 4 / 0.9216
n ≥ 4.34028
Since n must be an integer, the smallest value of n that satisfies this inequality is n = 5.
Therefore, the minimum value of n is 5.
5. The Central Limit Theorem states that as the sample size increases, the distribution of the sample mean approaches a normal distribution. Given that X is uniformly distributed on [h-3, h+3], we can use the Central Limit Theorem to construct a confidence interval for h.
The Z-score for a 95% confidence interval is approximately 1.96 (which can be found in a standard normal table).
The formula for the confidence interval is:
CI = [H - (Z * (σ / sqrt(n))), H + (Z * (σ / sqrt(n)))]
Since X is uniformly distributed on [h-3, h+3], the standard deviation (σ) of X can be calculated using the formula:
σ = (b - a) / sqrt(12)
where a = h - 3 and b = h + 3.
Substituting the values into the formula:
σ = (h + 3 - h + 3) / sqrt(12)
σ = 6 / sqrt(12)
σ = 6 / 2√3
Therefore, the 95% confidence interval for h is:
CI = [H - (1.96 * (6 / (2√3)) / sqrt(n))), H + (1.96 * (6 / (2√3)) / sqrt(n)))]
CI = [H - (6 * 1.96 / (2√3 * sqrt(n))), H + (6 * 1.96 / (2√3 * sqrt(n)))]