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=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) that we can assert/guarantee is:
Var(H)≤
unanswered
Calculate the smallest possible 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 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.
This minimum value of n is:
unanswered
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.
Normal Table
Show
[H−1.96⋅3−−−−−−√n−−√,H+1.96⋅3−−−−−−√n−−√]
[H−1.963n−−√,H+1.963n−−√]
[H−1.96⋅3–√n−−√,H+1.96⋅3–√n−−√]
[H−1.96⋅3n−−√,H+1.96⋅3n−−√]
unanswered
E[H]= 𝑯 = 1/𝒏∑𝒊=1𝒏𝑿𝒊
Given that the variance of X is at most 4, the smallest upper bound for Var(H) that we can assert/guarantee is:
Var(H)≤ 𝑽𝒂𝒓(𝑯) = 1/𝒏²𝑽𝒂𝒓(𝑿)
Calculate the smallest possible value of n such that the standard deviation of H is guaranteed to be at most 0.01.
This minimum value of n is: 𝟔𝟒𝟎𝟎
Using the Chebyshev inequality, we have:
P(|H - h| ≤ 0.02) ≥ 1 - (𝑽𝒂𝒓(𝑯) / (0.02)²)
We want to be at least 96% sure, so we set the inequality as:
1 - (𝑽𝒂𝒓(𝑯) / (0.02)²) ≥ 0.96
Simplifying, we get:
𝑽𝒂𝒓(𝑯) ≤ (1 - 0.96) * (0.02)²
We can approximate the variance of H to be 𝑽𝒂𝒓(𝑯) ≈ 4/𝒏, so:
4/𝒏 ≤ (1 - 0.96) * (0.02)²
Solving for n, we get:
𝒏 ≥ 400
This minimum value of n is: 400
Now, suppose X is uniformly distributed on [h-3,h+3], for some unknown h. Using the Central Limit Theorem, the most appropriate expression for a 95% confidence interval for h is:
[H - 1.96⋅(3/√n), H + 1.96⋅(3/√n)]
To estimate the mean h, we use the sample mean H = (1/n) * ΣXi.
1. Expected value of H (E[H]):
The expected value of H is equal to the mean of the random variable X, which is h. Therefore, E[H] = h.
2. Smallest upper bound for Var(H):
To find the smallest upper bound for Var(H), we use the fact that the variance of X is at most 4. The variance of H is given by Var(H) = Var((1/n) * ΣXi).
Since the samples Xi are i.i.d., we can use the property Var(aX) = a^2 * Var(X) for a constant a. Therefore, Var(H) = (1/n^2) * ΣVar(Xi).
Since the variance of X is at most 4, we have Var(H) ≤ (1/n^2) * n * 4 = 4/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 given by the square root of Var(H). So, we need to find the smallest value of n such that sqrt(Var(H)) ≤ 0.01.
From the previous part, we know that Var(H) ≤ 4/n. Taking the square root, we get sqrt(Var(H)) ≤ 2/sqrt(n).
Therefore, we need to find the smallest n such that 2/sqrt(n) ≤ 0.01.
Solving this inequality, we get sqrt(n) ≥ 200, which gives n ≥ 200^2 = 40,000.
4. Minimum value of n to be at least 96% sure that the estimate is within 0.02 of the true mean h:
Using the Chebyshev inequality, we can find the minimum value of n to achieve this level of confidence.
The Chebyshev inequality states that P(|X - E[X]| ≥ k * σ) ≤ 1/(k^2), where X is a random variable with mean E[X] and standard deviation σ.
Here, we want the probability of being within 0.02 of h, so we want to find the value of k such that P(|H - h| ≥ 0.02) ≤ 0.04 (1 - 0.96).
Since the samples Xi are i.i.d., the variance of H is Var(H) = Var((1/n) * ΣXi) = (1/n^2) * n * Var(X) = Var(X)/n.
We are given that Var(X) ≤ 4, so Var(H) ≤ 4/n. Therefore, σ = sqrt(Var(H)) ≤ 2/sqrt(n).
Using k * σ = 0.02, we can solve for k: k * (2/sqrt(n)) = 0.02.
Simplifying, we get k = (0.02 * sqrt(n))/2 = 0.01 * sqrt(n).
Substituting k into the Chebyshev's inequality, we have 1/(k^2) ≤ 0.04.
Therefore, we need to find the minimum value of n such that (0.01 * sqrt(n))^2 ≤ 0.04.
Solving this inequality, we get n ≥ 10,000.
5. 95% confidence interval for h when X is uniformly distributed on [h-3, h+3]:
When X is uniformly distributed on [h-3, h+3], the Central Limit Theorem states that the sample mean H follows approximately a normal distribution with mean h and standard deviation (3 * sqrt(n))/sqrt(12).
For a 95% confidence interval, we use the standard normal distribution table (Z-table) and the critical value for a 95% confidence level, which is 1.96.
Therefore, the most appropriate expression for a 95% confidence interval for h is [H - (1.96 * (3/sqrt(12)) * (1/sqrt(n))), H + (1.96 * (3/sqrt(12)) * (1/sqrt(n))).
To answer these questions, we need to use some basic concepts and formulas related to random variables, estimation, variance, standard deviation, confidence intervals, and the Central Limit Theorem.
1. Expected Value of H (E[H]):
The expected value of H is denoted as E[H] and represents the mean of the random variable H. Since H is the sample mean of n i.i.d. samples Xi, and the mean of X is denoted as h, we have:
E[H] = E[1/n * (X1 + X2 + ... + Xn)]
Since all Xi have the same distribution as X, we can rewrite this as:
E[H] = (1/n) * (E[X1] + E[X2] + ... + E[Xn])
Since E[Xi] = h for all i, we have:
E[H] = (1/n) * (nh) = h
Therefore, E[H] = h.
2. Upper Bound for Var(H) (Var(H)):
To find the upper bound for Var(H), we need to consider that Var(H) = Var(1/n * (X1 + X2 + ... + Xn)). Since the Xi's are independent, we can use the property Var(aX) = a^2 * Var(X), where a is a constant, to get:
Var(H) = (1/n^2) * (Var(X1) + Var(X2) + ... + Var(Xn))
Given that the maximum variance of X is 4 and the Xi's have the same distribution as X, we can write:
Var(H) ≤ (1/n^2) * (4 + 4 + ... + 4) = (4/n^2) * (n) = 4/n
Therefore, Var(H) ≤ 4/n.
3. Minimum Value of n for Standard Deviation of H ≤ 0.01:
The standard deviation of H is denoted as SD(H) and is equal to the square root of Var(H). We need to find the minimum value of n such that SD(H) ≤ 0.01. From the previous result:
SD(H) = √(Var(H)) ≤ √(4/n)
We want SD(H) ≤ 0.01, so:
√(4/n) ≤ 0.01
Squaring both sides and rearranging, we get:
4/n ≤ 0.01^2
4/n ≤ 0.0001
n ≥ 4/0.0001
n ≥ 40,000
Therefore, the minimum value of n is 40,000.
4. Minimum Value of n for 96% Confidence Interval:
To calculate the minimum value of n for a 96% confidence interval, we can use the Chebyshev inequality. The Chebyshev inequality states that for any random variable with mean μ and variance σ^2, and any k > 0, the probability that the random variable deviates from its mean by more than k standard deviations is at most 1/k^2.
In our case, we want to be at least 96% sure that our estimate is within 0.02 of the true mean h. Let's denote the standard deviation of H as SD(H). We need to find n such that:
P(|H - h| > 0.02) ≤ (1 - 0.96) = 0.04
Using the Chebyshev inequality, we have:
P(|H - h| > 0.02) ≤ Var(H) / (0.02^2)
Since Var(H) ≤ 4/n, we can write:
4/n ≤ Var(H) / (0.02^2)
Cross-multiplying and rearranging, we get:
n ≥ 4 / [(Var(H) / (0.02^2))]
Substituting Var(H) ≤ 4/n, we have:
n ≥ 4 / [(4/n) / (0.02^2)]
n ≥ 4 * (n / 4) * (1 / (0.02^2))
n ≥ n / 100
n ≥ 100
Therefore, the minimum value of n is 100.
5. Confidence Interval for h with Unknown Distribution:
When X is uniformly distributed on [h - 3, h + 3], we can use the Central Limit Theorem (CLT) to approximate the distribution of H. The CLT states that the distribution of H approaches a normal distribution as the sample size increases.
Given that H approaches a normal distribution, a 95% confidence interval for h can be calculated using the standard normal distribution (Z-table). The most appropriate expression for the confidence interval is:
[H - 1.96 * 3 / √n, H + 1.96 * 3 / √n]
This is because the 1.96 value corresponds to the 2.5th percentile (0.025) and the 97.5th percentile (1 - 0.025) of the standard normal distribution, which gives us the interval containing 95% of the data.
Note: The actual value of H is unknown, so it is replaced by h when calculating the confidence interval.
Therefore, the correct option from the provided choices is:
[H - 1.96 * 3 / √n, H + 1.96 * 3 / √n]