The sample mean
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=1 to n Xi.
1) Express your answers for this part in terms of h and n using standard notation.
a) E[H]=
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Given the available information, the smallest upper bound for Var(H) that we can assert/guarantee is:
b) Var(H)≤
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2) 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:
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3) 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:
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4) 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.
a) [H − (√1.96⋅3/√n),H + (√1.96⋅3/√n)]
b) [H− (1.96/√3n),H + (1.96/√3n)]
c) [H− (1.96⋅√3/√n),H + (1.96⋅√3/√n)]
d) [H− (1.96⋅3/ √n),H + (1.96⋅3/√n)]
a) E[H] = h
1) Express your answers for this part in terms of h and n using standard notation.
a) E[H] = h
The expected value of H is equal to the true mean h.
b) Var(H) ≤ 4/n
The upper bound on the variance of H is 4 divided by the sample size n.
2) Calculate the smallest possible value of n such that the standard deviation of H is guaranteed to be at most 0.01.
To find the smallest possible value of n, we need to set an upper bound for the standard deviation of H using the variance expression from part 1b.
Standard Deviation of H = √Var(H)
Since we want the standard deviation to be at most 0.01, we can set up the inequality:
√(4/n) ≤ 0.01
To solve for n, we can square both sides of the inequality:
4/n ≤ (0.01)^2
n ≥ 4/(0.01)^2
n ≥ 40000
Therefore, the minimum value of n is 40000.
3) 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.
The Chebyshev inequality states that for any random variable with mean μ and variance σ^2, the probability that the random variable deviates from its mean by more than k standard deviations is at most 1/k^2.
In this case, we want to be at least 96% sure that our estimate H is within 0.02 of the true mean h. This means we want the probability of H deviating from h by more than 0.02 to be at most (1 - 0.96) = 0.04.
Using the Chebyshev inequality, we can set up the inequality:
P(|H - h| > 0.02) ≤ 0.04
Since the standard deviation of H is equal to the square root of the variance from part 1b, we can rewrite the inequality as:
P(|H - h| > 0.02) ≤ σ^2 / (0.02)^2
To achieve a probability of at most 0.04, we want the upper bound on the right-hand side to be at most 0.04:
σ^2 / (0.02)^2 ≤ 0.04
Substituting the upper bound on the variance from part 1b:
(4/n) / (0.02)^2 ≤ 0.04
Simplifying the inequality:
4/n ≤ 0.04 * (0.02)^2
n ≥ 4 / (0.04 * (0.02)^2)
n ≥ 5000
Therefore, the minimum value of n that will achieve at least 96% confidence is 5000.
4) 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.
The Central Limit Theorem states that, under certain assumptions, the distribution of a sum or average of a large number of i.i.d. random variables will be approximately normal.
In this case, since X is uniformly distributed, the sample mean H will be approximately normally distributed according to the Central Limit Theorem.
A 95% confidence interval for the true mean h can be calculated as:
[H - (1.96 * (3/√n)), H + (1.96 * (3/√n))]
Therefore, the most appropriate expression for a 95% confidence interval for h is option a) [H - (1.96 * 3/√n), H + (1.96 * 3/√n)].
1) Express your answers for this part in terms of h and n using standard notation.
a) E[H] is the expected value of the sample mean. Since X is a continuous random variable, E[H] is equal to the true mean h. Therefore, E[H] = h.
b) Var(H) is the variance of the sample mean. Given that the variance of X is at most 4, we can use a well-known formula for the variance of the sample mean:
Var(H) = Var(X)/n
Since the variance of X is at most 4, we have Var(H) ≤ 4/n.
2) To guarantee that the standard deviation of H is at most 0.01, we need to find the smallest possible value of n that satisfies this condition.
The standard deviation of H is equal to the square root of the variance of H. So, we want to find the smallest n such that:
sqrt(Var(H)) ≤ 0.01
Substituting Var(H) = 4/n, we have:
sqrt(4/n) ≤ 0.01
Simplifying, we get:
2/sqrt(n) ≤ 0.01
Dividing both sides by 2, we have:
1/sqrt(n) ≤ 0.005
Taking the reciprocal of both sides and squaring, we get:
n ≥ (1/0.005)^2
n ≥ 40000
Therefore, the minimum value of n is 40000.
3) Using the Chebyshev inequality, we can find the minimum value of n required to be at least 96% sure that our estimate is within 0.02 of the true mean h.
The Chebyshev inequality states that for any random variable, the probability that the absolute difference between the random variable and its mean is greater than or equal to k standard deviations is at most 1/k^2. In our case, we want the difference between H and h to be less than or equal to 0.02, so k = 0.02.
Therefore, we want to find the minimum value of n such that:
1/k^2 ≤ 1 - 0.96
1/0.02^2 ≤ 0.04
2500 ≤ 0.04n
Dividing both sides by 0.04, we get:
62500 ≤ n
Therefore, the minimum value of n is 62500.
4) Suppose X is uniformly distributed on [h−3, h+3]. Using the Central Limit Theorem, we can identify the most appropriate expression for a 95% confidence interval for h.
The Central Limit Theorem states that for a large sample size, the distribution of the sample mean approaches a normal distribution. In this case, since X is uniformly distributed, the distribution of the sample mean H will approach a normal distribution.
For a 95% confidence interval, we want to find the range of values around the sample mean H that contains the true mean h with 95% confidence.
The appropriate expression for a 95% confidence interval is:
[H - (1.96 * σ/√n), H + (1.96 * σ/√n)]
Since X is uniformly distributed, the standard deviation σ is equal to the range of the distribution, which is 3.
Substituting σ = 3, the correct expression for a 95% confidence interval is:
[H - (1.96 * 3/√n), H + (1.96 * 3/√n)]
Therefore, the correct answer is d) [H − (1.96⋅3/ √n), H + (1.96⋅3/√n)].