1. n₁ to n₀, we can see that the sample size required to achieve the same level of confidence decreases when the value of € is reduced to half.
2. n₂ to n₀, we can see that the sample size required to maintain the same level of confidence increases when the probability δ is reduced to half.
To determine how the value of n recommended by the Chebyshev inequality changes in the following cases, we'll analyze each case separately:
Case 1: Value of € is reduced to half its original value
The Chebyshev inequality states that for a random variable X with mean μ and variance σ², the probability that X deviates from μ by more than kσ is less than or equal to 1/k². In this context, X represents the sample proportion of smokers (Mn), μ represents the true fraction of smokers in the population (f), and σ² represents the variance of the sample proportion.
When the value of € is reduced to half its original value, the Chebyshev inequality provides a narrower interval within which Mn is likely to fall. This means that we can achieve the same level of confidence (P(Mn-f≥6) ≤ 8) with a smaller sample size.
To quantify this change, let's denote the original sample size as n₀ and the reduced sample size as n₁. The Chebyshev inequality for the original sample size can be expressed as:
P(Mn₀-f≥6) ≤ 8
Applying Chebyshev's inequality:
1/k² ≤ 8
Solving for k:
k ≥ 1/3
Using the formula for variance of the sample proportion:
σ² = f(1-f)/n
Substituting into the Chebyshev inequality:
1/(1/3)² ≤ 8
64f(1-f)/n₀ ≤ 8
Solving for nn₀ ≥ 64f(1-f) / 8
When the value of € is reduced to half, the new Chebyshev inequality can be expressed as:
P(Mn₁-f≥3) ≤ 8
Applying Chebyshev's inequality:
1/k² ≤ 8
Solving for k:
k ≥ 1/3
Using the formula for variance of the sample proportion:
σ² = f(1-f)/n₁
Substituting into the Chebyshev inequality:
64f(1-f)/n₁ ≤ 8
Solving for n₁:
n₁ ≥ 64f(1-f) / 8
Comparing n₁ to n₀, we can see that the sample size required to achieve the same level of confidence decreases when the value of € is reduced to half.
Case 2: Probability δ is reduced to half its original value
Reducing the probability δ from 8 to 4 means we have a lower tolerance for the deviation of Mn from f. This implies that we need a larger sample size to maintain the same level of confidence.
To quantify this change, let's denote the original sample size as n₀ and the increased sample size as n₂. The Chebyshev inequality for the original sample size can be expressed as:
P(Mn₀-f≥6) ≤ 0.08
Applying Chebyshev's inequality:
1/k² ≤ 0.08
Solving for k:
k ≥ 2.5
Using the formula for variance of the sample proportion:
σ² = f(1-f)/n₀
Substituting into the Chebyshev inequality:
25f(1-f)/n₀ ≤ 0.08
Solving for n₀:
n₀ ≥ 1250f(1-f) / 32
When the probability δ is reduced to half, the new Chebyshev inequality can be expressed as:
P(Mn₂-f≥6) ≤ 0.04
Applying Chebyshev's inequality:
1/k² ≤ 0.04
Solving for k:
k ≥ 3
Using the formula for variance of the sample proportion:
σ² = f(1-f)/n₂
Substituting into the Chebyshev inequality:
9f(1-f)/n₂ ≤ 0.04
Solving for n₂:
n₂ ≥ 225f(1-f) / 16
Comparing n₂ to n₀, we can see that the sample size required to maintain the same level of confidence increases when the probability δ is reduced to half.
Question
In order to estimate f, the true fraction of smokers in a large population, Alvin selects n people at random. His estimator M, is obtained by dividing S, the number of smokers in his sample n, i.e. Mn Sn. Alvin chooses the sample size n to be the smallest possible number for which the = Chebyshev inequality yields a guarantee that:
P(Mn-f≥6) ≤ 8,
where e and 8 are some prespecified tolerances. Determine how the value of n recommended by the Chebyshev inequality changes in the following cases.
1. The value of € is reduced to half its original value.
2. The probability & is reduced to half its original value.