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Statistics II: Statistical Inference, Diapositivas de Estadística

Unit 7: Confidence Intervals Statistics II: Statistical Inference

Tipo: Diapositivas

2021/2022

Subido el 13/11/2023

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Statistics II:
Part III: Statistical Inference
Unit 7: Confidence Intervals
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Statistics II:

Part III: Statistical Inference

Unit 7: Confidence Intervals

Contents

1. Concept of confidence interval.

2. Confidence intervals for normal populations.

3. Confidence intervals for non-normal populations.

a) For the probability 𝑝 of a binomial 𝐵( 1 , 𝑝). b) For unknown distributions (Chebyshev).

  1. Determination of sample size

Concept of confidence interval If we are trying to estimate 𝜃, we will search that 𝑃 𝜙 1 𝑋 1 , … , 𝑋𝑛 ≤ 𝜃 ≤ 𝜙 2 𝑋 1 , … , 𝑋𝑛 = 1 − 𝛼. We will call 𝛼 the significance level and 1 − 𝛼 the confidence level.

Concept of confidence interval Notice that now, the random variables are the extremes of the confidence interval, 𝜙 1

1

𝑛 and 𝜙 2 𝑋 1 , … , 𝑋𝑛. On the contrary, 𝜃 is an (unknown) number. Once the (specific) sample has been selected, the random variables 𝑋𝑖 will take each a value 𝑥𝑖. As a consequence, the extremes of the interval will be the values 𝜙 1 𝑥 1 , … , 𝑥𝑛 and 𝜙 2 𝑥 1 , … , 𝑥𝑛.

Confidence intervals for normal populations

We will study the following confidence intervals

for a normal population 𝒩(𝜇, 𝜎):

a) For the expected value 𝜇 when 𝜎 is known. b) For the expected value 𝜇 when 𝜎 is unknown. c) For the standard deviation 𝜎 or the variance 𝜎 2 .

Confidence interval for 𝜇 When the population is normal , 𝑋 ∼ 𝒩 𝜇, 𝜎 , and we know 𝜎, we know that 𝑋 ∼ 𝒩 𝜇, 𝜎 𝑛

Given a significance level 𝛼, we search for 𝜙 1 , 𝜙 2 such that 𝑃 𝜙 1 𝑋 1 , … , 𝑋𝑛 ≤ 𝜇 ≤ 𝜙 2 𝑋 1 , … , 𝑋𝑛 = 1 − 𝛼. There are many ways to achieve this.

Confidence interval for 𝜇 The equation 𝑃 𝐾 1 ≤ 𝑍 ≤ 𝐾 2 = 1 − 𝛼 has multiple solutions. However, it can be proved that there is one unique solution which makes the interval [𝐾 1 , 𝐾 2 ] be as narrowest as possible: 𝐾 1 = −𝐾 2 ≡ 𝐾.

Confidence interval for 𝜇 We then have 𝑃 −𝐾 ≤ 𝑍 ≤ 𝐾 = 1 − 𝛼, and because of the symmetry of the standard normal, we deduce that 𝐾 = 𝑧𝛼 (^) Τ 2 , i.e., 𝐾 is the point in the standard normal distribution which leaves to the right a probability of 𝛼 2

. In other words 𝑃 𝑍 ≥ 𝐾 =

Confidence interval for 𝜇 In conclusion, the CI for 𝜇 (when 𝜎 is known) is: 𝑋 −

𝛼 2

𝛼 2

Confidence interval for 𝜇 (bis) When the population is normal , 𝑋 ∼ 𝒩 𝜇, 𝜎 , but we do not know 𝜎. Again, given a significance level 𝛼 , we search for 𝜙 1 , 𝜙 2 such that 𝑃 𝜙 1 𝑋 1 , … , 𝑋𝑛 ≤ 𝜇 ≤ 𝜙 2 𝑋 1 , … , 𝑋𝑛 = 1 − 𝛼. However, now we can’t work with 𝑍 = 𝑋−𝜇 ൗ 𝜎 𝑛

Confidence interval for 𝜇 (bis) We now search for numbers 𝐾 1 , 𝐾 2 such that 𝑃 𝐾 1 ≤ 𝑊 ≤ 𝐾 2 = 1 − 𝛼. Again, there are many possible choices, but it can be proved that there is one solution which makes the interval [𝐾 1 , 𝐾 2 ] be the narrowest: 𝐾 1 = −𝐾 2 ≡ 𝐾. We then have 𝑃 −𝐾 ≤ 𝑍 ≤ 𝐾 = 1 − 𝛼, and because of the symmetry of the Student’s t, we deduce that 𝐾 = 𝑡𝑛− 1 , 𝛼Τ 2.

Confidence interval for 𝜇 (bis) Thus, 𝑃 −𝐾 ≤

Confidence interval for 𝜎 When the population is normal , 𝑋 ∼ 𝒩 𝜇, 𝜎 , we know that W = 𝑛− 1 𝜎 2

1 2 ∼ 𝜒 𝑛− 1 2 (pivot). Again, given a significance level 𝛼 , we search for 𝜙 1

2 such that 𝑃 𝜙 1

1

𝑛

2

1

𝑛

Confidence interval for 𝜎 First, we search for numbers 𝐾 1

2 such that 𝑃 𝐾 1 ≤ 𝑊 ≤ 𝐾 2 = 1 − 𝛼. Again, there are many possible choices of 𝐾 1 and 𝐾 2. However, it is usually agreed to choose them so that the interval leaves to the right and to the left the same probability, i.e., 𝛼/ 2. (This solution is not the one that makes the interval be the narrowest)