Interval Estimation and Confidence Intervals

Lecture Note 9 ∗ Interval Estimation and Confidence Intervals MIT 14.30 Spring 2006 Herman Bennett 22 Interval Estimation Interval estimation is another approach for estimating a parameter θ. Interval estimation consists in finding a random interval that contains the true parameter θ with probability (1 − α). Such an interval is called confidence interval and the probability (1 − α) is called the confidence level. � � P A(X1 , ..., Xn ) ≤ θ ≤ B(X1 , ...Xn ) = 1 − α (71) Example 22.1. Assume a random sample from a N (µ, σ 2 ) population, where both para­ meters are unknown. Find the 90% (symmetric) confidence interval of σ 2 . ∗ Caution: These notes are not necessarily self-explanatory notes. They are to be used as a complement to (and not as a substitute for) the lectures. 1 Herman Bennett LN9—MIT 14.30 Spring 06 Example 22.2. Assume a random sample from a N (µ, σ 2 ) population, where the para­ meter µ is unknown and σ = 2. Find the 95% (symmetric) confidence interval of µ. How would your answer change if σ is unknown? � 1.96σ ¯ √ ,X n 1.96σ √ n � • The random interval X̄ − + contains the true parameter θ with 95% probability. It is wrong to say that θ lies in the interval with 95% probability...θ is not a RV! 23 23.1 Useful results t-student distribution A RV X is said to have a t-student distribution with parameter v > 0 (degrees of freedom) if the pdf of X is: X ∼ t(v) : f (x|v) = Γ( v+1 ) 1 1 2 √ , 2 v x Γ( 2 ) vπ (1 + ( v ))(v/2+0.5) for − ∞ < x < ∞ and v positive integer. 2 (72) Herman Bennett LN9—MIT 14.30 Spring 06 Let X ∼ N (0, 1) and Z ∼ χ2n be independent RVs. Then, the RV H is distritbuted t-student with n degrees of freedom. X H=� ∼ t(n) Z/n (73) • Symmetric distribution around 0, which implies that tα/2,n = −t1−α/2,n . • As n → ∞, t(n) → N (0, 1). (See attached graph and table). Example 23.1. Important result. Assume a normally distributed random sample. Then, √ n(X̄−µ) ∼ t(n−1) . Prove this result. S 23.2 F distribution A RV X is said to have an F distribution with parameters v1 > 0 and v2 > 0 if the pdf of X is: X ∼ F(v1 ,v2 ) : Γ( v1 +2v2 ) f (x|v1 , v2 ) = v1 Γ( 2 )Γ( v22 ) for 0 < x < ∞ and vi positive integer. 3 � v1 v2 �(v1 /2) x(v1 /2−1) , (1 + ( vv12 )x)(v1 /2+v2 /2) (74) Herman Bennett LN9—MIT 14.30 Spring 06 Let X ∼ χ2n and Z ∼ χ2m be independent RVs. Then, the RV G is distritbuted F with n and m degrees of freedom. G= 24 X/n ∼ F(n,m) Z/m (75) Constructing Confidence Intervals for θ In what follows, we consider 5 possible cases of limited information about the parameter(s) θ. For each case we study how to construct confidence intervals. 24.1 Case 1: θ̂ ∼ N (θ, V ar(θ̂)) and V ar(θ̂) known We just saw an example of this case in Example 22.2. Note that in this example θ = µ, θ̂ = X̄, and the V ar(θ̂) is known since σ is known. 24.2 Case 2: θ̂ ∼ N (θ, V ar(θ̂)) and V ar(θ̂) unknown Example 24.1. Assume as in Example 22.2 a normal random sample, but now both µ and σ2 are unknown. Construct a 95% confidence interval of µ. 4 Herman Bennett 24.3 LN9—MIT 14.30 Spring 06 Case 3: θ̂ not ∼ N ( ) but pmf/pdf known We just saw an example of this case in Example 22.1. Note that in this example θ = σ 2 , θ̂ = S 2 , and the pdf of θ̂ (a function of θ̂ in this case) is known and depends only on one parameter, σ 2 . 24.4 Case 4: θ̂ ∼ not Normal, pmf/pdf unknown, and n > 30 Example 24.2. Assume a random sample of size n from a population f (x), which is unknown. Construct a 99% confidence interval of µ = E(Xi ). 24.5 Case 5: θ̂ ∼ not Normal, pmf/pdf unknown, and n < 30 5