Math 362 Problem Set #9 â€“ Solution 27 April 2012 - Faculty web ...

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$Math 361 Problem Set #5 Due 5 October 2012 - Faculty web pages$

We are 100(1 − α)% confident that σ 2 lies between(n − 1)S2χ 2 Rand(n − 1)S2.χ 2 L(b) A random sample of 20 seniors at Edge City High School had a mean mathematicsSAT score of 455 with a sample standard deviation of 69. Assuming thatmathematics SAT scores are normally distributed, construct a 90% confidenceinterval for the standard deviation of all mathematics SAT scores at Edge CityHigh School.<strong>Solution</strong>: We first find χ 2 L and χ 2 R so thatPr(X < χ 2 L) = Pr(X > χ 2 R) = 0.05with 19 degrees of freedom. Our χ 2 table says that χ 2 L = 10.117 and χ 2 R = 30.144.The left end of our confidence interval for σ 2 19 × 692is ≈ 3000.90 and the right30.14419 × 692end is ≈ 8941.29. To get a confidence interval for the standard deviation10.117σ, we just take square roots.We are 90% confident that the standard deviation for all mathematics SAT scoresat ECHS lies between 54.78 and 94.56.3. The following data are from a normal distribution with mean µ and variance σ 2 :52.2, 46.4, 34.3, 43.8, 33.6, 32.5.Construct 95% confidence interval estimates for µ and σ.<strong>Solution</strong>: We find that the sample mean is about 40.46667 and the sample standarddeviation is about 8.156388.The sample size is 6, so our critical t-value has 5 degrees of freedom and an area of0.05 in two tails. We get t α/2 = 2.571, so the radius of our confidence interval for µ isE ≈ 2.571 × 8.156388 √6≈ 8.561.We are 95% confident that µ lies between 49.03 and 31.91.To construct a confidence interval for σ 2 , we find χ 2 L and χ 2 R so that P (X < χ 2 L) =P (X > χ 2 R) = 0.025 with 5 degrees of freedom. In the table, we find χ 2 L = 0.831 andχ 2 R = 12.833. Using the result from problem 2a, we get5 × (8.15639) 212.833≈ 25.9202 and 5 × (8.15639)20.831≈ 400.281
as the left and right endpoints of a 95% confidence interval for σ 2 . To find a confidenceinterval for σ, we take square roots.We are 95% confident that σ lies between 5.09 and 20.01.4. The probability density function for the Student t distribution with ν degrees of freedomis given by[ (Γν+1) ] ( ) −ν+12f T (t) = √ (πν Γν1 +2)t2 2.νFor a fixed ν, the quantity in brackets is a constant, and so the “core” of the distribution) −ν+1is the function t ↦→(1 + t2 2.ν(The quantity in brackets is there in order to make the PDF integrate to 1.)Show that, as ν → ∞, the core of the Student t distribution PDF approaches the coreof the standard normal PDF.What does this tell you about the Student t distribution?Γ ( )ν+12What does it tell you about lim √ (ν→∞ πν Γν2)?( ) −ν+1We need to evaluate lim 1 + t2 2, which we could write asν→∞ νlimν→∞[(1 + t2ν) ν )] −1(1 + t2 2.νUsing continuity and the fact that the limit of the product is the product of the limits,we getlimν→∞) ν+1(1 + t2 2ν==[limν→∞) ν ] −1(1 [ ( )]+ t2 2 −1lim 1 + t2 2νν→∞ ν[e t2] − 1 2· [1] − 1 2= e − t2 2 ,which is the core of the probability density function for the standard normal distribution.
Page 1: Math 362 Problem Set #9 - Solution
Page 5 and 6: (b) Now suppose Y follows a Gamma d

Math 362 Problem Set #9 â€“ Solution 27 April 2012 - Faculty web ...

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