independence of the random sample measurements, we have U = Z i ~ χ 2 (n) with σ / n 1. Now let W = σ 2. We then have σ 2 (x i µ + µ x ) 2 i =1 ( )

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1 MATH 48 Chi-Square Aalysis of a Normal Stadard Deviatio Dr Neal, WKU We ow shall use the chi-square distriutios to aalyze the stadard deviatio of a measuremet that is kow to e ormally distriuted The proof of the mai theorem elow requires the more advaced result that x ad S are actually idepedet of each other But we shall assume this result without proof Our mai result is the: Theorem Let { x 1, x,, x } deote the collectio of all radom samples of size from a ormally distriuted measuremet havig a o-zero variace Let S = 1 ( x 1 i x ) e the distriutio of all possile sample variaces The ( 1) S follows a χ ( 1) distriutio Proof Let Z i = x i µ ~ N(0, 1) for,,, The Z σ i ~ χ (1), ad y the idepedece of the radom sample measuremets, we have U = Z i ~ χ () with 1 mgf M U (t ) = / x µ Also, 1 t σ / ~ N(0, 1) ; so V = x µ ~ χ (1) with mgf σ / 1 M V (t) = 1/ ( 1) S Now let W = 1 t We the have W = ( 1) S = 1 (x i x ) = 1 (x i µ + µ x ) ( ) = 1 ( x i µ ) + (x i µ )(µ x ) + (µ x ) = x i µ + (µ x ) σ ( x i µ) = Z i + (µ x ) ( x µ ) + = χ () = χ () = U V (x µ ) ( x µ ) + (x µ ) = χ () ( x µ ) x µ σ / + (µ x ) σ = χ () χ (1)

2 ( 1) S x µ Thus, W + V = U, where W = ad V = are idepedet y our σ / assumed fact of the idepedece of x ad S Now sice the mgf of a idepedet sum is the product of the mgf s, we have M U (t ) = M W+V (t ) = M W (t ) M V (t) Thus, 1 M W (t ) = M U (t ) / M V (t ) = / 1 1/ 1 1 t 1 t = ( 1)/ 1 t So W = ( 1) S has the mgf of a χ ( 1) distriutio; thus, ( 1) S ~ χ ( 1) Applicatio I: Whe radom samplig o a ormally distriuted measuremet, we ca evaluate P(a S ), provided is kow, y P(a S ) = P(a S ) ( 1)a ( 1)a ( 1)S ( 1) χ ( 1) ( 1) Applicatio II: Let S e the sample variace from a radom sample of size of a ormally distriuted measuremet havig variace A (1 α ) 100% cofidece iterval for is give y ( 1)S R 1 ( 1)S L 1, where L 1 = χ (1 α/) ( 1) ad R 1 = χ α / ( 1) are the left ad right chi-square scores of the χ ( 1) distriutio that give 1 α proaility i the middle ( 1)S ( 1)S cofidece iterval for σ is the σ R 1 L 1 A To derive this cofidece iterval, we solve for the edpoits a ad such that ( ) = 1 α, with P( a) = α / ad P( ) = α / P a

3 To solve for a, we have ( ) = P 1 a 1 α = P a = P ( 1)S a ( 1)S ( 1)S a χ ( 1) So the value the have a = ( 1)S a ( 1)S R 1 must equal the right-tail chi square score R 1 = χ α / ( 1) We Similarly, we have ( ) = P 1 α = P σ 1 = P ( 1)S ( 1)S = P χ ( 1) ( 1)S, so that P χ ( 1) which gives = ( 1)S L 1 ( 1)S = 1 α / Thus, ( 1)S = L 1 = χ (1 α/) ( 1) Applicatio III: To test the ull hypothesis H 0 : σ = M for a ormally distriuted measuremet, we otai the sample deviatio S from a radom sample of size The ( 1) S ( 1) S test statistic is the x = = M which is compared with the χ ( 1) distriutio We compute the (left-tail) P -value P χ ( 1) x ad compute the (right-tail) P -value P χ ( 1) x ( ) for the alterative H a : σ < M, ( ) for the alterative H a : σ > M Applicatio IV: We have see that E[χ ()] = ad Var(χ ()) = Thus, with samples of size from ormally distriuted measuremets, we have Var ( 1)S = Var(χ ( 1)) = ( 1) Thus, ( 1) = Var ( 1)S = ( 1) σ 4 Var(S ) Hece, Var(S ) = σ 4 1

4 Example (a) Radom samples of size 46 are take from a measuremet that is N(100,15) What is P(13 S 17)? What is the stadard deviatio of all S? () From a ormally distriuted measuremet, a sample of size 0 yields S = 396 Fid a 98% cofidece iterval for the true stadard deviatio σ (c) From a ormally distriuted measuremet, a sample of size 5 yields a sample deviatio of 1396 Is there evidece to reject the hypothesis H 0 : σ = 15? Solutio (a) Covertig to S ( 1) S P( 13 S 17) = P( 13 S 17 ) = P ~ χ ( 1), we otai ( 1)13 ( 1)S ( 1) χ ( 1) ( ) P 338 χ (45) 578 (usig TI commad χ cdf(338, 578, 45) ) Also, Var(S ) = σ 4 1 = = 50; thus, σ S = () The 98% left ad right chi-square scores with 1 = 19 degrees of freedom are L 1 = 7633 ad R 1 = 3619 Thus, a 98% cofidece iterval for σ is σ or 8693 σ (c) For S = 1396, we use the alterative H a : σ < 15 The test statistic is x = ( 1) S = 15 = , ( ) , usig χ cdf(0, , 4) which we compare to the χ ( 1) = χ (4) distriutio The P -value for the leftsided alterative is P χ (4) If σ = 15 were true, the there would e a % chace of otaiig a sample deviatio of 1396 or lower with a sample of size 5 There is ot eough evidece to reject H 0

5 Exercises 1 Let X ~ χ (15) Fid (a) P(13 X 17), () P(X < 13) ad (c) P(X > 17) Show a graph for each (d) Fid the ouds that cotai 95% of the distriutio Adult heights are foud to e ormally distriuted with mea µ = 68 iches ad stadard deviatio σ = 35 iches Suppose various radom samples of size = 6 are collected Compute P(8 S 4) ad fid the stadard deviatio of all S 3 From a ormally distriuted measuremet, a sample of size 5 yields a sample deviatio of 1485 Fid a 95% cofidece iterval for the true stadard deviatio 4 From a ormally distriuted measuremet, a sample of size 16 yields S = 46 Is there evidece to reject the hypothesis H 0 : σ = 3? State iitial ad ull hypotheses, compute the test statistic ad P -value, the explai your coclusio i detail