Section 13.5 Test of Hypothesis for Population Proportion

Transcription

1 Secton 135 Test of Hypothess for Populaton Proporton As dscussed n Secton , nference about the probablty of success n ndependent Bernoull trals s same as the mean of ndependent observatons on bnary outcomes If we assume that random varable X for the th unt n the populaton s Bnary valued {0,1} [coded values for (H,T), (S,F) (Buy, not Buy) etc] wth P X 1 Then, E( X ), Var( X ) (1 ) The dfference from a general mean problem s that the populaton varance depends on the mean (proporton) We wsh to test the null hypothess H 0 : 0 about the probablty of success from a RS of sze n Bernoull trals, aganst smple alternatves or composte hypotheses ( ) 0, or ( ) 0 or ( ) 0 In ths experment, the suffcent statstc s X X = # of Successes out of n trals It has a Bnomal dstrbuton bxn ( ;, )[Appendx B] For ths problem, Exercse 11(Assgned problem n HW #5) gave a test for smple aganst smple alternatve s a onetaled test based on X For composte alternatves, the LR test gves the level crtcal regons as follows:

2 Reect H 0 f ( ) x k, where k s the smallest nteger for whch b( y; n ) ( ) x k, where ks the largest nteger for whch b( y; n ) ( ) x k or x k respectvely / / n yk k Example:1356, 1357, 1358, 1360 Show how to fnd P- value Bnomal Probablty Tables for n 1,, 0, gven n the Text Tables are avalable for n up to 100 Excel and Statstcal software packages can also be used to obtan these values For any n, one can fnd the P-value for the one-taled or two taled test for any 0 The level test reects the null hypothess, f P-value < However, for large n, e, when [ n 0 5 and n(1 0) 5 ] one can use the CLT to approxmate the P-value 1 For n large, CLT mples that X X s approxmately n normally dstrbuted wth EX ( ), VarX ( ) (1 )/ n Under H 0 : y0 0 0 Z X 0 X n0 (1 ) / n n (1 ) N(0,1) The test based on Z-statstc s the same as the test for normal mean dscussed n Secton 13

3 Therefore, level tests for these hypotheses are gven by: Reect H 0 f ( ) z z,( ) x zand ( ) z z/ or z z/respectvely Examples: 1363, 1366 Note: A better CLT approxmaton can be obtaned by usng the contnuty correcton ntroduced n Chapter 6, Example 65 But we wll not dscuss ths mprovement here, snce the test outcome rarely changes, unless the observed value s very close to the cut-off pont In that case, we may want to obtan more nformaton (larger n) anyway The remanng sectons n ths Chapter are devoted to one of the twenty mportant scentfc breakthroughs of the 0 th Century, accordng to the ournal Scence: the development of the ch-square statstcal test It s applcable to a large class of tests of hypotheses In each case, the dstrbuton of the statstc under the null hypothess, for large samples, can be approxmated by the v dstrbuton Its df ( k1 t) depends on the dmenson (k) of the data summary, and # of free parameters (t) estmated from the data It compares the observed data summares f wth the expected summares under the null model e If the observed summary vector ( f 1, f,, f k ) s not close to the expected summary vector ( e 1, e,, e k ), the ch-squared statstc k ( f e) s large, and the level test s of the form: 1 e Reect the Null f ;

4 Secton 136 -Tests for Two or More Proportons Example: or more (say, k) supplers of same components th producer s probablty of producng acceptable qualty components: Take a random sample of sze n from th populaton Let X = # of successes out of n Bernoull Trals X, 1,,, k, mutually ndependent random varables ~ Bn( n, ) Want to test the hypothess that all supplers provde same qualty product, e, H0 : 1 k = 0(known) H0: Not all 's are equal to 0 If sample szes (n s) are suffcently large, CLT allows us to approxmate the dstrbuton of X, 1,,, k, by normal dstrbutons, as dscussed n prevous secton Standardze these to ndependent N(0,1) random varables X n Z, 1,,, k n (1 ) From Theorem 88, we know that the sum of squares of k k ndependent N(0,1) random varables, Z, has a ch squared dstrbuton wth k df, e, Z 1 k 1 Under the null hypothess, all, 0 therefore, X n Z, where, Z, 1,,, k (1 ) k 0 1 n0 0 k

5 If one of ' s s not equal to, 0 then ths statstc s expected to be large Thus, a level- test s: Reect H 0, f, k However, ths test s not used much, snce n most practcal problems, the common value of these proportons s unknown (not a specfed value 0 ) It must be estmated from the data Under H 0, X s are ndependent samples from Bnomal wth unknown parameter : common prob of success k k Then the suffcent statstc X Bnn; 1 1 The mvue (mle), so called the pooled estmate of s X n n n Now, the ch-squared statstc s modfed: Replace 0 k ( X ) ( ) n n 1 n (1 ) (1 ), by Its dstrbuton s ch-squared wth (k-1) df The modfed ch-squared statstcs contans only (k-1) ndependent random quanttes, snce ( n ) 0 Example: Exercses 1370, 1371, page 49

6 Connecton to estmatng the dfference of two proportons Under H 0, the statstc for estmatng the dfference of two ( 1 ) proportons becomes z 1 1 (1 ) n1 n For k=, the two terms n statstcs nvolve ( ) n n ( ), ( ) ( ) n1 n n1 n Now, smple algebra shows that ( ) (1 ) n z Ch-squared (or z ) test s for two-sded alternatves The test based on z can be used for one-sded alternatves Ch-squared statstc va a k Table lstng observed number of falures and successes Another verson applcable to a large varety of tests of dfferent hypotheses: k ( f e ),where 1 1 e f : observed frequency n cell(, ), e : expected frequency n cell(, ) under H0 Example: Exercse 1371, and the common verson

7 Secton 137 Test of Hypotheses for r c Tables Same general procedure s applcable to () Problems nvolvng samples of a fxed sze from each of the r populatons Each response has c (c>) categores (X : Multnomal based on n ndependent trals, wth probablty vectorθ ( 1,,, c); 1,,, r ) Hypothess: The r populatons have same underlyng probablty vector H θ θ θ aganst the alternatve 0 : 1 r H 1 : at least one probablty vector s dfferent from others Example: Detergent Preferences, page 418 () Problems nvolvng a large sample of sze N from one populaton Categorze each ndvdual accordng to two attrbutes, wth r and c categores respectvely Let P(observaton falls n cell(,)), c P(observaton falls n row ) 1 r P(observaton falls n column ) 1 Null hypothess s that the two attrbutes are ndependent, e, H0 : for all (,) Example: Exercse 1377 Both types of problems use the same Ch-squared test procedure:

8 r c ( f e ),where 1 1 e f : observed frequency n cell(, ), e : expected frequency n cell(, ) under H0 The degrees of freedom for the ch-squared statstcs s ( r1)( c 1) The concluson of the level test s: Reect the Null f ; Case (): The common probablty vector under H 0, say, θ ( 1,,, c ), s estmated by / ; 1,,, where # f f c f total obs Snce the th row was based on a sample sze n, t follows that f n, and the expected count n cell (, ) s equal to e n f f / f Overall, there are r(c-1) ' s, and we estmated (c-1) parameters n the common probablty vector So df = # free parameters - # estmated parameters = r(c-1) - (c-1) = (r-1)(c-1) = If at least one probablty vector was dfferent from others, the observed ch-squared wll be large The level test reects null f ; Case(): The margnal row and column probabltes for multnomal are estmated by: f / f, 1,,, r; and f / f, 1,,, c Now, under H 0 :

9 Hence, the expected count n cell (,) s gven by e f f f f / f Expected counts are same as n case () Now, the df = # free parameters - # estmated parameters = (rc-1) (r-1) - (c-1)= (r-1)(c-1) = If the two attrbutes are not ndependent, the condtons would be volated n at least two cells, and the ch-squared statstc would be large The level test reects null f ; Note that the two test procedures are testng dfferent hypothess, and the Ch-squared statstcs and ts degrees of freedom are same, even though they follow very dfferent logc to arrve at the fnal soluton

10 Secton 138 Testng Goodness of Ft Gven a data set of sze n (large) We wsh to test the null hypothess that the data s a random sample from the dstrbuton px (, ) If the parameter s known, the null dstrbuton p( x, ) s unquely specfed If t s unknown, we wll need to fnd an estmate assumng that the data s a random sample from the model p( x, ) Now act as f the dstrbuton under the null s p p( x, ) Now assume the values of the random varable X can be parttoned nto m dsont subsets (bns) A1, A,, A m Let, f Observed counts n the bn A; p P{ X A p},and e n p denote the expected counts n bn A The ch-squared statstc for testng the ft of the data to the specfed model s gven by m ( f e) 1 e If the observed frequences are not close to the expected frequences n some bns, the chsquared statstc wll be large The level test reects the null f ;, m t 1 where m = # of bns and t = # of ndependent parameters estmated from the data

11 Example: Exercse The null hypothess s that the 160 tosses of 4 cons follow a Bn(4;5) Example: Exercse The null hypothess s that the data follows a Posson dstrbuton Example: Exercse The null hypothess s that the data follows a Normal dstrbuton Fnally, a CAUTIONARY NOTE: The concluson not to reect the NULL Hypothess does not mean that the NULL s true See, eg, We smply start wth the assumpton that the null s correct, and see f there s ample evdence aganst the null based on the data If not, we stay wth our assumpton But t could stll be untrue Future evdence mght stll dscover that the null hypothess s not true Note: Next few lectures wll be devoted to Non-parametrc statstcs, whch does not assume a parametrc model for the data