Section 11.5 Estimation of difference of two proportions
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1 ection.5 Estimtion of difference of two proportions As seen in estimtion of difference of two mens for nonnorml popultion bsed on lrge smple sizes, one cn use CLT in the pproximtion of the distribution of smple men. z / / /. Thus, e z z n. 4n 4n n e Therefore, n pproximte 00(- )% CI for the difference ( ) of two popultion proportions is given by ( ˆ ˆ ) z /,( ˆ ˆ ) z /, n n n n where, ˆ ˆ ( ˆ ) nd ˆ ˆ ( ˆ ) Exmple: Exercise.49 R of tourists t fmous tourist ttrction, 84 out of 50 men AND 56 out of 50 women bought souvenirs. Construct 95% CI for the difference between true proportions of men nd women who buy souvenirs t this plce ( ˆ ˆ ) (.376).336(.664) How does one find smple size necessry for specified bound on mx error e in the estimte of this difference? Assume equl smple sizes, nd mx vlue of vrinces. Then
2 ection.6 Intervl Estimtion of Vrinces Rndom mple of size n from Norml popultion with both Men nd Vrince unknown From Theorem 8., we hve lerned tht the smpling distribution of the mple Vrince is given s follows: ( n) o Y, where =(n-). Now using n intervl (,b) such tht P ( Yb), we obtin the probbility sttement ( n) P ( b). An esy choice of,b is to llocte / re to the intervls (0, ) nd (b, ) under the chi-squre density, so tht the re inside the intervl (,b) equls. Then /, nd b /,, the vlues which re tbulted in Tble V, pp 576 for different vlues of.05,.05,.0 nd.005 nd,30.. o Note: Other choices of, b might give shorter intervl, but tht s for dvnced level of decision mking. Note tht the distribution of Y is sme for ll vlues of, therefore Y is pivotl quntity, nd the two inequlities inside the bove probbility sttement cn be reversed in terms of upper nd lower confidence limits on,.., ie
3 ( n) ( n), nd ( n) ( n) b. b Combining the two inequlities, we get the CI for : Theorem.9: Given R of size n from norml popultion with unknown men, 00( )% CI for the popultion vrince is given by ( n) s ( n) s, b where s = n ( Xi X) is the smple vrince. n i To obtin the 00( )% CI for, just tke the squre root of the bove lower nd upper limits, i.e., ( n) s ( n) s. b Exmple: Exercise.53. Obtin 95% CI for the true vrince of the skull length of given species discussed in Ex..30. Normlly distribution of skull lengths, n=0, s=.9cm. = =.7 nd b 9.03 /,.975,9 /,.05,9 The 95% CI for the true vrince of skull length is given by 9(.9 ) 9(.9 ) 9(.084) 9(.084),, i.e., (0.04,0.9)
4 ection.7 Estimtion of the Rtio of Two Vrinces Why might we need this? o For the equlity of two mens for smll smple sizes, we need to ssume tht the two popultions hve equl vrinces. o In generl, if the unknown vrinces of two popultions ren t similr, compring their mens my be useless for clculting probbility of prcticlly meningful events. Obtining the CI for the rtio of vrinces: nd denote the smple vrinces for independent rndom smples of sizes n, n from norml popultions with unknown vrinces,. Theorem 8.5 proved tht the smpling distribution of / the U is rndom vrible with nedecor s F- / distribution with n, nd n d.f. o Keep in mind the d.f. n, nd n re stted in order corresponding to the numertor nd denomintor. Now choosing nd b such tht P ( Ub), we / obtin the probbility sttement P ( b). / o Agin, n esy choice of,b is to llocte / re to the intervls (0, ) nd (b, ) under f, density, so tht the re inside the intervl (,b) equls.
5 o Then f /,, v nd b f /,, v. [how Figure for these two cut-off points.] Note: Other choices of, b might give shorter intervl. Note tht the distribution of U is sme for ll vlues of (, ), therefore it is pivotl quntity. The two inequlities inside the bove probbility sttement cn be reversed in terms of upper nd lower confidence limits on the rtio : / ; / / b. / b Combining these two inequlities, we get the desired CI. Theorem.0: Given R of size n nd n from two norml popultions with unknown mens, 00( )% CI for the rtio of their popultion vrinces is given by. b The vlues f.05,, v nd f.0,, v for different combintions of (, v ) re tbulted in Tble VI, pp o Note tht these tbles do not suffice for obtining the vlues of nd b even for 98% nd 90% CIs.
6 o However, the min reson for not tbulting needed cut-off points is n interesting theoreticl fct: o The smpling distribution of V / Uhs the f, density. Therefore, P ( U) P(/ U/ ) PV ( / ). o Hence, f /,, v =. f/,, o Vlues of b cn be obtined directly from the Tble VI. It is esy to scertin tht the 00( )% CI for corresponds to the squre roots of the bove lower nd upper limits, i.e., b
7 Exmple: Exercise.59 Two norml popultions, smple sizes n =, nd n 5 Citrus trees of two vrieties, the mple stndrd devitions re:. nd.5 ince 90% CI for the rtio of two vrinces is required, we hve.05.. The rtio of two smple vrinces =.64.5 Now, for =, nd 4, b f.05,,4.56, nd f.95,,4 = f.05,4,.74 Therefore, the 90% CI for the rtio of vrinces (.64) (.64).74 or (.5,.7536). b.56 Note: This CI covers the vlue. o there is not sufficient evidence ginst the ssumption tht the two popultions hve equl vrinces, which ws needed to estimte the difference of the men heights of two vrieties of citrus trees.
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