FIU - MAT, STA 3123 - Class Notes - Week 2

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Unformatted text preview: MAC 3123: STATS. FOR BEHAVIORAL SCIENCES T Week 2: Ch. 3 Inferences based on 2 samples Chypothesis Testing. Test akout the diffennu eteen 2 population means M, Standard deviatica Lsar a variance Variance = (standard deviations Population Memns When Samples are independent samples : compare 2 independ sa Srps A and B When som ples are de pendent samplescom pace only 1 group before and after treatment Sample size mes vary Sample size ALWAYS Tre some Z test: BOTH sample SI zes 230 n. z 30 (n,230 t-test: n. 230 And n2430 n. 230 and nz 230 n, 30 and n, 230 Home =0 HAAFO right-tailed Independent Samples Hypothesis for miru, test: Houmaso Ho: mr-, 20 Hai - 2 > Ha. ruL0 2. ways of notation teft tailed Hou, Ho z Hame> HA , 42 2. tailed test M Ho - HA: 1. independent Ex i n=160 n2180 K, = 98 Determine whether there is a difference % -9 S. = 16.6 between the two population means using a 5 2 17. 6 1 0. level of significance. What we knowo: , Ho - o Hino (2 tails) 3. 2-tust vlen, 30 AND Obtain Test Stahshc: X- X, 98-06 z'Aro 16,65 * 0170= 1077 Simple 4. CD-0.05 Rejection Region: &/2 = 0.10/2 = 0.05 2 tailed ks+ divide a b 2 FAIL TO TEJECT Ho Le cause Ksr stehstic 0.5-0.os = 0.4500 Value is NOT within valle de redarf -CV z: - 1.64, 1.64 he rejcom on rgion ble 2 taited dont forget (f) ANDE) val 4 a = 0.05 cone Conclusion: There is NO sutricitat evidence to support the daim that u. Me at . FIVESTAS Ex 2'. Despite the fuck that business travelers are the airlines biggest customers, They pay the highest prices. In his study of corporate air travel costs, Richard Fox reportedl That 45% ot 56 bis user Arms and III of 146 small users have established in-house travels agency. Test the hypothesis for a significant dicheria in The proportion of brg ad small -Surs Thort have established -house travel agency. Use Q=OOL LUSE critical value methods. What we know: independent samples big users and sell users) 2. mean or proporton ? iz proportion , claimip Pa sus Somallars 3. n = 56 -0.45 P = = 0.76 x = (0.45 56 25 X= n2 = 146 Success 4. o.o O He: =P2 HA'S P* Pz (2 tails obtain Tes+ Stortishet proportons ALWAYS 2-test) ZER-P = , whert p = X11*2 = 25*10L = 0.673 and qat-p = 0.327 De-4 (4A) sahip z = 0.45-0.76 = -4.204 10-673-0.327 (St+1946) PRyuton Region: a/2 = 0.01 /2 = 0.005 oxo.cos 0.5-0.065 = 0.49 50 Recor He becam.se test stetste feils amin L2 - 2.57,+ 2.574 vejcetian region 0.5-0.0 AS 0.005 V Conclusion: There is suthurat w durue to support PI+Pz at a=o.ol. a claim thatFIVESTAT Ex 3: n = 6 nz 5 Does The data s-apest That the poplattor 8-5 x2 = 8.2 mean 1 is greater than the popuiten S, = 0.3668 si = . mtan 2 ? Use as o.os. O Claim: AL Hoi-AYO Ariu. - >o (right tail) e Obtain Test Statistic : Use t-test joecause sompile sizes (niLn, are 430 t: X- X, Spy L pool estanteria comburen en of the two yeri mnces (letter to use sp2 = (n - Dls,) + (ra-1) | ,') n n , - 2. = (6-) (0.3666)7 +(5-1) (0.0BII) - Orlo7 6 +5 -2 sp= 150 loi = 0.3327 t = 8.5 - 8.2 (0:3327) t-t 1.489 Rejretion Region = o.os z t-to-cs degrees of trudom :(nutn2-259. FAIL TO REJECT Lcorres pardin table velse! . 833 No Fellihin onclusion: Thuat is no sutkatot in due to support the claim that muata = o.os. FIVES HIVESTAR Ex 4: Independent randon sam puas were selected from 2 populations : Sample Sam ple 2 n = 12. n2 = 32 K - 17,8 x = 15.3 s2 = 74. 2 s. - KO.S Test the hypothesis mat tue population mean 1 is greater than the population mean 2 , ntn 1. level of signifiance (use crinal Valve method) Oclaim: M M2 Hola- H: M. -A2>o right tail) Obtain Test Stanste: Use t test brease une simple size (n) 430 t> ****, st , - In-156,5* +(n2-1)(sa) = (12-1) ( 79.2) + (32-1) bos) 12 + 32 - 2 t: 17.8-15.3 sp sp2 = 64.056 = 8.00 8.005 Tz ORyceron Region 1 0 2 0.01 - ta-to-ci; degru of hudom : .+n2-2 = 42 corresponding table value (to reerst degue of heedom) : 2.423 FAIL to rejcer Ho de

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