Unformatted text preview: MAC 3123: Statistics for Behavioral Sciences I Week 1: Chapter 2 Inferences based on one sample (Hypothesis Teshg) / parameter Pa u sed for populahaas statistic 1 x F S used for samples *Using sample statistic we can estimate the parameter Critical Value Method : Ex1 Claim the population mean is greater than 2,400 lu> 2,400) Sample information: n. 50 NEEDED values to be able to x = 2,460 estimate I wm bl GIVEN In the problem. sa 2002 Test the claim at a 107. level of significance. Ho: 42,400 HM> 2,400 right tailed Ho : , ,= H2, >, lotta FIVE STAT OLrain tes+ Statiste X -A For A n230 : use z, 2 CAR n=50 use z-test h 2 30 : use t ; t- - Z: A = 2460 - 2400 3R where s p z= 2:121 (NEVER vound alk3rd decimal point for multipke chuite) ZAT2.00 For , PE A Study Soup I There at two possible devisions : Reject the Ho OR fail to reject the Ho Need rejection region to mow which HAM>2,400 a = 0.10 (3avingnyeuses) 4TYG Tassig FIVE STAR H : Rejection for Filed this area. BE PORE referring to the table 0.5-0.10 = 0.4000 By looking up The 2-valve that gives lus the closest area to 0.4000 we obtain : z = 1.2.8 Our test stetisme value of z= 2.121 is greater than aur zs |.28 Therefore we ruect the Ho be cana The test statistic fills into the reaction vagion, CVO CN * For to tailed: 2 TF z value fills within the rejection region you rycot Ho if not Then fail to rajut the HO. STARHow to decide if you have Suffisert endinte : Conclusion: There is suffident evidence to support the claim that u>2,400 at a O.10. Does the claim corte in the condihen of equality? YES NO Coriginal claim does not crtalin e polity and becomes AAST Cor goodalanaray) vecut Ho Do you rejcet no 3 YES He "the sample YES derte Supports Do you the claim that... rejeit Ho 3 Loriginal claimy" NO NO fail to reject feril to rjeet "IyLis not sufficient em derul + Suppur+ Tue "There is not clorien That - original dim" Sumhui entendre to rejce+ Tha chim... Loriginale dei HO * There is Su kuat evidence to reject the claim...Coriginal t claims Ex 2% elain: The populatan propokon is at least 25%. (p>0.25) Sam ple information: h=200 Not always, Ps+ of success = X - 49 = 19 0.245 ize = = 0.245 tot, simple size X = 49 Test the claim at 1. level of significente O Ho: P>0.25 HA: P 40.25 , left tailed - Obtain Tes+ Statisha: n= 200 = vse z-test tee pe fochies ALWAYS a se z-values) z P-P -0.245 -0.25 -0.163 PL 0.25 * 0.75 200 > Hai PL 0.25 a = 0.01 FIVE STAR -CVO For lett Teiled test remember to manually assign (-) value!!! o.s-o.ol = 0.4900 By looking - Tu 2-value that gives us Tue olosest value to 0.4 900 we obtain Tue value 2= -- 2.33 ar test stahishe value of -0.163 is less Than our 2= - 2:33 Turfere we fail to rejeet The Ho been use it does not fall umhin Tue rejection region. O Conclusion: There is not sthuert evidence turejeet Tle daim at -o.ol. That pz 0.25EX 33 Claim: The population mean is 1.2 ( 1.2) Sem al Infurmatian: n=o. x=1.05 S=O.S Test Tu dain about su populehen ata 19. level of signih conce O Hos u=1.2 HA: M1 .2, two + ailed ed ks+ (divde s !!) 2 Ottorin Test Stensho: n=100 se z-score z=- _ 1.05-12 SHR0.5/ico 6 Ha. 1.2 = 0.01 /2 = 0.005 A=0.005 N =0.005 -CVO CV 0.5-0.005 = 0. 49 50 By looking up the z-value that gives us The closest value to O. 9950 we obtain tu value 25 -2.575 2.575 (we get tuo vales bloot Tre tuo tailed tet) Our tes+ stahiste of 2-3 falls within the riguekun megfor Turefore we should rejeet the Ho confusion. There is sufficient emiona to rejest tue dan That 1.2 a+ a-o.or.