################ ### OSMI CAS ### ################ #2. prop.test(x = c(52,120), n = c(68,130), conf.level = 0.99) # odbracujemo H_0 #4. 2*(1-pf(1.28, 120, 120)) #p vrednost #5. 2*(pf(0.7143, df1 = 24, df2 = 24)) pt(-0.433, df = 48) #6. A <- c(200,220,320,450,500,550,550,550,600,670,700,1000,1200,1200) B <-c(230,230,300,300,300,400,450,450,500,500,500,800,800,800,1000,1000,1000,1500,1500,1500,1500) # Prvo testiramo hipotezu o jednakosti disperzija var.test(A, B, ratio = 1, alternative = "two.sided") # p~0.2, pa mozemo smatrati da su disperzije jednake t.test(A, B, mu = 0, var.equal = TRUE) # p>0.05 => ne odbacujemo H0 # nema znacajnih razlika izmedju prosecnih plata u te dve firme # ILI A <- c(200,220,320,450,500,550,550,550,600,670,700,1000,1200,1200) B <-c(230,230,300,300,300,400,450,450,500,500,500,800,800,800,1000,1000,1000,1500,1500,1500,1500) #prvo testiramo hipotezu o jednakosti disperzija alfa <- 0.05 sn1 <- var(A) sn2 <- var(B) n1 <- length(A) n2 <- length(B) c1 <- qf((1/2)*alfa,n1-1,n2-1) #trazimo levu granicu fiserove raspodele c2 <- qf(1-(1/2)*alfa,n1-1,n2-1) #trazimo desnu granicu fiserove raspodele W <- c(c1,c2) #kriticna oblast TN <- (sn1)/(sn2) #ako TN upadne izmedju c1 i c2 upada u kriticnu oblast TN < c1 | TN > c2 #ako je true onda upada u kriticnu oblast i odbacujemo H0 tj. nama je FALSE pa su disperzije jednake TN1 <- (mean(A) - mean(B))/(sqrt(sn1/length(A) + sn2/length(B))) #ako upadne u kriticnu oblast odbacujemo H0, tj da su prosecne plate jednake Q <- (sn1/length(A) + sn2/length(B))^2/((sn1/length(A))^2/(length(A)-1)+(sn2/length(B))^2/(length(B)-1)) #stepeni slobode c <- qt((1/2)*alfa,Q) abs(TN1) < c #ne upada u kriticnu oblast. Prosecne plate su jednake, tj ne odbacujemo H0 #7. # Prvo testiramo hipotezu o jednakosti disperzija (F_kapa <- 1600/625) (c1 <- qf(0.1/2, 24, 24)) (c2 <- qf(1-0.1/2, 24, 24)) F_kapa < c1 | F_kapa > c2 # F_kapa jeste u kriticnoj oblasti => odbacujemo H0 # Disperzije nisu jednake # broj stepeni slobode za Studentovu raspodelu (Q <- (1600/25 + 625/25)^2/((1600/25)^2/24+(625/25)^2/24)) # Q>30, pa mozemo aproksimirati normalnom, inace bismo imali qt(0.05, df=40) (c <- qnorm(0.05)) (t_kapa <- (570-600)/sqrt(1600/25 + 625/25)) t_kapa < c # odbacujemo H0 tj. bolje je da uzme rezim rada B #7. (t_kapa <- 0.045/sqrt(0.0084)*sqrt(20)) (alfa <- 0.05) (c <- qt(1-alfa, 19)) t_kapa > c # realizovana vrednost T upada u kriticnu oblast => odbacujemo H0 #9. prvi <- c(3.4, 5.1, 3.5, 7.2, 7.5) drugi <- c(2.8, 5.4, 3.3, 7.2, 7.8) # pravimo uzorak razlika d <- prvi - drugi t.test(d, mu=0, alternative = "greater") # t.test za jedan uzorak # drugi nacin: t.test(prvi, drugi, mu = 0, paired = TRUE, alternative = "greater") # t.test za dva uzorka ali sa parametrom paired=TRUE sto sugerise na uparene podatke, # p>0.05 => ne odbacujemo H0 # nema razlika izmedju srednjih vrednosti cena u dva supermarketa