Sunday, June 21, 2020

Statistical computation - 550 Words

Statistical computation (Coursework Sample) Content: Studentà ¢Ã¢â€š ¬s Name:Tutorà ¢Ã¢â€š ¬s Name:Course:Date Due:Math Problem, MathematicsIntroductionQuestion oneStandard deviation=578 pounds,ÃŽ =0.05SolutionHo: Â=3000H1: Â3000 this will be right tailed testN=60 mean=3120 S=578 ÃŽ =0.05 Test Value= 1.61C.V= 1-0.05=0.95the shaded regionFrom the table Zcv= 1.65 C.V=1.65Test value: Z= mean-ÂS/à ¢Ã… ¡n= (3120-3000)/ 578/à ¢Ã… ¡60=1.61Reason= the test value doesnà ¢Ã¢â€š ¬t fall within the shaded region hence no evidence that the average production of peanuts has increased. (C.V=1.65)Question 8SolutionAverage=$59.93H0= $59,593 H1=$59,593 (this will be the claim) -1.65 0Critical value= ÃŽ 0.0500-1.65Test value= Z= mean-ÂS/à ¢Ã… ¡nZ= (58,800-59593) / 1500/à ¢Ã… ¡30Z=-2.90Analysis= the test value is less than critical valueWe reject the H0Conclusion= the state employe es earn less than the federal employeesQuestion 11H0: 500 à Ã†â€™=42. à Ã†â€™=0.01H1: 476Z= mean-ÂS/à ¢Ã… ¡nZ= (500-476) /42/ à ¢Ã… ¡50=4.04Analysis = the test value is more than the critical valueWe accept the H0,The mean differs from 500.Question 16SolutionH0: Â= 52H1: à ¢ 52 Z= mean-ÂS/à ¢Ã… ¡nZ= (56.3-52) / 3.5 / à ¢Ã… ¡50Z=8.69 -1.96 0 1.96Reason= NO, the Z value is into the rejected region. Therefore we reject his claim.Section 8-3Question 8SolutionH0 : ÂH1: Â25.4 ÃŽ=0.1, df=n-1 t=(22.1-25.4)/ 5.3/à ¢Ã… ¡25 =-3.11 CV: t0.1= -1.38.Comparing: -3.11 -1.38 (outside CV).Conclusion: reject H0 suggesting that the average commute is indeed shorter.Question 14SolutionClaim Â= 110 calories, if false  110 calories.H0: Â= 110 calories. H1 : 110 calories. Â0=110 ÂÂ0p- Value= 0.000309 hence reject H0. There is sufficient evidence to reject the claim that the average content of calories is not more than 110.Question 17SolutionAt ÃŽ= 0.05, n=20, mean = 3.85 and à Ã†â€™=2.519. d.f= 19.H0: Â= 5.8 H1: Âà ¢ 5.8(claim) t= mean-ÂS/à ¢Ã… ¡n= (3.85-5.8)/ 2.519/à ¢Ã… ¡20= -3.46.p- Value0.01.if p-value =ÃŽ, 0.01=0.05 hence true, we reject the null hypothesis. There is no enough evidence to support the claim that the mean is not 5.8Section 8.4Question 10Ho: Â=500H1: Â=420à Ã†â€™=0.05ÃŽ=0.86Z= mean-ÂS/à ¢Ã… ¡nZ= (500-420) /0.86/0.05à ¢Ã… ¡3801.13The proportion does not differ from the national percentage.Question 14SolutionHo: p=0.517Ha: pà ¢0.517 n=200, x=115, p^=115/200=0.575Z = (0.575-0.517)/ ((0.517)(0.483)/200)=1.64P-value = 2P (Z1.64) =0.101Conclusion: since the p-value (0.101) 0.05Don ´t reject H0The evidence supports the c laim there is no enough evidence to say that the percentage has changed.