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Final Exam 11)y=87.767 + 2.157*x
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1b)
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(b) Given a=0.04, the critical value is Z(0.04)= -1.75 (from standard normal table)
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(c)Z=(phat-p)/√(p*(1-p)/n) = (151/403-0.31)/√(0.31*(1-0.31)/403)= 2.81(d) Fail to reject Ho.(e)B2b)
3) (a) B(b) r=0.335(c) positive4) Given a=0.01, |Z(0.005)|=2.58 (from standard normal table)(a) n=(Z/E)^2*p*(1-p)=(2.58/0.03)^2*0.42*(1-0.42)= 1801.666
Take n=1802(b) n=(Z/E)^2*p*(1-p)=(2.58/0.03)^2*0.5*(1-0.5)= 18495) Qualitative6) (a) 0.16*0.16=0.0256(b) 1-(1-0.16)^2= 0.2944
7) (a) P(male or nursing major)= P(male)+P(nursing major)-P(both) =1110/3431 + 694/3431 -94/3431=0.498397(b) P(female or not nursing) = P(female)+P(not nursing) - P(both) =2321/3431+2737/3431-1721/3431=0.9726027(c) P(not female or nursing) = P(males)+P(not nursing)-P(both) = 1110/3431+2737/3431 - 1016/3431=0.8251239(d) C
8) AA
9) (a) xbar= 908.8(b) s=305.4(c) Given a=0.05, |t(0.025,df=n-1=11)|= 2.2 (from student t table)So 95% CI is
xbar+/- t*s/√n--> 908.8+/- 2.2*305.3892/√12--> (714.9, 1102.7)10) Bt=(xbar-µ)/(s/√n) = (72.7-71)/(2.9/√23) =2.811Given a=0.01, the critical value is |t(0.01,df=n-1=22)|= 2.508 (from student t table)The p-value= P(t with df=22 >2.811)= 0.0051 (from student t table)B
Final Exam 2
1) (a) range = max -min =46.9-22.2=24.7(b)range = max -min = 63.1-22.2=40.9(c) D