Question

Consider the accompanying observations on bond strength. 9.5 5.5 11.4 12.0 9.8 9.4 7.7 6.4 6.5 6.9 13.6 16.9 5.8 5.2 5.0 4.9 5.1 10.5 15.1 8.4 4.4 3.8 4.0 3.7 3.7 3.2 20.5 25.7 14.0 12.8 13.0 8.8 8.0 10.9 14.4 7.8 5.0 5.7 4.9 5.1 5.4 4.9 3.9 3.7 3.6 3.4 3.5 3.8 (a) Estimate true average bond strength in a way that conveys information about precision and reliability. (Hint: [ *; = 387.2 and < x2 = 4,250.3.) (Round you answers to three decimal places.) A 95% CI for u = the true average strength is (6) Calculate a 95% CI for the proportion of all such bonds whose strength values would exceed 10. (Round your answers to three decimal places.)

Answer 1

	x̅=Σx/n=		8.07
	s12=(Σx2-(Σx)2/n)/(n-1)=		23.98
	s1=√s12                                =		4.90

a)

sample mean 'x̄=		23.976
sample size    n=		48
std deviation s=		4.8966
std error ='sx=s/√n=4.8966/√48=		0.7068

for 95% CI; and 47 df, value of t=		2.012		from excel: t.inv(0.975,47)
margin of error E=t*std error    =		1.422
lower bound=sample mean-E =		22.554
Upper bound=sample mean+E =		25.398
from above 95% confidence interval for population mean =(22.554<μ<25.398)

b)

sample success x =		13
sample size n=		48
pt estiamte p̂ =x/n=		0.2708

for 95 % CI value of z=

1.960

from excel:normsinv((1+0.95)/2)

margin of error E=z*std error   =		0.1257
lower bound=p̂ -E                       =		0.1451
Upper bound=p̂ +E                     =		0.3965
from above 95% confidence interval for population proportion =(0.145,0.397)

(please try 0.166 , 0.410 , if this comres wrong and reply)