Question

A soft-drink manufacturer purchases aluminum cans from an outside vendor. A random sample of 70 cans is selected from a large shipment, and each is tested for strength by applying an increasing load to the side of the can until it punctures. Of the 70 cans, 59 meet the specification for puncture resistance.

a) Find a 95% confidence interval for the proportion of cans in the shipment that meet the specification. Round the answers to three decimal places.
b) Find a 90% confidence interval for the proportion of cans in the shipment that meet the specification. Round the answers to three decimal places.
c) Find the sample size needed for a 95% confidence interval to specify the proportion to within ±0.06. Round up the answer to the nearest integer.
d) Find the sample size needed for a 90% confidence interval to specify the proportion to within ±0.06. Round up the answer to the nearest integer.
e) If a 90% confidence interval is computed each day for 300 days, what is the probability that more than 280 of the confidence intervals cover the true proportions? Round the answer to three decimal places.

Answer 1

a)

95% confidence interval for the proportion =0.758 ; 0.928

b)

for 90 % CI value of z=				1.645
margin of error E=z*std error                            =				0.0715
lower confidence bound=sample proportion-margin of error				0.772
Upper confidence bound=sample proportion+margin of error				0.915

90% confidence interval for the proportion = 0.772 ; 0.915

( please try 0.771 ; 0.914 if above comes wrong)

c)

here margin of error E =	0.060
for95% CI crtiical Z          =	1.960
estimated proportion=p=	0.843
required sample size n =	p*(1-p)*(z/E)2=	142.00

d)

here margin of error E =	0.060
for90% CI crtiical Z          =	1.645
estimated proportion=p=	0.843
required sample size n =	p*(1-p)*(z/E)2=	100.00

e)

her expected inerval contain true proportion =np=300*0.9=270

and std deviation=sqrt(np(1-p))=5.196

probability that more than 280 of the confidence intervals cover the true proportions=P(X>280)

=P(Z>(280.5-270)/5.196)=P(Z>2.02)=0.022