Question

a. What is the minimum sample size required to estimate the overall mean weight of boxes of Captain Crisp cereal to within 0.02 ounces with 95% confidence, if the overall (population) standard deviation of the weights is 0.23 ounces? Enter an integer below. You must round up.

b. What is the minimum sample size required to estimate the overall proportion of boxes that have less than 16 ounces of cereal to within 5% with 95% confidence, if no guess as to the value of this proportion is available?

c. What is the minimum sample size required to estimate the overall proportion of boxes that have less than 16 ounces of cereal to within 5% with 95% confidence, if it is known that this proportion is at most 0.4?

Answer 1

a)

for 95 % CI value of z=	1.960
standard deviation σ=	0.23
margin of error E =	0.02
required sample size n=(zσ/E)2                                         =	509.0

b)

here margin of error E =	0.05000
for95% CI crtiical Z          =	1.960
estimated proportion=p=	0.500
required sample size n =	p*(1-p)*(z/E)2=	385.00

c)

here margin of error E =	0.05000
for95% CI crtiical Z          =	1.960
estimated proportion=p=	0.400
required sample size n =	p*(1-p)*(z/E)2=	369.00