Question

2. It is estimated that the standard deviation of the height of American adult males is 4 inches. Suppose we survey a sample of 50 men and find a sample mean of 69 inches.

(a) For a confidence level of %90, what is the margin of error and the confidence interval?

(b) For a confidence level of %95, what is the margin of error and the confidence interval?

(c) If we sample 500 men instead, and still insist on %90 confidence level, what is the new margin of error and the new confidence interval?

(d) If we wanted a 95% confidence level and we wanted a margin of error less than 0.1, how many men would we need to survey (what must our sample size be)?

Answer 1

a)

sample mean	x̄=	69.0000
sample size	n=	50.00
population std deviation	σ=	4.000
standard errror of mean =	σx=σ/√n=	0.5657

for 90 % CI value of z=				1.645
margin of error E=z*std error                            =				0.930
lower confidence bound=sample mean-margin of error=				68.070
Upper confidence bound=sample mean +margin of error=				69.930

b)

for 95 % CI value of z=				1.960
margin of error E=z*std error                            =				1.109
lower confidence bound=sample mean-margin of error=				67.891
Upper confidence bound=sample mean +margin of error=				70.109

c)

standard errror of mean =

σx=σ/√n=

0.1789

for 90 % CI value of z=				1.645
margin of error E=z*std error                            =				0.294
lower confidence bound=sample mean-margin of error=				68.706
Upper confidence bound=sample mean +margin of error=				69.294

d)

	for95% CI crtiical Z          =	1.96
	standard deviation σ=	4.000
	margin of error E =	0.1
required sample size n=(zσ/E)2                  =		6147