Question

SHOW WIL OOST Out of 500 people sampled, 200 had kids. Based on this, construct a 90% confidence interval for the true population proportion of people with kids. Give your answers as decimals, to four places <p Get Help: License Points possible: 1 This is attempt 1 of 2. Submit © 2020 XYZ Homework I Privacy Policy I ADA Compliance PDF | Contact Us xyzHomework is powered by IMathAS © 2006-2020 by David Lippman

Answer 1

1)

sample success x =		200
sample size n=		500
pt estiamte p̂ =x/n=		0.4000
se= √(p*(1-p)/n) =		0.0219

for 90 % CI value of z=

1.645

from excel:normsinv((1+0.9)/2)

margin of error E=z*std error   =		0.0360
lower bound=p̂ -E                       =		0.3640
Upper bound=p̂ +E                     =		0.4360

from above 90% confidence interval for population proportion =(0.3640<p< 0.4360)

2)

sample mean 'x̄=		29.100
sample size    n=		24
std deviation σ=		16.60
std error ='σx=σ/√n=16.6/√24=		3.388

for 90 % CI value of z=		1.645	from excel:normsinv((1+0.9)/2)
margin of error E=z*std error =		5.5735
lower bound=sample mean-E=		23.53
Upper bound=sample mean+E=		34.67
from above 90% confidence interval for population mean =(23.526 <µ<34.674)