Question

A website that sells movie tickets can tell what devices are used to access their site and make purchases. The manager of the website is approached by members of a movie production company who want to partner in the development of a phone app. To estimate what proportion of ticket buyers make purchases using their smart phones, 400 users of the website were randomly sampled from the existing database and 180 were found to have used their smart phones.

   Assuming the true proportion were thought to be 0.45, what would be the probability that a majority of the purchases were made using a smart phone?

   What would be an 85% confidence interval for the proportion of website visitors who purchased tickets using a smart phone?

Answer 1

1)

for normal distribution z score =(p̂-p)/σp
here population proportion=     p=	0.4500
sample size       =n=	400
std error of proportion=σp=√(p*(1-p)/n)=	0.0249

probability that a majority of the purchases were made using a smart phone :

probability =P(X>0.5)=P(Z>(0.5-0.45)/0.025)=P(Z>2.01)=1-P(Z<2.01)=1-0.9778=0.0222

2)

sample size          n=		400
sample proportion p̂ =x/n=		0.4500
std error se= √(p*(1-p)/n) =		0.0249
for 85 % CI value of z=		1.44
margin of error E=z*std error   =		0.0358
lower bound=p̂ -E                       =		0.4142
Upper bound=p̂ +E                     =		0.4858
from above 85% confidence interval for population proportion =(0.414,0.486)