The First National Bank of Wilson has 680 checking account
customers. A recent sample of 60 of these customers showed 21 have
a Visa card with the bank.
Construct the 90% confidence interval for the proportion of
checking account customers who have a Visa card with the bank. (Use
z Distribution Table.) (Round your answers to 3
decimal places.)
sample success x= | 21 | |||
sample size n= | 60 | |||
sample proportion p̂ =x/n= | 0.3500 | |||
population size N= | 680 | |||
std error Se =(√(p*(1-p)/n))*(√((N-n)/(N-1))= | 0.0588 | |||
for 90 % CI value of z= | 1.645 | |||
margin of error E=z*std error = | 0.097 | |||
lower confidence bound=p̂-E= | 0.253 | |||
upper confidence bound=p̂+E= | 0.447 |
0% confidence interval for the proportion =(0.253 , 0.447)
Get Answers For Free
Most questions answered within 1 hours.