Question

Of 44 bank customers depositing a check, 23 received some cash back.

(a) Construct a 90 percent confidence interval for the proportion of all depositors who ask for cash back. (Round your answers to 4 decimal places.)
  
The 90% confidence interval is from  to

(b) May normality of p be assumed?
  

• Yes

• No

Answer 1

a)

sample success x =		23
sample size          n=		44
sample proportion p̂ =x/n=		0.5227
std error se= √(p*(1-p)/n) =		0.0753
for 90 % CI value of z=		1.645
margin of error E=z*std error   =		0.124
lower bound=p̂ -E                       =		0.3988
Upper bound=p̂ +E                     =		0.6466

The 90% confidence interval is from 0.3988 to 0.6466

b)

No, since np and n(1-p) both are greater than 5