Question

You want to build an interval estimate for the proportion of adult Hoosiers who favor legalizing Sunday sales of alcoholic beverages in Indiana. You survey a random sample of 820 adult Hoosiers, of whom 574 favor Sunday liquor sales.
									29	What is the margin of error for an interval with a 90% level of confidence?
									a	0.026
b	0.032
c	0.038
d	0.043
30	The interval estimate with a 5% error probability is,
a	0.680	0.720
b	0.674	0.726
c	0.669	0.731
d	0.659	0.741
31	If you wanted to build a 95% interval estimate with a margin of error of ±2 percentage points, how many persons would you have to include in your sample? Use 0.6 for the planning value.
a	1998
b	2006
c	2155
d	2305
32	Suppose someone else reported a 95% interval estimate shown below:
			0.688	0.752
	What is the point estimate for this interval?
a	0.69
b	0.71
c	0.72
d	0.73
33	What is the margin of error for this estimate?
a	0.032
b	0.029
c	0.026
d	0.024
34	What is the sample size used to obtain this estimate?
a	722
b	757
c	798
d	805

Answer 1

29)p=574/820=0.7

for 90% CI ; critical z=1.645

margin of error for an interval with a 90% level of confidence =z*sqrt(p(1-p)/n)=0.026

30)

interval estimate with a 5% error probability is:

0.669 ; 0.731

31)

here margin of error E =	0.020
for95% CI crtiical Z          =	1.960
estimated proportion=p=	0.600
required sample size n =	p*(1-p)*(z/E)2=	2305.00

32)

point estimate=(0.752+0.688)/2= 0.72

33)

margin of error=(0.752-0.688)/2=0.032

34)

here margin of error E =	0.032
for95% CI crtiical Z          =	1.960
estimated proportion=p=	0.720
required sample size n =	p*(1-p)*(z/E)2=	757.00