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	0.038
b	0.032
c	0.026
d	0.015
30	The upper end of the interval estimate with a 5% error probability is,
a	0.731
b	0.726
c	0.720
d	0.714
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	2456
b	2305
c	2155
d	2006

Answer 1

29)

n = 820 ....... Sample size

x = 574 .......no. of successes in the sample

Let denotes the sample proportion.

     = x/n   = 574 / 820= 0.7

For 90% confidence , = 1.645

the margin of error is given by

E = /2 *  

= 1.645 * [(0.7* (1 - 0.7)/820]

=  0.026

Answer :  0.026

30)

Upper end =  ( + E) = 0.7 + 0.026 = 0.726

Answer : 0.726

31)

For 95% confidence , = 1.96

E = 2% = 0.02

p = 0.6

1 - p = 1 - 0.6 = 0.4

The sample size for estimating the proportion is given by

n =

= (1.96)² * 0.6 * 0.4 / (0.02²)

= 2305

Answer : 2305