Question

5. A random sample of 200 Girl Scouts found 180 who felt they could earn national recognition from their volume of Snickerdoodle sales to college administrator Bovina Loca. Let a = proportion of cookie-confident Girl Scouts, and use the continuity correction for the Normal Approximation To Binomial for these questions. a. Find a 98% confidence interval for a. b. With what level of confidence can we claim î is within 0.04 of 7? c. What conservative sample size is needed to estimate a within a 99% error margin of 0.5%? Recall: Use a = 0.5. (120)

Answer 1

Answer:

Given,

sample n = 200

x = 180

sample proportion p^ = x/n = 180/200 = 0.90

a)

Here at 98% CI, z value is 2.33

98% CI = p^ +/- z*sqrt(p^(1-p^)/n)

substitute values

= 0.90 +/- 2.33*sqrt(0.9(1-0.9)/200)

= 0.90 +/- 0.05

= (0.85 , 0.95)

b)

E = 0.04

consider,

E = z*sqrt(p^(1-p^)/n)

0.04 = z*sqrt(0.9(1-0.9)/200)

z = 1.89

since from standard normal table

Confidence interval = 94%

c)

Here z at 99% CI is 2.58

E = z*sqrt(p^(1-p^)/n)

0.005 = 2.58*sqrt(0.9(1-0.9)/n)

On solving we get

sample n = 23963.04

n = 23963