Question

In an opinion poll involving n = 100 college students, 58 answered "YES" and the remaining said "NO". Let p denote the proportion of people in the entire voting population that are "NO" responding individuals.

(a). Construct a 99% confidence interval for p.

(b). In a 95% confidence interval, what should the sample size n be to keep the margin of error within 0.1?

Answer 1

a) we have X =number of students voting No = 42

n = number of students involving = 100

Let p^ = Sample proportion = X/n = 42/100 = 0.42

The 99% confidence interval for population proportion p

p^ - E < p < p^ + E

Where E = Z​​​​​​a/2* Sqrt [ p^ *(1-p^)/n ]

For a = 0.01, Z​​​​​​0.005 = 2.58

E = 2.58* sqrt [ 0.42*0.58/100]

E = 0.13

0.42 - 0.13 < p < 0.42 + 0.13

0.29 < p < 0.55

b)

For margin of error E = 0.1

c = confidence level = 0.95

p^ = estimate of proportion = 0.42

Sample size n = p *( 1-p) *( Z​​​​​​a/2/E)2

For a = 0.05 , Z​​​​​​0.025 = 1.96

n = 0.42 *0.58*(1.96/0.1)2

n = 94