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

Solution :

Point estimate = sample proportion = = x / n = 58/100=0.58

1 -   = 1-0.58 =0.42

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z  0.005 = 2.576 ( Using z table )

  Margin of error = E = Z / 2    * (((( * (1 - )) / n)

= 2.576* (((0.58*0.42) /100 )

E = 0.127

A 99% confidence interval for p is ,

- E < p < + E

0.58 - 0.127 < p < 0.58 + 0.127

0.453 < p < 0.707

(0.453 , 0.707)

b

= 0.58

1 - = 1 - 0.58 = 0.42

margin of error = E = 0.1

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025= 1.96 ( Using z table )

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

= (1.96 / 0.1)2 * 0.58 * 0.42

= 93.58

Sample size =94 rounded