Question

A random sample of 500 students were selected and it was found that 345 of them were satisfied with their field. Construct a 95% confidence interval for the true proportion of students who are satisfied with their field. (i)Identify n, p̂, q̂, and ??/2. ? = p̂ = q̂ = ??/2 =

(ii)Calculate the margin of error, E.

(iii)Construct the 95% confidence interval

Answer 1

Solution :

n = 500

x = 345

= x / n = 345 / 500 = 0.690

q̂ = 1 - = 1 - 0.690 = 0.310

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.960

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

=1.960 * (((0.690 * 0.310) / 500 )

= 0.040

A 95 % confidence interval for population proportion p is ,

- E < P < + E

0.690 - 0.040 < p < 0.690 + 0.040

0.650 < p < 0.730