Question

Example 2: A survey of 1200 Canadians indicated that 64% would support online voting in the next federal clection if Election Canada offered a safe an secure Internet platform. Construct a 95% confidence interval of the true proportion of Canadian online-voting support. Solution:

please solve it step by step

Answer 1

Solution

given : n = 1200, P=0.64

confidence level (C) = 0.95

a=1-C=1-0.95 = 0.05

a _ 0.05 = 0.025

Refer Z-table or use excel function "=NORM.S.INV((1-0.025))" to find the Z-value.

..29 = 1.960

formula : Confidence interval for the true proportion, p

$= \hat p \pm z_{\frac{\alpha }{2}}*\left \{ \sqrt{\frac{\hat p*(1-\hat p)}{n}} \right \}$

= 0.64 +1.960 * { 0.64 *(1-0.64) 1200

= 0.64 +1.960 + { 10.64 + 0.36 1200

= 0.64±0.0272

Lower limit = 0.64 -0.0272 = 0.6128

Upper limit = 0.64 +0.0272 = 0.6672

We are 95% confident that the true proportion of Canadian online-voting support is from 0.6128 to 0.6672.