# How to find Confidence Level?

A survey of 215 students on a college campus reveals that 54% of them support the construction of a parking garage with a 5% margin of error.

What is the confidence level of the results of this survey?

I know how to find the Confidence Interval, but I have no idea how to find the Confidence Level.

Relevance
• 8 years ago

Is confidence level = confidence coefficient? (1-α)100%.

(a) The margin of error associated with an estimate of x/n, proportion of success, is 100d%, where

d = z(α/2) / (2√n), with CI of (1- α)100%.

Sample size

n = [z(α/2) / 2d]^2. [=(1/4) {z(α/2)/d}^2]

Here we initially assumed p=0.50=1/2, and pq=0.25= 1/4, hence

d = z(α/2) / (2√n) = z(α/2) √(pq)/√n = z(α/2)/(√(n/pq)), since √(pq)=1/2, or 2=1/√(pq).

For CI=96%, α/2 = 0.02, z(α/2)=2.054, d=0.03

n=(2.054/2*0.03)^2 = 1172

(b) When an estimate for p is available from, say a pilot study,

n = [z(α/2) (√pq) /d]^2 = pq / [d/ z(α/2)]^2 or pq*[z(α/2)/d]^2

Example: Take 95% CI, α/2 = 0.025. For p=0.3, q=0.7, d=0.10, z(0.025)=1.96,

n= (0.3)(0.7)(1.96/0.10)^2 = 81.

Here 215=(0.54)(0.46)[z(α/2)/0.05]^2

z(α/2)=1.47, α/2 = 0.0708, α=0.1416

Confidence level = 86%

• 8 years ago

Margin of error = Z(alpha/2)*SE = Z(alpha/2)*Sqrt{0.54*0.46/215} = 0.05

Z(alpha/2)*0.034 = 0.05

Z(alpha/2)= 0.05/0.034 = 1.47

Therefore alpha = 0.07x2 = 0.14