Question

In the run-up to the 2016 election, a random sample of 100 people was taken to determine a confidence interval for the population proportion who would vote in favor of Proposition 61 to legalize marijuana.  The confidence interval was found to be (.402 to .598).  

What was the confidence level, (1-alpha)*100%, associated with this confidence interval?

95%
		90%
		We would need to know the number in the sample who said they would vote for Proposition 61.  If we knew this number, we could answer the question
		Even if we know the number in the sample who said they would vote for Proposition 61, there is not enough additional information to answer the question
		It is possible to answer the question with the information that is given, but neither A) nor B) above is correct

Answer 1

Given CI =  (.402 to .598).

margin of error = length of CI / 2

margin of error = ( 0.598 - 0.402 ) / 2

margin of error = 0.196/2

margin of error = E = 0.98

p = X/N = 61/100 = 0.61

z_c = 0.0498

which is neither 95% nor 90%

It is possible to answer the question with the information that is given, but neither A) nor B) above is correct