Question

You intend to estimate a population proportion with a confidence interval. The data suggests that the normal distribution is a reasonable approximation for the binomial distribution in this case.

While it is an uncommon confidence level, find the critical value that corresponds to a confidence level of 91.1%.
(Report answer accurate to three decimal places with appropriate rounding.)

Answer 1

Solution:
If we will use the normal distribution to approximate the binomial distribution than the confidence interval
so confidence interval can be calculated as
p^+/- Zalpha/2*sqrt(p^ * q/n)
Here confidence level is 91.1%
So level of significance or alpha = 1 - 0.911 = 0.089
alpha/2 = 0.089/2 = 0.0445
So lower boundary alpha/2 = 0.0445
Upper bound alpha/2 = 0.911+0.0445 = 0.9555
From Z table we found Zalpha/2 = 1.701 this is for the upper boundary
Or you can use Excel formula to Find Z-Critical value
i.e. NORM.INV(Probability,Mean, Standard deviation) =  NORM.INV(0.9555,0,1) = 1.701
So Critical value that corresponds to a confidence level of 91.1% is 1.701