# Calculate alpha level of a given confidence interval for relative frequencies

I have a (for me tricky) statistic exam question and can't find the answer. All I know is the information which I post here. Here we go ...

A research company asked 1000 participants which party they would vote in the next election. The calculated CI is [31,53;36,47]. Unfortunately, the research company forgot to tell the company which α level they used. "Calculate" the α level (or do whatever necessary to find the α ;) )

The answer is 10% but I have no idea how to get it ...

• Please add the self-study tag and tell us what you have tried so far so we can see where you are missing something. Oct 6, 2016 at 7:15
• Well, thx for the answer. I think I'm pretty familiar with this topic but in this case I have no idea how to start. I think that this problem is not too difficult - it is from a statistic exam for beginners. Oct 6, 2016 at 7:22
• Hint: how many people voted for the party? Oct 6, 2016 at 7:26
• I would say between 315 and 360.. Oct 6, 2016 at 8:46

You could also use R's pbinom() function.

> pbinom(364.7, 1000, 0.34)
 0.9484374


0.34 is the binomial probability. You get this by computing the middle of the confidence interval (34% = 0.34).

1000 is the number of trials (= participants).

364.7 is the upper limit of the confidence interval, expressed in terms of the number of participants expected to answer in the "yes" direction.

The result, 0.948, tells us that 5% of the distribution lies to the right of the confidence interval. Double that to get 10%, the total amount of the distribution that lies outside the confidence interval.

From the confidence interval, you know the point estimate: 34. You also know that the radius of the confidence interval 36.47-34 = 2.47 is linked to the standard error of the estimator and the level of the test, $\alpha$. In this case, an estimate of the standard error of the estimator is equal to $$\hat \sigma = \sqrt(34*(100-34)/1000) = 1.5$$ When the (two-sided) test is at 5%, we usually say that the radius of the CI is roughly twice this standard error, twice corresponding to $Q(1-.05/2) = 1.959964$, where $Q(.)$ is the quantile function of the normal distribution. Here $Q(1-.10/2) = 1.644854$ which happens to be very close to 2.47/1.5 = 1.646667.