The Austin-American Statesman asked an SRS of 100 Austinites for their opinion about Mayor Adler's performance. Using the results of the sample, the C% confidence interval for the proportion of all Austinites who approve of the mayor's job performance is .565 to .695. What is the value of C?

A) 82

B) 86

C) 90

D) 95

E) 99

To do this, I set up the equation to get the confidence interval:

Confidence interval = $.63 \pm z \cdot \sqrt{\dfrac{0.63 \cdot 0.37}{100}}$

Solving for z gives around 91%...

Am I doing this right?

  • 2
    $\begingroup$ I think so, but given only 100 people you might need to get the confidence interval/formula for a t-distribution, not the normal distribution. $\endgroup$
    – eSurfsnake
    Commented Feb 7, 2019 at 3:53
  • $\begingroup$ @eSurfsnake I think so. So the t-score would be 1.346, and this corresponds to 82% $\endgroup$
    – JenkinsMa
    Commented Feb 7, 2019 at 4:13
  • 1
    $\begingroup$ @JenkinsMa Why you get so different answers by using the normal distribution(91%) and the t-distribution (82%). It doesn´t make sense. $\endgroup$ Commented Feb 7, 2019 at 12:58

1 Answer 1


In R, pnorm((.695-.63)/sqrt(.63*.37/100)) returns 0.9108973, as you say, but that is not quite the end of the story.

This implies that the upper confidence value cuts about 9% from the upper tail. Similarly, the lower value cuts about 9% from the lower tail. So the confidence level is about 82%.

Furthermore, qnorm(c(.09, .91)) returns $\pm 1.340755$ as the z-values to use in an 82% CI of the form $$\hat p \pm 1.341\sqrt{\frac{\hat p(1-\hat p)}{n}}.$$

Then we have $(0.565, 0.695)$ as an 82% confidence interval for the binomial success probability $p;$ the computation using R as a calculator is shown below:

p.hat = .63;  n = 100;  c = 1.341;  pm = c(-1,1)
p.hat + pm*c*sqrt(p.hat*(1-p.hat)/n)
[1] 0.5652559 0.6947441

I agree with @callculus that appeal to a t-distribution in this context makes no sense. However, with $n = 100$ and 99 degrees of freedom, there is little difference between Student's t distribution and standard normal. So your Comment is on the right track toward an 82% interval.

Note: A Jeffreys 82% Bayesian probability interval is $(0.564, 0.692).$ This interval, based on a non-informative prior is often numerically similar to the frequentist Wilson interval. Both the Jeffreys interval (used as a frequentist CI) and the Wilson interval have more accurate coverage probabilities than the Wald interval shown above, especially for $n$ as small as 100.

qbeta(c(.09,.91), 63.5, 37.5)
[1] 0.5637247 0.6923295

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