Finding confidence level given confidence interval and sample sizeThe 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) 82B) 86C) 90D) 95E) 99To do this, I set up the equation to get the confidence interval:Confidence interval = .63 ± z ⋅ 0.63 ⋅ 0.37 100 Solving for z gives around 91%...Am I doing this right?

Question

Finding confidence level given confidence interval and sample sizeThe Austin-American Statesman asked an SRS of 100 Austinites for their opinion about Mayor Adler&#039;s performance. Using the results of the sample, the C% confidence interval for the proportion of all Austinites who approve of the mayor&#039;s job performance is .565 to .695. What is the value of C?A) 82B) 86C) 90D) 95E) 99To do this, I set up the equation to get the confidence interval:Confidence interval =  .63  ±  z  ⋅                              0.63          ⋅          0.37                100            Solving for z gives around 91%...Am I doing this right?

Nelson Santana · Accepted Answer

Step 1In R,   p  n  o  r  m  (  (  .695  −  .63  )      /        .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   ±  1.340755 as the z-values to use in an 82% CI of the form            p      ^        ±  1.341                                          p            ^                          (        1        −                              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  .  h  a  t  =  .63  ;  n  =  100  ;  c  =  1.341  ;  ±  =  c  (  −  1  ,  1  )    p  .  h  a  t  +  ±  ⋅  c  ⋅      p    .    h    a    t    ⋅    (    1    −    p    .    h    a    t    )          /        n      [  1  ]  0.5652559  0.6947441Step 2However, with   n  =  100 and 99 degrees of freedom, there is little difference between Student&#039;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.  q  b  e  t  a  (  c  (  .09  ,  .91  )  ,  63.5  ,  37.5  )    [  1  ]  0.5637247  0.6923295

Answered question

Answer & Explanation

New Questions in College Statistics