I have been confused by two separate questions (Stock & Watson - introduction to econometrics ch.3), using different values for standard errors.

The first: In a survey of 400 voters, 215 respond to vote for the incumbent, 185 for the challenger. Let p denote the fraction of all likely voters who preferred the incumbent at the time of the survey and $\hat p$ be the fraction of survey respondents that prefer the incumbent.

Now for the variance it is given by $\hat p(1-\hat p)/n$ and when calculating the $SE(\hat p)$ we have to take the square root of the variance to get $0.0249$, and I am fine with this.

The second question: In a given population 11% of voters are African American. A survey using a random sample of 600 landline telephone numbers finds 8% African Americans. Is there evidence the survey is biased?

Now when calculating the t-statistic we use the null hypothesis with $p=0.11$, but it then states that $se(\hat p)=\hat p(1-\hat p)/n$.

Why do we no longer have to take the square root of the value above to find the standard error? I imagine it must be to do with knowing the population variance?

  • 1
    $\begingroup$ it might be useful if you could edit the textbook names into your question (for future reference) too. Thanks! $\endgroup$ May 16, 2013 at 22:01

1 Answer 1


I suspect that the second example's description of the standard error is incorrect -- I've never seen the normal approximation approach to getting a standard error for a proportion reported without using the square root.

[see e.g. http://www.stats.org.uk/statistical-inference/Newcombe1998.pdf for a description of different approaches to confidence interval calculations for proportions, including the normal approximation or asymptotic method as described in your question.]

EDIT: Just confirming my suspicions...

The first and second examples are items 3.3 and 3.7 respectively in the exercises for chapter 3 of Stock & Watson - Introduction to Econometrics, the answers available online at this site.

and for item 3.7, their answer of $t = 2.71$ has been calculated using the standard error as $\sqrt{\frac{p(1-p)}{n}}$ although their answer gives the standard error as $SE(\hat{p}) = \hat{p}(1-\hat{p}) /n$

  • $\begingroup$ Thank you very much for the help - this is one of the issues when textbooks make errors, as when learning it can be impossible to verify! $\endgroup$
    – Tom
    May 16, 2013 at 21:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.