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?
Best Answer
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 $SE(\hat{p}) = \hat{p}(1-\hat{p}) /n$