Population standard deviation for a binomial(?) distribution?

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?

• it might be useful if you could edit the textbook names into your question (for future reference) too. Thanks! – James Stanley May 16 '13 at 22:01

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