3
$\begingroup$

Am struggling to understand part of the answer to a question have done-

Qu- In a given population, 11% of the likely voters are African American. A survey using a simple random sample of 600 landline telephone numbers finds 8% African Americans. Is there evidence that the survey is biased?

To answer the question I found it quite simple. Set up H0: Survey is random H1: Survey is biased $ \hat p=0.08$ & $p=0.11 $

Calculated my t value using $ t=(\hat p - p)/SE(\hat p)$

where $SE(\hat p)= (\hat p(1− \hat p)/n)^{1/2} \\$

and got a t value of $t=2.72$ and rejected the null as the p value was less than 1%. According to the answers my method is correct, however it is also stated:

An alternative formula for $SE(\hat p )$ is $0.11(1-0.11)/n$ which is valid under the null hypothesis that p=0.11)

I imagine that the lack of square root there is just a mistype, however am I correct in assuming that what they've done is calculate the standard error using the population data rather than the sample data? Is that acceptable, because obviously it would produce a different t value. I'm aware that in most questions this wouldn't be possible, but in bernouilli distributions it is.

Thanks

$\endgroup$

1 Answer 1

9
$\begingroup$

The idea of a hypothesis test is that you come up with a statistic whose distribution you know if the null hypothesis is true.

The most well-known case is the t-statistic, where you divide sample mean minus mean under the null by the square root of sample standard deviation divided by $n$. Some mathematical statistics then shows that this t-statistic follows a t-distribution with $n-1$ degrees of freedom if the null is true and we sample from a normal population.

Now, if the null is true, computing the standard error from $p=0.11$ is correct, because you are using the right $p$.

Then, we can write your test statistic as $$ t=\frac{\sqrt{n}(\hat p - p)}{\sqrt{\hat p(1− \hat p)}} $$ By the CLT, because $p=E(X_i)$ ($\hat p=1/n\sum_iX_i$) and assuming random sampling, $$\sqrt{n}(\hat p - p)\to_dN(0,Var(X_i))$$ But $Var(X_i)=p(1-p)$ for such a Bernoulli random variable, so that the test statistic converges in distribution to $$\sqrt{n}\frac{(\hat p - p)}{\sqrt{p(1-p)}}\to_dN(0,1),$$ i.e., it will behave like a standard normal r.v. in large samples, if the null is true.

Now, by the law of large numbers, $\hat p\to_pp$, it is also correct to use $\hat p$, as replacing the true $p(1-p)$ by a consistent estimator of this quantity does not alter the asymptotic distribution.

(This result is known, at least in econometrics, as Slutzky's theorem, which says that a product of two sequences, one of which converges in distribution and one of which converges in probability to a constant will converge in distribution to the product of the limits - "the weaker convergence mode dominates".)

$\endgroup$
1
  • $\begingroup$ Thanks this makes things very clear and aids my understanding of the actual process. Appreciated $\endgroup$
    – Nik-D
    Aug 25, 2015 at 12:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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