So the question is as follows (paraphrased):

A population votes for/against a political candidate. In a sample of 200 voters, let $X$ be the fraction of people who vote for this candidate.

Population size is unknown

Population proportion who vote for is known to be $0.65$

What is the standard deviation of the samples proportions, i.e. the standard deviation of $X$?

My approach:

$X \sim B(200,p)$


$\rightarrow p=0.65/200= 0.00325$

$\sigma_X=\sqrt{200\cdot 0.00325\cdot0.99675}=0.8049$

Then I read somewhere that the standard deviation of a sampling proportions is $\sqrt{\displaystyle\frac{pq}{n}}$, which isn't the same as the one in my approach. Is this because $\sqrt{\displaystyle\frac{pq}{n}}$ is used for estimating the true population proportion when it's unknown (which isn't the case for my problem)?

EDIT: I've edited the question so the variables are clearer.

  • 2
    $\begingroup$ $X$ has two definitions in your question. 1)X be the fraction of people who vote for this candidate. 2) X∼B(200,p), It means X is number of voters who vote for this candidate. So please clarify. $\endgroup$
    – user158565
    Oct 20 '18 at 18:03
  • 3
    $\begingroup$ Now can $np = 0.65$ when $n=200$ and $p=0.65$? $\endgroup$
    – jbowman
    Oct 20 '18 at 18:08
  • 1
    $\begingroup$ First p: " p=0.65"; second p: "p=0.65/200=0.00325" $\endgroup$
    – user158565
    Oct 20 '18 at 19:04
  • $\begingroup$ @a_statistician excuse my clumsiness; I've edited the question so that p only denotes the probability that X=0.65. I hope this also clears up the confusion regarding X, which denotes the fraction (i.e. proportion) of people in the sample who vote for the candidate $\endgroup$
    – user98937
    Oct 20 '18 at 22:55
  • $\begingroup$ @jbowman sorry for the confusion; I've edited the question. Please see my comment above this one $\endgroup$
    – user98937
    Oct 20 '18 at 22:57

I think the end of this story is to use the number $X$ of people in favor of the candidate out of $n = 200$ people interviewed in order to get a 95% confidence interval for $p,$ the proportion of people in the population who favor the candidate. I agree with @a.statistician (+1) that you are confusing $SD(X)$ and $SD(p),$ but it seems to me you are also confusing $p$ and $\hat p,$ so there is a little more to this before you are ready to do statistical analysis.

Let $X \sim \mathsf{Binom}(n = 200, p).$ Then $E(X) = np = 200p,\, Var(X) = npq = 200pq,\,$ and $SD(X) = \sqrt{npq} = \sqrt{200pq},$ where $q = 1-p.$

Then suppose you take a poll and you observe $X = 130$ people in favor of the candidate out of $n = 200$ people interviewed. Then the estimate $\hat p$ of the population $p$ in favor of the candidate is $\hat p = \frac{X}{n} = \frac{130}{200} = 0.65.$ Then $\hat p$ is also a random variable, based on a binomial distribution, but not itself binomially distributed. [Notice that this addresses an error pointed out by @jbowman, that you never corrected in your Question.]

We can find $E(\hat p) = E(X/n) = \frac 1nE(X) = \frac 1n np = p,$ $V(\hat p) = V(X/n) = \frac{1}{n^2}V(X) = \frac{1}{n^2}npq = \frac{pq}{n}.$ Then finally, $SD(\hat p) = \sqrt{\frac{pq}{n}}.$

Then if you want a 95% confidence interval for $p$ based on $\hat p,$ you can assume (for sufficiently large $n$) that $\hat p$ is normally distributed. Upon standardization, this leads to saying that $Z = \frac{\hat p - p}{SD(\hat p)}$ is approximately standard normal. Thus $$P\left(-1.96 < \frac{\hat p - p}{SD(\hat p)} < -1.96 \right) \approx 0.95,$$ from which we get that a 95% CI for $p$ would be something like $\hat p \pm 1.96\sqrt{\frac{pq}{n}}.$ But since we don't know the values of $p$ and $q,$ we need to estimate them. Thus we obtain the approximate 95% confidence interval of the form $$\hat p \pm 1.96\sqrt{\frac{\hat p\hat q}{n}},$$ where $\hat q = 1 - \hat p.$ For $n = 300$ and $X = 130,$ the 95% CI computes to $(0.377, 0.489).$

Note: Because this CI has two approximations (a) normal approximation to binomial and (b) using $\sqrt{\frac{\hat p\hat q}{n}}$ for $\sqrt{\frac{pq}{n}},$ it has been shown not to be very useful for smaller values of $n$. Agresti and Coull have shown that the following adjustment improves accuracy of a 95% CI for $p:$ Use $\check p = \frac{X+2}{n+4},\, \check q = 1 - \check p,$ and $\check n = n+4,$ so that the adjusted CI becomes

$$\check p \pm 1.96\sqrt{\frac{\check p\check q}{\check n}}.$$

For $n = 300, X = 130,$ the adjusted 95% CI computes to $(0.378, 0.490).$ For $n$ as large as $300,$ the adjustment is relatively minor.


Basically, you mixed up two variances together. In statistics, if you say "variance" alone, it has no meaning. You need to specify whose variance you speak of.

In your case, there are two variances, one for random variable $X$, who follows Binomial distribution with parameter $p$, and $n$, which is $npq$, i.e., $\mathrm{Var}(X)=npq$. Another variance is for the estimate of $p$, i.e., $\hat p = \frac Xn$. And its variance is $\mathrm{Var}(\hat p) = \mathrm{Var}(\frac Xn)= \frac {pq}n$. (Where $q=1-p$.)


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