I'm having troubles with understanding a problem which is basically asking to calculate a sample size (in order to estimate what portion of population has income more than 40K) with 95% confidence level and margin of error (E) of 1%. The standard deviation is not given and it says that I should take a maximum possible value for that. In addition, it says that 2 should be used as the quantile of the normal distribution of 0.975.
Two things confuse me here: how to find maximum standard deviation and how quantile is different from a z-score?
If quantile and a z-score are the same thing, then I thought that it should be 0.475 => 1.96 (and not 0.975 => 2)? (Since the confidence interval is 95% the alpha value is going to be 0.05 and the tails will have an area of 0.025 each. Having this the critical z-value is going to be 1.96 (z of 0.475 (0.5 - 0.025)), right?)
Any feedback is appreciated, thanks in advance

  • 1
    $\begingroup$ Since this problem is routine bookwork, as one might do in a homework exercise, you should add the self-study tag and read its tag wiki info. $\endgroup$
    – Glen_b
    Jun 15, 2014 at 8:44

1 Answer 1


Your assertion that the correct value of the quantile should be 1.96 (if we assume the normal approximation is accurate) is completely correct.

However, the suggested value of 2 is a common approximation; it's only 2% larger, and that small additional margin by rounding the value up may be a good idea since several approximations are involved in doing the calculations.

That is, while you know how to compute 1.96 correctly, just use 2 anyway, like it says.

Secondly, if I told you the population proportion, could you compute the standard deviation?

Edit after chat discussion with OP:

As you figured out, the standard error is maximized when $p=0.5$. By changing the scale on the y-axis (a simple monotonic transformation), it's perhaps easier to see:

enter image description here

What remains is to figure out the margin or error in terms of the standard deviation.

Your question says to take it to be twice the standard error, so all you have left to do is find the smallest value of $n$ that has $2\sqrt{0.25/n}\leq 0.01$. Even if you're not able to do the algebraic manipulation to solve for $n$, you can find this by trial and error.

[n=1 will be too wide. What happens at n=10? 100? 1000? etc ... if you go past it, you can then try the middle of the interval until you hit it exactly or you get two consecutive numbers that give a margin of error either side of the right answer]

  • $\begingroup$ I've found a formula called Standard Error in this regard, but it needs a sample size which I don't have... $\endgroup$
    – moriesta
    Jun 15, 2014 at 9:47
  • $\begingroup$ Yes, of course you don't have the sample size! The entire point is to calculate the sample size you need in that formula to get the desired margin of error. $\endgroup$
    – Glen_b
    Jun 15, 2014 at 10:26
  • $\begingroup$ The sample size is what you have to figure out. You can do this with algebra or by looking in tables. You have been told what size error is permitted. $\endgroup$
    – Peter Flom
    Jun 15, 2014 at 10:26
  • $\begingroup$ If you knew the population proportion and the sample size, could you compute the standard deviation? $\endgroup$
    – Glen_b
    Jun 15, 2014 at 10:27
  • $\begingroup$ sqrt[p*(1-p)/n] ? $\endgroup$
    – moriesta
    Jun 15, 2014 at 10:29

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