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
Solved – Calculating sample size with unknown standard deviation
normal distributionsample-sizeself-studystandard deviation
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I agree with Glen_b on this. Maybe I can add a few words to make the point even clearer. If data come from a normal distribution (iid situation) with an unknown variance the t statistic is the pivotal quantity used to generate confidence intervals and do hypothesis testing. The only thing that matters for that inference is its distribution under the null hypothesis (for determining the critical value) and under the alternative (to determine power and sample). Those are the central and noncentral t distributions, respectively. Now considering for a moment the one sample problem, the t test even has optimal properties as a test for the mean of a normal distribution. Now the sample variance is an unbiased estimator of the population variance but its square root is a BIASED estimator of the population standard deviation. It doesn't matter that this BIASED estimator enters in the denominator of the pivotal quantity. Now it does play a role in that it is a consistent estimator. That is what allows the t distribution to approach the standard normal as the sample size goes to infinity. But being biased for any fixed $n$ does not affect the nice properties of the test.
In my opinion unbiasedness is overemphasized in introductory statistics classes. Accuracy and consistency of estimators are the real properties that deserve emphasis.
For other problems where parametric or nonparametric methods are applied, an estimate of standard deviation does not even enter into the formula.
This sounds like you use normal approximation interval which is not optimal in any case and especially unsuited for probalities close to 0 and 1 (e.g. 97.5%).
Look at the following graph.
For the first histogram a normal distribution would work fairly well. In the second case you can see that the distribution has considerable skew, which would makes the normal distribution inappropiate.
In either case, there is no need to use normal approximations for confidence intervals as more exact answers can be derived (in contrast to other more complex statistics, where sometimes a normal approximation is needed). Better options to construct confidence intervals for binomial proportions are described in the link above as well.
Best Answer
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:
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]