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Obtained a sample size from standard deviation and another sample size from proportion, which one should I use if I need to be mindful not to propose a more costly study than necessary?

$n = 74$ (from standard deviation) and $n = 358$ (from proportion)

such that $37\%$ is the population proportion the standard deviation of annual income in this population is around $22000$. $95\%$ confidence interval $\pm0.05$ for the proportion with a post-secondary credential $95\%$ confidence interval with width no more than $\$5,000$ for the average income in the population.

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  • $\begingroup$ Your question is unclear. Are you saying that you obtained estimates of standard deviation and mean from two DIFFERENT samples? (which is unusual), and you want to know what sample size to use to get a confidence interval? If you gathered two sets of samples, you could always combine them and get an overall estimate of standard deviation which is what you need to determine how many samples to use for a prescribed confidence interval (assuming distribution is Gaussian). $\endgroup$ Commented Sep 19, 2015 at 19:18
  • $\begingroup$ There are numbers in the problem that might lead to two different kinds of analysis. As a result, I think you have gotten confused trying to use two different formulas at once for the for sample size $n$. Please read my Answer. If it helps clarify the issues, fine. If not, try to clarify your question and maybe someone can help. $\endgroup$
    – BruceET
    Commented Sep 19, 2015 at 21:01
  • $\begingroup$ Goal: To produce estimates of both average annual income and proportion with PSC in a target population. A simple random sample from the population will be used. Informed pre-study guesses are that (i) 37% of this population has a PSC, and (ii) the standard deviation of annual income in this population is around $ 22000. $\endgroup$
    – Tree
    Commented Sep 19, 2015 at 22:14
  • $\begingroup$ The boss desires a 95% confidence interval ±0.05 for the proportion with PSC and a 95% confidence interval with width no more than $5,000 for the average income in the population. Using the pre-study guesses, and being mindful of not proposing a more costly study than necessary, what sample size do you recommend? Should I calculate the sample size in terms of p(1-p) or standard deviation? $\endgroup$
    – Tree
    Commented Sep 19, 2015 at 22:14

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I think you may be trying to find the sample size necessary to achieve a certain margin of error in a confidence interval of the type

$$\text{Parameter Estimate} \pm \text{Margin of Error}.$$

(1) Suppose you are going to have $n$ observations from a normal population with unknown population mean $\mu$ and known population standard deviation $\sigma_0.$ Then a 95% confidence interval (CI) is

$$\bar X \pm 1.96 \sigma_0/\sqrt{n},$$

where $\bar X$ is the sample mean and $1.96 \sigma_0/\sqrt{n}$ is the margin of error. If you want to have a specific margin of error $E$ in your CI, then you set $E = 1.96 \sigma_0/\sqrt{n}.$ Everything but $n$ is known. Solve for $n$ and you know how many observations to take.

(2) Suppose you are doing a poll to see how popular Caidate X is in the weeks before an election. Then you want to estimate the population proportion $p$ in favor on Candidate X. You will estimate this is $\hat p = X/n$, where $X$ is the number of interviewed people currently favoring Candidate X, and $n$ is the number of people interviewed. Then a 95% CI for $p$ takes the form $$\hat p \pm 1.96\sqrt{\hat p (1 - \hat p)/n}.$$

Here the margin of error is $1.96 \sqrt{p(1-p)/n}$, but you don't know $p$. So, for planning purposes you might use $p = 1/2$ and set your desired margin of error $$E = 1.96 \sqrt{p(1-p)/n} = 1.96\sqrt{.5(1-.5)/n} \approx 1/\sqrt{n}.$$ Then you can solve for $n$ and you will know how many subjects to interview.

Undoubtedly, you will get different answers for $n$ depending on whether you use the formula in (1) or the formula in (2). And there are other kinds of formulas for other kinds of problems.

So before you try to find $n$ for a particular experiment or survey, you have to make sure you are thinking about the correct kind of statistical analysis and have the correct formula for $E$ in order to get a meaningful value of $n$.

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