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Student's $t$-test requires the sample standard deviation $s$. However, how do I compute for $s$ when only the sample size and sample average are known?

For example, if sample size is $49$ and sample average is $112$, I will then attempt to create a list of $49$ identical samples with values of $112$ each. Expectedly, the sample standard deviation is $0$. This will create a divide-by-zero problem in the $t$ test.

ADDITIONAL DATA:
The average income of ACME North Factory workers is $\$200$. It is reported that a random sample of $49$ workers in ACME South Factory had an annual income of $\$112$. Is this difference statistically significant?

Am I correct in saying that the population mean is $\$200$?

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  • $\begingroup$ What problem are you trying to solve? It would help us help you if you told us more. $\endgroup$ – pmgjones Aug 18 '10 at 2:08
  • $\begingroup$ Sure. I added a sample problem. $\endgroup$ – Kit Aug 18 '10 at 2:58
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This may surprise many, but to solve this problem you don't necessarily need to estimate s. In fact, you don't need to know anything about the spread of the data (although that would be helpful, of course). For instance, Wall, Boen, and Tweedie in a 2001 article describe how to find a finite confidence interval for the mean of any unimodal distribution based on a single draw.

In the present case, we have some basis to view the sample mean of 112 as a draw from an approximately normal distribution (namely, the sampling distribution of the average of a simple random sample of 49 salaries). We are implicitly assuming there are a fairly large number of factory workers and that their salary distribution is not so skewed or multimodal as to render the central limit theorem inoperable. Then a conservative 90% CI for the mean extends upwards to

$$112 + 5.84\ |112|,$$

clearly covering the true mean of 200. (See Wall et al formula 3.) Given the limited information available and the assumptions made here, we therefore cannot conclude that 112 differs "significantly" from 200.

Reference: "An Effective Confidence Interval for the Mean With Samples of Size One and Two." The American Statistician, May 2001, Vol. 55, No. 2: pp. 102-105. (pdf)

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    $\begingroup$ Yes it is! That's why it's well worth studying: challenges to our intuition are exceptionally educational. I first learned of this from a clear paper on Carlos Rodriguez' (SUNY Albany) Web page but I couldn't find it this morning: it appears the server is down. Try Googling "carlos rogriguez statistics" later. (His paper is supposed to be at omega.albany.edu/8008/confint.html , but this might be an old URL.) $\endgroup$ – whuber Aug 18 '10 at 14:51
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    $\begingroup$ Amazing. I didn't know that. Thanks for the reference. $\endgroup$ – Rob Hyndman Aug 19 '10 at 0:29
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    $\begingroup$ Thanks - any chance this is the Rodriguez paper you're thinking of? arxiv.org/abs/bayes-an/9504001 $\endgroup$ – ars Aug 20 '10 at 6:07
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    $\begingroup$ This is great. However, I'm curious why you applied formula (3) (which comes from Edelman), which Wall et al describe as "wider than necessary". Toward the end of the paragraph immediately prior to mentioning (3) they use 4.84 (exactly 1 smaller than 5.84) for a 90% interval, which comes from their equation (4). No doubt I missed something. $\endgroup$ – Glen_b -Reinstate Monica Jan 25 '14 at 0:19
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    $\begingroup$ @Glen_b On the contrary, most likely I missed something. I'll pay attention to that the next time I need this paper, but in the meantime the difference in constants does not affect the analysis here. $\endgroup$ – whuber Jan 25 '14 at 1:29
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This does look to be a slightly contrived question. 49 is an exact square of 7. The value of a t-distribution with 48 DoF for a two-sided test of p<0.05 is very nearly 2 (2.01).

We reject the null hypothesis of equality of means if |sample_mean - popn_mean| > 2*StdError, i.e. 200-112 > 2*SE so SE < 44, i.e. SD < 7*44 = 308.

It would be impossible to get a normal distribution with a mean of 112 with a standard deviation of 308 (or more) without negative wages.

Given wages are bounded below, they are likely to be skew, so assuming a log-normal distribution would be more appropriate, but it would still require highly variable wages to avoid a p<0.05 on a t-test.

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Suppose there are 999 workers at ACME north factory each making a wage of 112, and 1 CEO making 88112. The population mean salary is $\mu = 0.999 * 112 + 0.001 * 88112 = 200.$ The probability of drawing the CEO from a sample of 49 people at the factory is $49 / 1000 < 0.05$ (this is from the hypergeometric distribution), thus with 95% confidence, your population sample mean will be 112. In fact, by adjusting the ratio of workers/CEOs, and the salary of the CEO, we can make it arbitrarily unlikely that a sample of 49 employees will draw a CEO, while fixing the population mean at 200, and the sample mean at 112. Thus, without making some assumptions about the underlying distribution, you cannot draw any inference about the population mean.

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    $\begingroup$ (1) I think you meant to write that the sample, not the population mean, is 112 with 95% probability, not confidence. (2) Your point is well taken--it could apply to any question--but isn't it stated a little extremely? First, the question doesn't ask for an inference about the population mean: we are told it is $\$$200. Thus, conditional on the assumptions, we can estimate the population mean with certainty! Second, even if we were asked to estimate the population mean from the sample, there's still some trivial information we could offer (e.g., it doesn't exceed $10^11 per annum). $\endgroup$ – whuber Aug 19 '10 at 16:56
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    $\begingroup$ (1) good catch. (2), yes, I can make the problem setup asymptotically perverse for fixed results, post hoc. my bad. however, I am no longer sure what the OP is trying to test. If they know the population mean is 200, why are they trying to test it? $\endgroup$ – shabbychef Aug 19 '10 at 17:13
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    $\begingroup$ BTW, evidently a CEO salary/least paid salary ratio of 400 is not considered extreme in the US. 800 is a bit perverse, though. $\endgroup$ – shabbychef Aug 19 '10 at 17:15
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I presume you are referring to a one sample t test. Its goal is to compare the mean of your sample with a hypothetical mean. It then computes (assuming your population is Gaussian) a P value that answers this question: If the population mean really was the hypothetical value, how unlikely would it be to draw a sample whose mean is as far from that value (or further) than you observed? Of course, the answer to that question depends on sample size. But it also depends on variability. If your data have a huge amount of scatter, they are consistent with a broad range of population means. If your data are really tight, they are consistent with a smaller range of population means.

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