Quick question: Are test statistics like the "F" and "t-statistic" the number of standard deviations on an F and t-distribution, respectively? For example, if I got a "t" test statistic of 4.13, this means a value (like a mean) is 4.13 standard deviations from the center of a t-distribution?
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https://stats.stackexchange.com/search?q=%22test+statistic%22 – whuber Nov 19 '21 at 18:07
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At the end of my post at https://stats.stackexchange.com/a/130772/919 I provide a general, conventional definition of "test statistic." The thousands of hits in the site search linked in my first comment suggest you can find your question answered in various ways through applications and examples. – whuber Nov 19 '21 at 18:09
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I'll take that as a "No". – Nate Nov 19 '21 at 18:16
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Not "no," but not "yes" either. In *some* cases the test statistic is a number of standard deviations of something. In other cases, it's not. For instance, we have lots of textbook questions about uniform distributions with unknown endpoints. They involve test statistics like the minimum, maximum, or range of the data. No standard deviations are involved; nor are F or t distributions. That's why you need to read a general account of what a test statistic is. – whuber Nov 19 '21 at 18:19
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I think you misunderstood my question. But no matter, this helps too. So, a t-statistic is likely the number of standard deviations because it's related to t-distribution (a form of the normal distribution with equally spaced deviations on either side), while an f-statistic is a ratio, I believe, and therefore different it seems. – Nate Nov 19 '21 at 18:24
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Conceptually it works in the other way. The t-statistic is a *standardized* statistic and, under certain specific assumptions, it just *happens* to follow a Student t distribution. From a fundamental conceptual perspective, there's nothing important about the fact the F-ratio statistic is computed as the ratio of two quantities. Again, these issues are addressed in some excellent answers in the duplicate threads. – whuber Nov 19 '21 at 18:40
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Ok, that’s all I needed. Thank you, I’ll check the others out too. – Nate Nov 19 '21 at 19:04