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My first intuition was to use a t-test, but after looking in my memories and then in wikipedia I could not discern a conclusive answer...

In my course I want to talk about the rationale of the t-test in comparision to the chi-square (with which I compare the distribution of men/women in my sample with an expectation/theoretical distribution). So I want to ask: are the empirical means of body-heights for men and women the same as the expected/theoretical means in the population? and thought I could simply carry over that rationale of the chi-square to the t-test. But after reading some simple sources I can't yet see, how I could apply that test. Or do I need another thing?

(disclaimer: I've no question about the t-test as a test for the significance of the difference of means of two groups in my sample, say: whether the means of body-heights of men and women are significantly different in an assumed population - that's not my question)

[Update]: According to Peter I could t-test the likelihood of equality of the means between sample-groups and expectations for the men and for the women separately. However, then I have two probabilities, but where I want to get one (for the combined result).
Moreover,the focus of the motivating question is more the conceptual one: in an introductory course I want to step from the explanation of the chi-square as a test, where I compare the empirical frequency table with an expected/theoretical one, towards the same concept concerning the means instead of the frequencies. So to say: to introduce a measure for the likelihood that a set of parameters (here: means) in the sample is the same as in the expectation/population. I thought, the t-test would be the "natural" candidate for this - but that's the reason, why I asked in the title "what is the required test (...)?" - I assumed the t-test were the "natural" candidate here...

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  • $\begingroup$ I don't think the t-test is a natural candidate here, because your question as formulated is testing two observed means against two population means. A t-test usually only tests between two sets of means. Another alternative approach would be model testing under ANOVA... but it suffers from the same issue I've mentioned before... i.e. to 'see if the means are equal to the population' will essentially require that your endorse the null hypothesis, and does not yield a sensible question. It should be clear that a chi-square does not provide an answer about the question, "does O=E". $\endgroup$
    – russellpierce
    Commented Feb 16, 2013 at 19:45
  • $\begingroup$ In short, the premise of the question is flawed because the "proportion/distribution of means agrees with the population expectation" is not a conclusion that you can reach with hypothesis testing. It is endorsing the null. You can't do it in a chi^2 either. Failing to reject the null is not the same as endorsing the null. $\endgroup$
    – russellpierce
    Commented Feb 16, 2013 at 19:48

2 Answers 2

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You can do this with the t-test, for men and women separately. Say the assumed population height for men is 70 inches and your vector of heights for men is MHt. Then, in R

Meq70 <- t.test(MHt, mu = 70)

is all you need, and similar for women. If you don't have R, you can just subtract 70 from each male height and do the usual one sample t-test to see if the result of that is 0.

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  • $\begingroup$ Thanks Peter! But is a significance test, that the proportion/distribution of means agrees with the expectation really satisfied with the two separate tests? (and how would I combine the two significances that I would obtain by the separate tests?) $\endgroup$
    – Gottfried Helms
    Commented Feb 16, 2013 at 13:28
  • $\begingroup$ If you wanted to combine the two, you could look at multivariate t. In your question (the italicized bit) you asked about means. If you want to compare something other than the means, there are different methods depending on what it is you want to compare. But then you would need to know the distribution (not just the mean) of the population. $\endgroup$
    – Peter Flom
    Commented Feb 16, 2013 at 13:31
  • $\begingroup$ Hmm, I'm thinking to compare the sum-of-centimeters (understood as frequencies) of men/womean between sample and theoretical values using Chi-square... but somehow this feels a bit "adventurous"... ;-) $\endgroup$
    – Gottfried Helms
    Commented Feb 16, 2013 at 13:34
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    $\begingroup$ I think you need to first state more precisely what it is you are trying to test, then someone here (maybe me) can come up with some test of that. $\endgroup$
    – Peter Flom
    Commented Feb 16, 2013 at 13:37
  • $\begingroup$ Hmm, I'll come back to this later and I'll update my question. Thanks so far... $\endgroup$
    – Gottfried Helms
    Commented Feb 16, 2013 at 13:57
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Er. As far as I know there is no classical statistical method to assess the probability that a set of values is the same as in the population. This claim is equivalent to trying to assert a null hypothesis.

If you care to update your hypothesis to be about whether the probability of a set of values falls within a certain range around the stated population value... then (I think) things start to get possible again using confidence intervals. However, I'm making this up based on what seams reasonable to me, so perhaps I am getting it wrong.

  1. Assess what proportion of the sampling distribution of the means for males falls in the range around the population mean for males
  2. Assess what proportion of the sampling distribution of means for females falls in the range of the population mean for females.
  3. Multiply these proportions to satisfy the AND part of your statement.
  4. Interpret the multiplied proportions to say (something like):
    • Frequentist: if you ran the experiment N times this proportion of times the means observed would satisfy these conditions
    • Bayesian: there is an probability chance that both really do fall in the ranges provided

There is also an element of your question that looks like it is about distributions. Distributions of means are reflected above, but if you are interested in individual distributions, then you might want a Kolmogorov-Smirnov test to compare the distributions. But again, your assertion that they are the same will be based on endorsing the null which is still not great.

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  • $\begingroup$ I think the original poster needs to clarify what he wants to investigate. Although I did not read his question the way you did, I think your reading of it is also sensible $\endgroup$
    – Peter Flom
    Commented Feb 16, 2013 at 15:10
  • $\begingroup$ Hmm, the formulation "whether a set of measures falls within a certain range around the population values(s)" describes very well what I want to relate to, I should perhaps take this into my question. But my problem is also a paedagogical: I hoped I could derive the rationale of the t-test from that of the chi-square test, which I'm to introduce before and with which I make a great deal about comparing an empirical (discrete) distribution (frequencies of a for instance dichotome variable) with an expected one. $\endgroup$
    – Gottfried Helms
    Commented Feb 16, 2013 at 17:21
  • $\begingroup$ The rationale is pretty simple. The numerator in both equations is an index of how far we've deviated from expectation. The denominator is a scaling factor telling us how notable the observed deviation is given. The $t_{critical}$ distribution is the expected distribution of t-values if the null is true. So the decision being made is about comparing an observed value (be it $\bar{x} - \mu$ or $\bar{x}_1 - \bar{x}_2$) versus what is expected under $H_0$. $\endgroup$
    – russellpierce
    Commented Feb 16, 2013 at 19:41

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