I am trying to compare the means of two groups (male/female) for different attitudes. Since my data sometimes violate the assumptions of normality and heteroscedasticity, I was planning to do a Mann-Whitney U Test. However, my professor insisted that I still use a parametric test and told me that a One-way ANOVA is more robust to such violations that a regular t-test. Now, my questions are:

  • Is a One-way ANOVA more robust than an independent samples t-test?
  • Can I report differences in means for just two groups, using ANOVA?

I have a copy of the SPSS guide by A. Field and would like to make such a statement: "On average, participants experienced greater anxiety to real spiders (M = 47.00, SE = 3.18) than to pictures of spiders (M = 40.00, SE = 2.68). This difference was not significant t(22) = −1.68, p > .05."

It is from the chapter on t-tests (which makes sense!), but now I am confused whether or not I can make a similar statement based on an ANOVA (for two groups only). Thanks!

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    $\begingroup$ Your professor told you incorrect information. $\endgroup$
    – Dave
    Sep 21, 2023 at 21:39
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    $\begingroup$ One way ANOVA on two groups is equivalent to the corresponding two-sided t-test. You should get the same p-value every time. $\endgroup$
    – Glen_b
    Sep 22, 2023 at 3:00
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    $\begingroup$ The exceptions would be if there is a Welch test performed (default behavior in R) or if the calculation winds up doing a paired test. $\endgroup$
    – Dave
    Sep 22, 2023 at 3:05

1 Answer 1


For two samples, the standard (two-sided) two-sample t-test and an ANOVA are equivalent (unless the t-test is run with a correction for unequal variances). ANOVA is not more robust in any way. (One can easily show that $F=t^2$ where $F$ is the ANOVA test statistic and $t$ is the one of the two-sample t-test.)

There is no reason to prefer ANOVA for two groups. Note that the two-sample t-test can be run without issues in a one-sided way (for example testing $\mu=0$ against $\mu>0$) whereas the ANOVA F-test cannot (because due to $F=t^2$ it doesn't differentiate between whether $t$ is larger or smaller than 0).

  • $\begingroup$ Awesome, thanks! I assume in that case, the result for a Welch t-Test and Welch-ANOVA is also identical? (I just ran a test and both results were identical.) $\endgroup$
    – ber1495
    Sep 22, 2023 at 8:04
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    $\begingroup$ @ber1495 I'd think so (haven't done the math). $\endgroup$ Sep 22, 2023 at 9:14

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