confidence intervalp-valuespsst-test
This is a question regarding the t-test in SPSS.
I have two groups and I want to test if the two means are equal. I am using the t-test with bootstrapping. In the end I got a p-value<0.005, which would generally cause me to reject the null hypothesis that the means of the two populations are equal but in my case the zero lies within the 95% BCa bootstrap confidence intervals based on 1000 samples.
Do I still reject the hypothesis of equal means?
Yes, there are some simple relationships between confidence interval comparisons and hypothesis tests in a wide range of practical settings. However, in addition to verifying the CI procedures and t-test are appropriate for our data, we must check that the sample sizes are not too different and that the two sets have similar standard deviations. We also should not attempt to derive highly precise p-values from comparing two confidence intervals, but should be glad to develop effective approximations.
In trying to reconcile the two replies already given (by @John and @Brett), it helps to be mathematically explicit. A formula for a symmetric two-sided confidence interval appropriate for the setting of this question is
$$\text{CI} = m \pm \frac{t_\alpha(n) s}{\sqrt{n}}$$
where $m$ is the sample mean of $n$ independent observations, $s$ is the sample standard deviation, $2\alpha$ is the desired test size (maximum false positive rate), and $t_\alpha(n)$ is the upper $1-\alpha$ percentile of the Student t distribution with $n-1$ degrees of freedom. (This slight deviation from conventional notation simplifies the exposition by obviating any need to fuss over the $n$ vs $n-1$ distinction, which will be inconsequential anyway.)
Using subscripts $1$ and $2$ to distinguish two independent sets of data for comparison, with $1$ corresponding to the larger of the two means, a non-overlap of confidence intervals is expressed by the inequality (lower confidence limit 1) $\gt$ (upper confidence limit 2); viz.,
$$m_1 - \frac{t_\alpha(n_1) s_1}{\sqrt{n_1}} \gt m_2 + \frac{t_\alpha(n_2) s_2}{\sqrt{n_2}}.$$
This can be made to look like the t-statistic of the corresponding hypothesis test (to compare the two means) with simple algebraic manipulations, yielding
$$\frac{m_1-m_2}{\sqrt{s_1^2/n_1 + s_2^2/n_2}} \gt \frac{s_1\sqrt{n_2}t_\alpha(n_1) + s_2\sqrt{n_1}t_\alpha(n_2)}{\sqrt{n_1 s_2^2 + n_2 s_1^2}}.$$
The left hand side is the statistic used in the hypothesis test; it is usually compared to a percentile of a Student t distribution with $n_1+n_2$ degrees of freedom: that is, to $t_\alpha(n_1+n_2)$. The right hand side is a biased weighted average of the original t distribution percentiles.
The analysis so far justifies the reply by @Brett: there appears to be no simple relationship available. However, let's probe further. I am inspired to do so because, intuitively, a non-overlap of confidence intervals ought to say something!
First, notice that this form of the hypothesis test is valid only when we expect $s_1$ and $s_2$ to be at least approximately equal. (Otherwise we face the notorious Behrens-Fisher problem and its complexities.) Upon checking the approximate equality of the $s_i$, we could then create an approximate simplification in the form
$$\frac{m_1-m_2}{s\sqrt{1/n_1 + 1/n_2}} \gt \frac{\sqrt{n_2}t_\alpha(n_1) + \sqrt{n_1}t_\alpha(n_2)}{\sqrt{n_1 + n_2}}.$$
Here, $s \approx s_1 \approx s_2$. Realistically, we should not expect this informal comparison of confidence limits to have the same size as $\alpha$. Our question then is whether there exists an $\alpha'$ such that the right hand side is (at least approximately) equal to the correct t statistic. Namely, for what $\alpha'$ is it the case that
$$t_{\alpha'}(n_1+n_2) = \frac{\sqrt{n_2}t_\alpha(n_1) + \sqrt{n_1}t_\alpha(n_2)}{\sqrt{n_1 + n_2}}\text{?}$$
It turns out that for equal sample sizes, $\alpha$ and $\alpha'$ are connected (to pretty high accuracy) by a power law. For instance, here is a log-log plot of the two for the cases $n_1=n_2=2$ (lowest blue line), $n_1=n_2=5$ (middle red line), $n_1=n_2=\infty$ (highest gold line). The middle green dashed line is an approximation described below. The straightness of these curves belies a power law. It varies with $n=n_1=n_2$, but not much.
The answer does depend on the set $\{n_1, n_2\}$, but it is natural to wonder how much it really varies with changes in the sample sizes. In particular, we could hope that for moderate to large sample sizes (maybe $n_1 \ge 10, n_2 \ge 10$ or thereabouts) the sample size makes little difference. In this case, we could develop a quantitative way to relate $\alpha'$ to $\alpha$.
This approach turns out to work provided the sample sizes are not too different from each other. In the spirit of simplicity, I will report an omnibus formula for computing the test size $\alpha'$ corresponding to the confidence interval size $\alpha$. It is
$$\alpha' \approx e \alpha^{1.91};$$
that is,
$$\alpha' \approx \exp(1 + 1.91\log(\alpha)).$$
This formula works reasonably well in these common situations:
Both sample sizes are close to each other, $n_1 \approx n_2$, and $\alpha$ is not too extreme ($\alpha \gt .001$ or so).
One sample size is within about three times the other and the smallest isn't too small (roughly, greater than $10$) and again $\alpha$ is not too extreme.
One sample size is within three times the other and $\alpha \gt .02$ or so.
The relative error (correct value divided by the approximation) in the first situation is plotted here, with the lower (blue) line showing the case $n_1=n_2=2$, the middle (red) line the case $n_1=n_2=5$, and the upper (gold) line the case $n_1=n_2=\infty$. Interpolating between the latter two, we see that the approximation is excellent for a wide range of practical values of $\alpha$ when sample sizes are moderate (around 5-50) and otherwise is reasonably good.
This is more than good enough for eyeballing a bunch of confidence intervals.
To summarize, the failure of two $2\alpha$-size confidence intervals of means to overlap is significant evidence of a difference in means at a level equal to $2e \alpha^{1.91}$, provided the two samples have approximately equal standard deviations and are approximately the same size.
I'll end with a tabulation of the approximation for common values of $2\alpha$. In the left hand column is the nominal size $2\alpha$ of the original confidence interval; in the right hand column is the actual size $2\alpha^\prime$ of the comparison of two such intervals:
$$\begin{array}{ll} 2\alpha & 2\alpha^\prime \\ \hline 0.1 &0.02\\ 0.05 &0.005\\ 0.01 &0.0002\\ 0.005 &0.00006\\ \end{array}$$
For example, when a pair of two-sided 95% CIs ($2\alpha=.05$) for samples of approximately equal sizes do not overlap, we should take the means to be significantly different, $p \lt .005$. The correct p-value (for equal sample sizes $n$) actually lies between $.0037$ ($n=2$) and $.0056$ ($n=\infty$).
This result justifies (and I hope improves upon) the reply by @John. Thus, although the previous replies appear to be in conflict, both are (in their own ways) correct.
The problem with tests of binomial proportion is that the tests used are generally approximate (since the exact "Clopper-Pearson" test is ridiculously conservative). Therefore, it's not clear that the procedure used to get the CI is the same as that used to test the hypothesis. Theoretically, either approach should lead to the same conclusion if you are using just one CI and one test.
You have a border case. Either statistic is telling you that your observation is not all that common under the null hypothesis. Remember: there is nothing special about 5% significance...its a cultural artifact from Ronald Fisher in the 1930's. Its a guideline.
For what it's worth, I'd conclude that its unlikely that the true success probability is as high as 0.75.
Per @John
In a strict hypothesis testing situation, you're stuck with your 0.05, so you would reject the null hypothesis under that criteria. However, I wouldn't run to the press quite yet ;-)...hypothesis tests can really destroy any nuance in inference.
Best Answer
Caveat: This answer assumes that the question is about interpreting bootstrapped p-values and CIs. A comparison between a traditional p-value (not bootstrapped) and a bootstrapped CI would be a different issue.
With a traditional (not bootstrapped) t-test, the 95%CI and the p-value's position relative to the .05 cutoff for significance will always tell you the same thing. That's because they're both based on the same information: the t-distribution for your degrees of freedom and the mean and standard error observed in your sample (or difference between means and standard error, in the case of a two-sample t-test). If your CI doesn't overlap with 0, then your p-value will necessarily be < .05 --- unless, of course, there's a bug in the software or a user error in implementation or interpretation of the test.
With a bootstrapped t-test, the CI and p value are both calculated directly from the empirical distribution generated by the bootstrapping: the p value is simply what percent of bootstrapped group differences are more extreme than the original observed difference; the 95%CI is the middle 95% of bootstrapped group differences. It is not impossible for the p-value and the CI to disagree about significance in a bootstrapped test.
Do you accept or reject the null hypothesis?
In the context of a bootstrapped test, the p-value (as compared to the CI) more directly reflects the spirit of the hypothesis test, so it makes the most sense to rely on that value to decide whether or not to reject the null at your desired alpha (generally .05). So in your case, where the p value is less than .05 but the 95%CI contains zero, I recommend rejecting the null hypothesis.
All of this skips over the big ideas about how important "significance" really should be and whether or not null hypothesis significance testing is actually that useful of a tool. Briefly, I always recommend complimenting any significance testing analysis with estimation of effect sizes (for a two-sample t-test, the best effect size estimate will probably be Cohen's d), which can provide some additional context to help you understand your results.
Related helpful post: What is the meaning of a confidence interval taken from bootstrapped resamples?