New answers tagged sample-size
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Unless you strike really lucky I think 50 is too small.
MacCallum, R. C., Widaman, K. F., Zhang, S. and Hong, S.
in a paper entitled "Sample size in factor analysis" published in Psychological Methods, in 1999 volume 4, 84-99. doi:10.1037/1082-989X.4.1.84 discuss this in much detail as there is no simple answer.
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The question makes sense. The estimate from the trivial model likely over estimates sigma, which will in turn mean your confidence interval will be narrower than you expect. No matter, the calculation will ensure the resulting CI will have width no larger than $w$.
EDIT: It might be worth examining where this equation comes from. If you write down the ...
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Good question, answered by Lachin in his book Biostatistical Methods.
Assuming your AB test is in terms of a binary outcome, the sample size formula is
$$ N = \Big( \dfrac{Z_{1-\alpha} \phi_0 + z_{1-\beta}\phi_1} {\Delta} \Big)^2$$
Here,
$\Delta$ is the expected difference between groups.
$\phi_0 = \pi (1-\pi)\times 6.25$ is the variance component under the ...
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However, since I'm manually counting to get actual object count, I selected only 30 images.
30 is one of the two conditions of the Central Limit Theorem for a sample mean. If you sample more than 30 cases and there are no extreme outliers, the sampling distribution of the mean is expected to be nearly normal. In the following analysis, we assume that the ...
4
Elaborating a bit on Jeremy's answer, let's think for a minute about what a power analysis is. The purpose is to determine how many participants one would need to "detect" an effect of a specific size. So in discussing the results of your experiment vis a vis the sample size you originally designed, and what the pandemic (unforseen circumstances) ...
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I would just explain what happened. You powered for N, and you got N*. It's not the first time this has happened (and won't be the last).
Post hoc power would not be especially useful (as you have realized.)
1
Strangely, power.prop.test gives you the power for the Pearson $\chi^2$ test of proportions, and yet you have used the Fisher Exact Test to calculate a p-value. Of course you can "analyze these data" if there are no events (or just 1 event) in group 1. The Fisher Exact Test does not require an evaluation of the information under the alternative ...
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First off, ANOVA is often surprisingly robust to strange distributions. What is important is not that model residuals are normally distributed, but that parameters are (approximately) normally distributed. Here, that would be the group means. You can take a look at this by bootstrapping each group mean. I suspect that the bootstrapped means will be nicely ...
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I will try to address 3 and also a few others affected by choice of transformation. You can try a log (base of your choice, though natural log or $log_{2}$ might be best) transform, but if any $Y = 0$ this is not optimal as you'd have to add a small amount to each value. Square-root transform will work better in this case. If you chose a transformation, try ...
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There is no need for that in general. The rationale might be the cost of both types of errors is the same. Nonetheless, it is very common to use, for example, one-sided $\alpha=0.025$ and $1-\beta$ equal 0.8 or 0.9.
Something interesting happens though if you do use $\alpha=\beta$ when you are testing $H_0:\theta \le 0$ with a postulated value of $\theta_1$. ...
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Statistical power is not usually the issue when doing prediction, but a loss of statistical power is related to the precision of predictions and to the volatility of the found form of the predictive model. Data splitting harms all of these. That's why 100 repeats of 10-fold cross-validation, or the bootstrap, are recommended resampling procedures to do ...
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Generally, the p-value can be less than 0.05 even if the observed value of the estimated parameter is much smaller than the postulated value. There is a simple formula for the smallest observed value you need to have significant result. Of course, that assumes all your assumptions are correct and that the estimate is normally distributed. If you are doing ...
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