# Tag Info

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When you do this kind of analysis you should pick a significance level $\alpha$ that will help you decide wether to reject the null hypothesis (models have same variance, etc.). Generally this value is set to 0.05 and you will reject the null hypothesis if p < $\alpha$. In this case $0.57 >> 0.05$ so I would conclude that there is no statistically ...

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I think you shouldn't even delete duplicate entries before feature selection because they're not real dups and data belongs to different patients. You're manually changing the data distribution by doing so. So, let alone after the feature selection, you shouldn't do it even before it.

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Unless the totals in the other direction ($B$, not-$B$) are also very highly unbalanced, the chi-squared value will be largely driven by the two numbers in the smaller category (i.e. in the not-$A$ row). It looks like it will have about 11 values total. If the $B$ categories are close to evenly split you'd need the smaller of the two not-$A$ values to be 0 ...

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Assuming normal data, under $H_0: \sigma_2^2/\sigma_1^2,$ the ratio $F =s_2^2/s_1^2 = 60.18^2/42^2 = 2.053 \sim \mathsf{F}(9,999).$ For a one-sided test against $H_a: \sigma_2^2/\sigma_1^2 >1,$ one rejects if the F-statistic exceeds the the 95th percentile $1.8892$ of $\mathsf{F}(9,999),$ called the critical value. So you do (just barely) reject at the 5%...

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Conversion rates without sample sizes (i.e. counts of those who convert vs. those who do not) are not useful. 25% could mean 25 out of 100 or 2500 out of 10,000. The precision of the estimate in the latter is greater than that of the former. If your design is a classic two groups-binary outcome, you can use any number of the tests I write about here. ...

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