7

By inspection, it is pretty clear that Cat is under-represented in the second database. Let's see how that plays out in a chi-squared test of your $2\times 4$ contingency matrix. db1 = c(22000, 2300, 42009, 106000) db2 = c( 380, 30, 7, 260) MAT= rbind(db1,db2); MAT [,1] [,2] [,3] [,4] db1 22000 2300 42009 106000 db2 380 30 ...


4

A $\chi^2$-test would be the obvious choice, especially since you do not seem to have a problem with small cell counts.


1

This cannot be answered without knowing the actual counts. Hypothetically, if you had 693, 731, 749, and 720 successes (under four different respective conditions) out of 1000 trials. then you would have the contingency table TBL below. [Please understand that 1000 is just a guess to be able to illustrate how to run the test in R. You need to use the actual ...


1

One method is to use a chi-squared test for homogeneity. Let's compare A+ and B+. The test is a bit difficult because counts for 'Stage5' and 'Unknown' are so small for B+. The null hypothesis is that the (proportions of) 'Stages' are similarly distributed in A+ and B+. This test requires a contingence table such as TAB below: A = c(124, 45, 234, 64, 163) B =...


1

Depending on your software the "chi-squared test of independence" will be implemented using the chi-squared distribution as an approximation. That will not work with a 6 x 3 table and only 30 observations. You should turn that of and use simulation instead. If using R you could use the argument simulate.p.value = TRUE with the chisq.test command. ...


1

What you propose looks OK. Since there is no estimated parameters, the number of degrees of freedom should be $10000 - 1$. Then you can find p-values from the $\chi^2_{9999}$ distribution. But your last paragraph indicated the setting is different: I can get alerts when a drive is above a threshold number of fragments relative to the number of fragments ...


1

A goodness of fit test with only 12 observations is borderline --- and would only be able to detect gross departures from the null. For testing the null of lognormality i would use some normality test, for instance the Shapiro-Wilk test, on the logarithm of the observations. With your data, using R I get shapiro.test(log(lnd)) Shapiro-Wilk normality ...


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