Search Results

To begin, I have checked your original chi-squared analysis of these data. The P-value of your chi-squared test is $0.02 < 0.025 = 2.5\%;$ you have a result significant at the 2.5% level. Thus, there …

Comment continued: Perhaps the 2-by-2 table of my comment is as follows Met Goal Gender Y N Total ---------------------------- M 10 15 25 W 40 35 …

Compare observed and expected counts in the cell of the table for 'NoPaper' by 'Churn'. Alternatively, look at Pearson residuals. Here is the chi-squared test in R for your data. MAT = matrix(c(102, …

According to your Comment, I'm subtracting the high-income subjects from the population before doing the chi-squared test in R: TBL = rbind(c(260,480,50)-c(50, 95, 8), c(50, 95, 8)) chisq.test(TBL) …

A chi-squared test is OK because $n = 26$ is large enough for the chi-squared statistic to have approximately the distribution $\mathsf{Chisq}(\nu=1),$ giving the P-value $0.00009 < 0.001 = .1\%.$ (Se …

Here is my analysis. I think it is best to start by looking at all of the data. First, put the data into a contingency table, with columns 'Positive' and 'Negative'. DTA = matrix(c(87,96-87, 112,124 …

Goodness-of-fit statistic. Suppose you want to test whether a die is fair by rolling it 600 times. Then you would expect, on average, to see each face $E = 100$ times. If the observed counts for faces …

With possible winnings from 0 to 25, it seems there is room for considerable variability. Because you show no information about frequencies of individual payouts, I don't see a rigorous way to test wh …

Suppose I have 10,000 observations precisely from $\mathsf{Gamma}(shape=5, rate=0.1).$ Generate them in R as follows: set.seed(2020); x = rgamma(10^4, 5, .1) (1) Then, because we know the distribu …

Here are some fragmentary answers based on what you have told us about your data and analysis. If $X \sim \mathsf{Chisq}(\nu = k),$ then $E(X) = k$ and $Var(X) = 2k.$ [See Wikipedia or your text or cl …

Here are stripcharts of your control and treatment observations (Lower plot for control.) Using R: ctrl = c(239.473, 247.451, 259.486, 254.329, 240.195) trtm = c(252.568, 313.24, 292.972, 269.032, 260 …

The rationale for using the test statistics of the form $Q = \sum_i\frac{(X_i-E_i)^2}{E_i}$ is that counts $X_i$ in level $i$ of a categorical variable (univariate or mulivariate) are roughly Poisson- …

For a possibly useful general discussion of 'Diversity indexes' you might want to look at Wikipedia, perhaps starting with the Simpson index $$\lambda = \sum_{i=1}^R p_i^2,$$ where $R$ is the number o …

Because of the round numbers in the first column (group), I wonder if all eight numbers are counts of subjects randomly chosen from two populations. If you randomly assigned 250 subjects to one Metho …

In Minitab statistical software the printout for the z-test is shown below: Essentially, your are comparing $\hat p_S = 0.135$ in your sample with $\hat p_C = 0.007$ in the complement of the sample. S …