New answers tagged

1

I would suggest using a mixed model to analyse these data: I assume that you're interested more in the specific ability of each candidate rather than the properties/bias of the reviewers. In this setting, a simple model would be $$ \mathrm{Grade}_{ij} = \alpha_i + \beta_j + \varepsilon_{ij}, $$ where $i$ ranges over candidates and $j$ over reviewers; $\...


1

It is hard to gauge how any method would work without even seeing sample data, but the possibilities include summarizing the last so many values, by a median, or mean, or more generally a trimmed mean (in this case, consider downweighting according to time elapsed) summarizing each day, ditto, and just leaving blank days without measurements. Whatever ...


8

In the example of the video, people are classified as "Hipster" or "Non-Hipster". That is a nominal scale level. Nominal data can be counted but not added. Without addition, there is no computing means. The mean of a "Hipster" and a "Non-Hipster" is not "Non-Hipster and a half". You can count people and ...


0

Suppose the running median was intended to be the median of 7 values, observed at noon every day of the past week. With irregular observations, one alternative is to take the median of 168 values over all the hours of the past week. The question is then how to fill in the values at the many unobserved hours. Depending on the KPI and the measurement procedure,...


1

Central Limit Theorem applies to the number of observations, not number of repeated draws. You draw 62 observations - this is the number that has to tend to infinity for the theorem to apply. To see why that is the case imagine that you are drawing 1 observation, instead of 62, and repeating it 100,000 times - would you expect this distribution to approach ...


1

The Central Limit Theorem is a large sample result. In your setting, 62 observations do not seem to be large enough for this. The question is: why do you require anything to be normal? Or was it just for curiosity?


1

There is another formula for the variance of a random variable derived from the first and second moments. $Var(X)=E((X-\mu)^2)=E(X^2)-E(\mu)^2=E(X^2)-\mu^2$. As it turns out, you have all the information you need to estimate this.


0

Depends on the standard deviation of the data. If the data are noisy, you will typically need more data than if they were not. It also depends on the size of the difference you hope to detect. Smaller differences require more data. Here is a plot demonstrating this relationship. You can see holding the difference in means constant but increasing the ...


1

Consider three samples (groups) a,b, and c, each of size $n=20$ from exponential distributions with different rates $\lambda_i,$ where means are reciprocal rates $\mu_i = 1/\lambda_i$ We can find the mean of the grand sample of all sixty observations by averaging the the three group means (weighted average if sample sizes differed), but it is not possible to ...


0

I have known what exactly I have done wrongly. The exact solution in R is as follows: dif=(New-Old) m=mean(dif) st=sd(dif) CI=m+c(-1,1)*qt(0.975,4)*st *sqrt(1/5) ExpCI=exp(CI) # equals [0.09 2.49] MEAN=mean(ExpCI) # equals 1.29


0

As others have pointed out, whether it is correct to calculate the mean and the standard deviation of percentages depends on your intended use. For you use, at least as I understand it, it seems to be incorrect. As I understand from your question and comment, you are trying to do anomaly detection. You are basically asking: Is the number of missed ...


0

The same problem occurs in image processing with mean filtering or other convolution operations, and it is solved in different ways by different "border treatments", which can be, e.g., repeat, reflect, zero, or wrap around. As the igraph library function running_mean does not support a "border treatment" option, you must implement it ...


0

Results from R: Consider a random sample of size $n=40$ from the distribution $\mathsf{Norm}(\mu=100, \sigma=15).$ Begin with numerical descriptive statistics: set.seed(2021) x = rnorm(40, 100, 15) summary(x); sd(x); length(x) Min. 1st Qu. Median Mean 3rd Qu. Max. 70.54 82.79 99.45 98.58 113.54 125.95 [1] 17.13621 # sample standard ...


3

You calculated the mean of logarithms of ratios, not the mean of ratios themselves. To obtain ratios from their logarithms, you have to raise $e$ to power of them. In Python and NumPy: import numpy as np logs = [1.304, -0.768, -0.473, 1.237, 2.405] # Logarithms of ratios ratios = np.exp(logs) # You omitted this! np.mean(...


3

If you have datapoints $x_1, \ldots, x_n$ and $y_1, \ldots, y_n$ and you want to find the mean ratio $\frac{1}{n} \sum_{i=1}^n \frac{x_i}{y_i}$, this is not equal to the exponentiated mean of the log ratios. In other words, this is not the same as computing $\exp\left(\frac{1}{n}\sum_{i=1}^n \log \frac{x_i}{y_i}\right)$, which is what it sounds like you are ...


0

Thanks Frank, we also found the following paper that may help. Would you recommend following either of the methods explained in 3.1 and 3.2, if not the ANCOVA, due to strong assumptions in the latter. https://www.ncbi.nlm.nih.gov/pmc/articles/PMC6297128/#appsec1


2

A parallel-group randomized trial is design to compare parallel groups, not to compare change from baseline. You need to be using ANCOVA with Y=post score and X=baseline score. Change from baseline assumes linearity and a slope of 1.0 for post-pre. In depression studies I've seen strong nonlinearity. The most general analysis that is powerful is a ...


2

Assuming treatment and control 3-month differences are nearly normally distributed, you want a power and sample size computation for a two-sample t test. In order to get the sample size for each group you need the following: Approximate variance of data in the two groups, Significance level of the test [5% ?], Size of difference you want to detect, power (...


1

Assuming $\mu, \sigma$ for mean and deviation respectively, you have two equations of the following form: $P(X\leq x)=p$. When standardised, it becomes $$P(X\leq x)=P\left(\frac{X-\mu}{\sigma}\leq \frac{x-\mu}{\sigma}\right)=P\left(Z\leq \frac{x-\mu}{\sigma}\right)=\Phi\left(\frac{x-\mu}{\sigma}\right)$$ where $\Phi$ represents the CDF of the standard normal ...


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