New answers tagged mean
0
You can calculate it through:
Fieller's method
The Taylor method, also called Delta method: it's easier than Fieller's but will fail if the denominator approaches zero.
The Hwang–bootstrap method, a bootstrap technique that does not result in unbounded confidence limits.
Here you can find a thorough description and comparison of these methods.
2
You can resort to the bootstrap version of Student's t-test.
It works as follows:
Compute the sample mean and standard deviation for each group and
label the results $X_1$ and $s_1$ for group 1, and $X_2$ and $s_2$ for group 2. Set $d_1=\frac{s_1^2}{n_1}$ and $d_2=\frac{s_2^2}{n_2}$, where $n_1$ and $n_2$ are the sample sizes.
Generate a bootstrap sample ...
1
Utilizing your reproducible example, I have worked up what appears to be the idea behind VIM using nls.
library(matrixStats)
set.seed(101)
df <- data.frame( "ID" = paste("ID", seq(1,100,1), sep = ""),
"x1" = sample(90:220, size = 100, replace = T),
"x2" = sample(90:220, size = 100, replace = T),
"x3"...
3
This notation is describing a statistical test that was applied in order to determine whether or not there is a "statistically significant" association between two variables. Here, the specific question is whether "content related to losing fat or weight" is associated with what site you are on (Thinsipration vs. Fitspiration). The null (default) hypothesis ...
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Welcome to Cross-Validated. :)
Sample mean is $\mu = \frac{\sum_i^n x_if_i}{\sum_i^n f_i}$, and sample variance is $\sigma^2 = \frac{\sum_i^n (x_i-\mu)^2f_i}{\sum_i^n f_i}$, where $f_i$ is the sample frequency of the $i^{th}$ interval, and $x_i = \frac{x^{upper}_i+x^{lower}_i}{2}$ is the average of the upper and the lower limits of the $i^{th}$ interval. ...
1
No, the sample mean of an independent sample from a Poisson distribution is NOT Poisson distributed. It is clear from the fact that a Poisson random variable can only have integer values, but the mean of such a sample does not need to be an integer.
But, the sum of the sample ($n$ times the mean) do have a Poisson distribution, and all wanted probabilities ...
1
I love the discussion here - the trimmed mean is a powerful tool to get a central tendency estimate concentrated around the middle of the data.
The one thing I would add is that there is a choice to be made about which "metric" to use in the cases of small and large sample sizes. In some cases we talk about
means in the context of large samples because of ...
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While it is correct that mathematically mean and expectation value are defined identically, for a skewed distribution this naming convention becomes misleading.
Imagine you are asking a friend about the housing prices in her city because you really like it there and actually think about moving to that city.
If the distribution of housing prizes were ...
0
If it is a continuous variable, then the statement $P(X=x)$ is not correct. It should be $P(X\leq x)$. This is also called the CDF or cumulative distribution function of a normal distribution.
If that is the case then, you are looking at $P(X\leq 10111) = 0.3$ and $P(X\leq 10840) = 0.5$ etc. My take on this is as follows
Use the CDF formula (you can ...
0
Assuming you meant probabilities of the sort $P(X\leq x)$, here is a numerical result using R
optim(par=c(1e4,1e2),
fn=function(x){
(qnorm(0.3,x[1],x[2])-10111)^2+
(qnorm(0.5,x[1],x[2])-10840)^2+
(qnorm(0.6,x[1],x[2])-10948)^2
},method="L-BFGS-B",lower=c(1,1),upper=c(1e5,1e5))$par
[1] 10734.506 1123.457
which results ...
1
One way you could approximate this is to draw a large random sample from your target distribution, find the nearest neighbors of your fixed samples, and replace those "nearby" random samples with the fixed samples. This would require a large enough sample to ensure that each fixed point has a neighbor that's "reasonably" close, how close that needs to be ...
2
I think your question is incomplete. You write that you are interested in "[the dollar value] significance while properly accounting for the customer and date" but you haven't specified a particular hypothesis.
Are you interested in seeing if there are differences between customers?
Are you interested in assessing date effects?
Are you interested in ...
1
$$\sum{(x_i-\overline{x})^2} = \sum{(x_i-\overline{x})x_i} - \sum{(x_i-\overline{x})\overline{x}} = \sum{(x_i-\overline{x})x_i} - \overline{x}\sum{x_i} + n\overline{x}^2 = \sum{(x_i-\overline{x})x_i} - \overline{x}n\overline{x} + n\overline{x}^2 = \sum{(x_i-\overline{x})x_i}$$
$\sum{x_i} = n\overline{x}$
1
It is incorrect to simply make a paired $t$-test because animal ID induces a clustering (blocking) to the measurements at hand. A random effects model for paired testing is potentially more natural in this scenario.
The immediate issues I can see are three:
Certain animals might have lower or higher average levels of turning under condA or condB or both. ...
answered Jan 5 at 0:28
usεr11852 says Reinstate Monic
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2
First, you would expect that the mutation and no mutation groups would have the same BMI since the presence of the mutation is completely independent from BMI and all other variables in the data. The fact that you got a significant difference using the first comparison is a fluke; if you use a different seed, (e.g., 9999), you get a nonsignificant difference....
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