21
No, $\bar x$ has its own sampling distribution. Take, for example, the variances of $\bar x$ and $x_i$, in which the former is always lower ($\leq$) than the latter, which means $\bar x$ is not sampled from $p_\theta(x)$.
20
Good examples so far but consider $$X_i \sim Bernoulli(.5)$$
In that case the distribution of the data will only have support on 0 and 1. But the sample mean will have an ever decreasing probability of taking a value of 0 or 1 as the sample size gets larger and larger. That alone should show that the mean isn't being sampled from the original distribution....
12
As an even more pathological example, consider a sample from the distribution which is uniform on the union of $[0,1]$ and $[3,4]$. As the sample size increases, the mean will tend to 2 which isn't even in the support of the distribution. Another similar example is the uniform distribution on the boundary of the unit sphere (in any number of dimensions)
12
No, it is only valid in cases as the Cauchy distribution, the means of samples of the Cauchy follow the same Cauchy dstribution.
9
No. Suppose you have $X_1, X_2 \sim N(0,1)$. Then,
$$
\bar{X} = \dfrac{X_1 + X_2}{2} \sim N\left(0, \dfrac{1}{2} \right)\,.
$$
But $N(0,1) \ne N(0, 1/2)$.
2
You could look at $2 \times 2$ table with columns Y and O for age
groups and rows P and N for whether a purchase was made.
Y O Tot
P 35 43 88
N 98 82 180
Tot 143 125 268
To see
if the proportions of sales for Y and O groups are significantly different, you have your choice of several kinds of ...
2
Well the probability that you are asking for is not computable, at least not using frequentist statistics. Observe that in general, a conditional probability is computed as
$$P(A|B)=\dfrac{P(A\cap B)}{P(B)}$$
But here, $P(B)=$ I flipped the coin n times seeing x occurrences of heads. And for this probability we would need to know the value of $p$.
You ...
2
There are many (interrelated) questions here and not enough space to pursue all their implications. Let's therefore focus on a central idea, which I wish to state rigorously and generally, so I will begin with some definitions that cover the examples in the question (and much more).
First, we need to capture the idea of a "distribution" in some interval ...
1
This experiment seems to have been planned on the spur of the moment (if there was any planning at all). Different numbers of swabs were used in
different rooms before and after. The likely effect is that
the smaller number of swabs in each comparison is the number
that matters.
You are correct that sample sizes here are quite
small---so small that ...
1
At $\alpha$ significance level, an irrelevant variable should only be significant in a fraction $\alpha$ of random samples, regardless of the sample size. I think you were simply unlucky with your random variable generation. If your repeated it 100 times, you should only find about 5 instances where the variable is significant at 5% level.
I indeed ran it ...
1
This site works best with one question at a time. You have
asked several. I will try to answer the ones that
might illustrate general principles.
(1) If your goal is to estimate the population mean $\mu$
from which a random sample of size $n = 60$ is available,
then the best estimate comes from analyzing the undivided
sample.
Suppose you split the sample ...
1
Let $X=(X_1, \dotsc, X_n)^T$ which has covariance matrix $\Sigma=\sigma^2 R$ where the correlation matrix $R$ has diagonal elements $R_{ii}=1$ and off-diagonals $R_{ij}=a, i\not = j$ (here $-\frac1{n-1}\le a \le 1$, see Intuition for near-decorrelation through centering.)
Then $\bar{X}=\frac1n 1^T X$ (where $1$ is a column vector of all ones.) The rest is ...
1
To make the problem easier to read, I am going to use slightly different notation to you. I will confine attention to the case where the set of interest $\mathscr{X}$ is a finite set. Without loss of generality, take $\mathscr{X} \equiv \{ 1,...,m \}$ and let the values of interest have a categorical distribution:
$$X_1,...,X_n \sim \text{IID Cat}(\mathbf{...
1
No.
For the average to be a sample of the distribution, it must belong to the support of the distribution.
Below are two examples where that is not the case (which is sufficient to show that the statement is not true in general).
Discrete
The distribution p(x=1) = 0.5; p(x=-1) = 0.5 has support $$S=\{-1,1\}$$ but average $0\notin S$.
Continuous
The ...
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