Unlock the Secrets of Contingency Table with Expert Guidance and Real-World Application

A survey was carried out to find out if the occurrence of different kinds of natural disasters varies from on part of the town to the other. The town was divided into two as center town and town outskirts, and the natural disasters were divided into tornadoes, lighting, and fires.The survey showed these results:
Tornadoe Lightning Fires
Town Center 10 220 120
Town Outskirts 20 150 200
Do the above findings provide enough proof at 95 confidence level to indicate that the happening of natural disasters depends on the part of town?

why doubling the number in a contingency table changes the p-value?
i am doing a facts trouble, which is checking out if the evaluation of someone is impartial of the individual's intercourse. i'm given a contingency desk, I calculated the expected fee for each access and calculated the chi-rectangular value then I got a p-fee.
Then the query requested me to do the same thing after doubling all entries in the contingency desk, I were given a p-price smaller than the only I got earlier than. Why does this take place? Can all and sundry supply me an cause of the distinction?

How do I generate doubly-stochastic matrices uniform randomly?
A doubly-stochastic matrix is an n×n matrix P such that
$\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{p}_{ij}=\underset{j=1}{\overset{n}{\sum <!--\; \sum \; -->}}{p}_{ij}=1$
where $p_{ij}\ge <!--\; \ge \; -->0$. Can someone please suggest an algorithm for generating these matrices uniform randomly?

Over a period of 100 days, weather forecasters issue forecasts for rain on a daily basis. Based on the forecasts and actual observations, you are supposed to help them finding out if the forecasts were any good. A forecast is simply one of the following three possible statements: “no rain”, “maybe rain”, or “certainly rain”. The forecasters record the number of days they issued each of these three forecasts, and also the corresponding number of days it rained. The results are collected below:
Forecast: "No rain" , Days with rain: 4, Days without rain: 18
Forecast: "Maybe rain" , Days with rain: 25, Days without rain: 22
Forecast: "Rain" , Days with rain: 25, Days without rain: 6
Show that $n_{ij}$ has a multinomial distribution with unknown probabilities $p_{ij}$, which are the unknowns parameters in this problem. Write down the likelihood of the data $n_{ij}$ as a function of $p_{ij}$ and N.
This is a past exam paper and I have no clue how to answer this in a way that could get me marks! I've written the multinomial distribution:
$\frac{n!}{X_1!...X_n!}{p}_1^{x_1}\ast <!--\; \ast \; -->p_2^{x_2}...\ast <!--\; \ast \; -->p_n^{x_n}$
But I can't work out how to do this question for the life of me! Can anyone help?

Mary is certainly one of medical doctor Brown's patients. She has carried out a domestic pregnancy take a look at which has given a high quality end result. what's the possibility that the pregnancy take a look at used by doctor Brown in his surgical procedure will say Mary is pregnant given that the home check became nice?
Home pregnancy test is 85% accurate
Doctor Brown pregnancy test is 95% accurate
20 females are pregnant and 80 females are not

I have two binary matrixes, of the same size (e.j. 5000x5000). Those matrixes represent the same area, divided in cells of the same size. Each cell of one matrix can be true or false, meaning some property is present or not in this cell. One matrix represents the presence of a property A, and the other one the property B.
Therefore, I can easily build a 2x2 contingency table using as variables the presence/absence of A and B:
A=1 A=0
B=1 a b
B=0 c d
a= number of cells where both A and B are present
b= number of cells where only B is present
etc.
And
I can apply a chi-square test on this table, building an "expected" contingency table, to assess the independency of both properties.
But I also need to assess if the number of cells that "overlap" (cells that are are true in both matrixes, i.e. where both A and B are present) is higher or lower than expected if both properties were independent. Of course I can compare real and expected value of a in the real and the expected contingency tables, but what I need is some thing like a probability or a measure of how much overlap is higher or lower than expected. In some way, it can also be seen as a measure of the "correlation" between both properties? I know if I had a smaller number of cells I could use Fisher's exact test, where obtained p-value will indicate the "direction" of the relationship between A and B. But as Fisher's exact test implies factorials, it is not possible.

If 99% of all new applicants tell the truth on their applications, then are submitted to a polygraph test which is 90% accurate what is the probability that:
for an applicant who did not lie his test will confirm this
for an applicant who did lie, the test will confirm this?
for an applicant who did lie he will pass the polygraph
for an applicant who is truthful will pass the polygraph
for an applicant who failed the polygraph,
lied on the application for an applicant who passed the polygraph,
was fully truthful on his application
How do I set up a 2-way contingency table or a tree diagram for these?

i am seeking to solve a query regarding contingency table as a ways as I know contingency desk display matter no longer densities,and i am having difficult time comprehending this simple desk. My tries turned into basically calculating the marginal distribution however the possibilities failed to sum to one. for example I tried solving the first question via:
$$P(A,B)=\int <!--\; \int \; -->(P(A,B,C)dC$$
but I'm missing something is dC probabilities or just count?

Can a contingent argument be proved invalid with natural deduction?
I know that when it is a non-contingent contradiction it can be refuted, but for example in this reasoning:
P⊢
P∧Q
I don't know how to refute it.
EDIT: What I mean is how someone can figure out that for example the fallacy of Affirming the consequent P→Q,Q⊢P is invalid reasoning without making truth tables. A mathematician could use a contingency in his reasoning inadvertently, so how could he be aware of his error?

Problem:
In a fictional stats class, 40% of students are female, and the rest are male. Of the female students, 30% are less than 20 years old and 90% are less than 30 years old. Of the male students, half are less than 20 years old and 70% are less than 30 years old.
(a) Make a contingency table to describe these two variables
(b) Find the probability that a randomly selected studet is 30 years or older
(c) If a student is 20 years or older, what is the probability that the student is female?
(d) If a student is less than 30 years old, what is the probability that the student is 20 years or older?
My Thoughts:
(b) P(<30 years) = 1 - 0.78 = 0.22
(c) What I first did was find P(S2 given 'not A1'), but the answer doesn't make sense because the denominator ended up being smaller than the nominator.
(d) Do I solve this problem by doing 'not 20 years'?

Using percentages to apply Fisher's exact test
I have a $2\times <!--\; \times \; -->2$ contingency table, but the total sample size is too large to be able to directly apply Fisher's exact test (as it involves factorials, so I'll obtain NaN or infinity). Data are like that:
A=1 A=0
B=1 10000 6900
B=0 89333 120033
I know I could use chi.square instead, but I wanted to provide Fisher's exact test results. Among other reasons, because calculating left and right p-values of Fisher's exact test I can have the probabilities of positive or negative associations between variables.
Could it be appropiate and acceptable to firstly transform the contingency table in percentajes, so sample size=100, then apply Fisher's exact test?

Is responses in statistics the equivalent to random variables in probability?
The focus of this class is multivariate analysis of discrete data. The modern statistical inference has many approaches/models for discrete data. We will learn the basic principles of statistical methods and discuss issues relevant for the analysis of Poisson counts of some discrete distribution, cross-classified table of counts, (i.e., contingency tables), binary responses such as success/failure records, questionnaire items, judge's ratings, etc. Our goal is to build a sound foundation that will then allow you to more easily explore and learn many other relevant methods that are being used to analyze real life data. This will be done roughly at the introductory level of the first part of the required textbook by A. Agresti (2013), which covers a superset of A. Agresti (2007)
in which, is responses here (statistics) the equivalent to random variables in probability
another page in that site says
Discretely measured responses can be:
Nominal (unordered) variables, e.g., gender, ethnic background, religious or political affiliation
Ordinal (ordered) variables, e.g., grade levels, income levels, school grades
Discrete interval variables with only a few values, e.g., number of times married
Continuous variables grouped into small number of categories, e.g., income grouped into subsets, blood pressure levels (normal, high-normal etc)
We we learn and evaluate mostly parametric models for these responses.
are variables and responses interchangeable here?

Is my answer correct? Are these two events independent?
$\begin{array}{|ccc|}\hline & A& B\\ C& 78& 520\\ D& 156& 56\\ \hline\end{array}$
This is a contingency table and the question is if D is independent of A.
Now I know that if they are, then $P(A\cap <!--\; \cap \; -->D)=P(A)\cdot <!--\; \cdot \; -->P(D)$
So in my case, $P(A\cap <!--\; \cap \; -->D)={\displaystyle \frac{156}{810}}$
$P(A)={\displaystyle \frac{234}{810}}$
$P(D)={\displaystyle \frac{212}{810}}$
$P(A\cap <!--\; \cap \; -->D)=0.19$
$P(A)\cdot <!--\; \cdot \; -->P(D)=0.07$

Why is the expected frequency during a chi square dependence test calculated the way that it is?
I understand the chi square test for testing whether or not a certain model is appropriate. I understand the process based upon which we pick the expected values. But, when it comes to the dependence test (the one where we use a contingency table), I don't understand why the expected frequency is calculated from the observed frequencies in the contingency table using (row total x column total)/grand total.
Someone please explain.

the use of a fact table, is the sentence below a tautology, contradiction or contingency?
$(P\Rightarrow <!--\; \Rightarrow \; -->Q)\iff <!--\; \iff \; -->(⌝<!--\; ⌝\; -->P\vee <!--\; \vee \; -->Q)$
Also, I am not sure what the double headed arrow is supposed to mean. I know a single headed arrow means "implies" but I am not sure about the double headed one.

For Chi-Squared test on contingency tables there is a proof to get from: $\sum <!--\; \sum \; -->\frac{(O_i-<!--\; -\; -->E_i{)}^2}{E_i}$ which equals $\frac{N(ad-<!--\; -\; -->bc{)}^2}{(a+b)(c+d)(a+c)(b+d)}$Can anyone explain the steps in the proof i know how to get from one to other but not sure why certain steps happen!
Below ill put the proof if anyone wants to see it or can explain it?

this is probably going to be a very dumb question, so my apologies in advance for that.
besides, i'm seeking to decide if there may be a large difference inside the quantity of testimonies that mention sweet, that point out fruit, or that mention neither candy nor fruit, among (a) the instances and the put up, (b) the times and the herald, and, (c) the publish and the usher in.
I ran this Chi-Square test with all three newspapers (from socscistatistics.com):
Chi-Square contingency table
However, is that a correct approach to make these three separate conclusions?
Or do I need to - instead - run three separate Chi-Square tests: (1) the Times and the Post (2) the Times and the Herald (3) the Post and the Herald

If ${\mathrm{\chi <!--\; \chi \; -->}}^2$=0 for a dataset, are the frequencies of the values in the contingency table all the same?
Could I say that if the ${\mathrm{\chi <!--\; \chi \; -->}}^2$ value of a dataset is 0, then the frequencies of the values of the cells in the contingency table are all the same? I have noticed that if I change the frequency of any of these values to be more than the other, that ${\mathrm{\chi <!--\; \chi \; -->}}^2$ stops being 0.
$$\begin{array}{lcc}& 1& 2\\ 1& 8& 8\\ 2& 8& 8\end{array}$$
${\mathrm{\chi <!--\; \chi \; -->}}^2=0$