P-Value Uncovered: Discover the Secrets with Plainmath's Expert Help

Let $X\sim <!--\; \sim \; -->Binomial(n,p)$
$H+0:p\ge <!--\; \ge \; -->.5$
$H_1:p<.5$
How do we evaluate the p-value?
ie. $P(X\ge <!--\; \ge \; -->x\mid <!--\; \mid \; -->p>=.5)$
I have seen settings where this is just interpreted as $P(X\ge <!--\; \ge \; -->x\mid <!--\; \mid \; -->p=.5)$, but I'm not sure how this is the same thing.

In the hypotehses testing, I would like to see an intuition how the operator $min$ and the constant $2$ in the computation of $p$-value of a two side alternative $p=2min(F(T_0),1-<!--\; -\; -->F(T_0))$ can be derived and understood for a statistics $T_0$. Here $F$ is the c.d.f.

The data generating process is
$f(x\mid <!--\; \mid \; -->\mathrm{\theta <!--\; \theta \; -->})=\frac{1}{\mathrm{\theta <!--\; \theta \; -->}},0\le <!--\; \le \; -->x\le <!--\; \le \; -->\mathrm{\theta <!--\; \theta \; -->}$
Let $H_1:\mathrm{\theta <!--\; \theta \; -->}=5$ and $H_2:\mathrm{\theta <!--\; \theta \; -->}\ne <!--\; \ne \; -->5$. Prior probabilities of $\frac{1}{2}$ are assigned to the two hypotheses, and a sample of size $5$ yields a maximum value of $4$. Find the $p$-value here.
According to the definition of $p$-value,
$p=\mathbb{P}(LR\le <!--\; \le \; -->L{R}^{\ast <!--\; \ast \; -->}\mid <!--\; \mid \; -->H_1),$
and $L{R}^{\ast <!--\; \ast \; -->}={\left(\frac{4}{5}\right)}^5$. I am not sure what to do next. Furthermore, intuitively, if the maxima is equal to $5$ exactly, shall we get an extremely big $p$ or a small one?

I am trying to create an algorithm that could calculate the p-value given the chi-square statistic and the degrees of freedom. Can anyone please point me in the right direction on how I could go about to evaluate the formula and what prerequisites I need to learn before I could do it.

A representative of a potential supplier claims that at least 90% of the equipment manufactured and sold by the company has no defects. You have obtained a sample of equipment usage records from a firm that uses these machines. Of 121 machines in the records, 105 machines have no recorded defects. What is the probability of this event if the supplier’s claim is correct (i.e. The p -value). On the basis of this data, do you believe the representative’s claim?
My thoughts:
The significance level of the test: α = 0.1 (10%)
The percentage of 121 machines that have no recorded defects is 86.7%.
I am not sure how to arrive at the p-value from this data.

How do I calculate the $p$ value of the following?
Students' height is approximately normal with s.d = $4$ inches, sample = $10$,mean height = $68$ inches.
Calculate the $p$ value corresponding to the following null hypotheses.
$H_0$ = Avg. height is $70$ inches
$H_1$ = Avg. height is not $70$ inches

A poll from a previous year showed that $10\mathrm{\%<!--\; \%\; -->}$ of smartphone owners relied on their data plan as their primary form of internet access. Researchers were curious if that had changed, so they tested $H_0:p=10\mathrm{\%<!--\; \%\; -->}$ versus $H_a:p\ne <!--\; \ne \; -->10\mathrm{\%<!--\; \%\; -->}$ where p is the proportion of smartphone owners who rely on their data plan as their primary form of internet access. They surveyed a random sample of $500$ smartphone owners and found that $13\mathrm{\%<!--\; \%\; -->}$ of them relied on their data plan.
The test statistic for these results was $z\approx <!--\; \approx \; -->2.236$, and the corresponding $P$-value was approximately $0.025$.
Assuming the conditions for inference were met, which of these is an appropriate conclusion?
a) At the $\mathrm{\alpha <!--\; \alpha \; -->}=0.01$ significance level, they should conclude that the proportion has changed from $10\mathrm{\%<!--\; \%\; -->}$.
b) At the $\mathrm{\alpha <!--\; \alpha \; -->}=0.01$ significance level, they should conclude that the proportion is still $10\mathrm{\%<!--\; \%\; -->}$.
c) At the $\mathrm{\alpha <!--\; \alpha \; -->}=0.05$ significance level, they should conclude that the proportion has changed from $10\mathrm{\%<!--\; \%\; -->}$.
d) At the $\mathrm{\alpha <!--\; \alpha \; -->}=0.05$ significance level, they should conclude that the proportion is still $10\mathrm{\%<!--\; \%\; -->}$.
The correct answer is $c$ but why could it not have been $b$? Why is it $c$?

Intend to calculate the first and third quartiles of a lognormal distribution with mu and sigma (two lognormal parameters) equal to -0.33217492 and 0.6065058. The expected value and the standard deviation gave me 0.8622153003153145 and 0.6191622375133721.
I intend to calculate the first and third quartiles and found this formula:
$F^{-<!--\; -\; -->1}(p)=exp(\mathrm{\mu <!--\; \mu \; -->}+\mathrm{\sigma <!--\; \sigma \; -->}{\mathrm{\Phi <!--\; \Phi \; -->}}^{-<!--\; -\; -->1}(p))\text{ with }0<p<1$
to calculate the quartiles but I didn't quite understand how to apply it and how to calculate the p-value. How can I find out the first and third quartiles of this distribution?

Calculation of $p$-value: Fabric water properties, $H_0:\mathrm{\mu <!--\; \mu \; -->}\le <!--\; \le \; -->55\mathrm{\%<!--\; \%\; -->}$ versus $H_A:\mathrm{\mu <!--\; \mu \; -->}>55\mathrm{\%<!--\; \%\; -->}$;
$n=15,\stackrel{¯<!--\; ¯\; -->}{x}=59.81\mathrm{\%<!--\; \%\; -->},s=4.94\mathrm{\%<!--\; \%\; -->}$
I understand that the hypothesized mean is $55$ samples mean is $59.81$ sample standard deviation is $4.94$ sample size is $15$. And I use the $t$-test, I have gotten $3.77106$. In this circumstance, how can I get the $p$-value, and what is different between critical value and $p$-value?

The summary of that regression shows that one of my variable has a big p-value (0.705). Should I include that variable when writing the the y hat equation?

Suppose that for every α∈(0,1) we have a size α test with rejection region Rα. Then,
$\text{p-value}=inf\{\mathrm{\alpha <!--\; \alpha \; -->}:T(X^n)\in <!--\; \in \; -->R_{\mathrm{\alpha <!--\; \alpha \; -->}}\}$
That is, the p-value is the smallest level at which we can reject $H_0$.
what is this exponent $n$. What does $X^n$ mean?

$X_i\sim <!--\; \sim \; -->Bern(p)$, $Y=\underset{i=1}{\overset{N}{\sum <!--\; \sum \; -->}}{X}_i$ and $H_0:p\ge <!--\; \ge \; -->0.5$ vs $H_1:p<0.5$ with $\mathrm{\alpha <!--\; \alpha \; -->}=0.05$.
I have to find the $p$-value at $n=30$ and $y=5$.
I'm not sure if this correct, but I figured the $p$-value was simply:
$P(Y\le <!--\; \le \; -->5)=\underset{i=1}{\overset{5}{\sum <!--\; \sum \; -->}}(\genfrac{}{}{0ex}{}{30}{y})(0.5{)}^y(0.5{)}^{30-<!--\; -\; -->y}$
which works out to be $0.00016$.
If this is correct, then why have I been given $\mathrm{\alpha <!--\; \alpha \; -->}$?

Suppose a coin with probability $p$ of heads is repeatedly flipped until $10$ heads appear. Call this number $X$. Find the p-value of $H_0:p=0.5$ versus the alternate $H_a:p\ne 0.5$ using the test statistic $|X-20|$ if $X=27$.

A data between rating of the restaurant and the approximate price of food in the restaurant. After doing Hypothesis testing. The pearson correlation coefficient was found to be 0.33 whereas p value was calculated much lesser than .05. Which means we have to accept our Alternate hypothesis as the finding is statistically significant. Our Alternate Hypothesis was that there is some relation between the two variables. Which is contrary to what we got as the coefficient i.e 0.33 which says there is no relation between the two.

From a collection of objects numbered $\{1,2,....,K\}$ objects are picked and replaced. We want to test $H_0:K=100000$ against $H_1<100000$, with the highest ranking number $M$ of our sample as test statistic. We find for our realisation for $M$ the value $81115$.
What is the $P$ value?
The correct answer is: $0.015$
I know that the definition of the $p$-value is:
The p-value is the probability of getting the observed value of the test static or a value with even greater evidence against $H_0$, if the hypothesis is actually true
or in formula form $P(T\ge <!--\; \ge \; -->t)$
I have the following questions:
What are $T$ and $t$?
I think that distribution is uniform, but how do I calculate the $p$-value?

$H_0\mathrm{\beta <!--\; \beta \; -->}=0,H_a\mathrm{\beta <!--\; \beta \; -->}\ne <!--\; \ne \; -->0$.
The test itself is a generalized likelihood ratio. The test statistic is the ratio LR:
$LR=\frac{L(\mathrm{\beta <!--\; \beta \; -->}=0)}{argma{x}_{\mathrm{\beta <!--\; \beta \; -->}\in <!--\; \in \; -->R}L(\mathrm{\beta <!--\; \beta \; -->})}$
Where $L$ is the likelihood function.Then $-<!--\; -\; -->2ln(LR)$ follows a ${\mathrm{\chi <!--\; \chi \; -->}}_1^2$.
I am trying to calculate the p-value for a specific value of LR. Say I look for at a ${\mathrm{\chi <!--\; \chi \; -->}}_1^2$ table (or online calc) and find out that
$P({\mathrm{\chi <!--\; \chi \; -->}}_1^2\ge <!--\; \ge \; -->-<!--\; -\; -->2ln(LR))=\mathrm{\alpha <!--\; \alpha \; -->}$
Is $\mathrm{\alpha <!--\; \alpha \; -->}$ my final $p$-value?
Or do I need to account for the fact that the test is two-sided and set the $p$-value $=2\ast <!--\; \ast \; -->\mathrm{\alpha <!--\; \alpha \; -->}$?