Improve Your Understanding of College Level Statistics Problems

Is P value Type I error in hypothesis testing?
I'm confused about the interpretation of P value in hypothesis testing. I know that we set significance level as 0.05 which is the threshold we set for this test so that it won't suffer from Type I error by 5%.
And we are comparing P to significance level, does it mean P is the probability of making type I error based on the sample?

How do you calculate the sample range, sample mean, sample median, sample variance, and sample standard deviation of 18, 19, 34, 38, 24, 18, 22, 51, 44, 14, 29?

Suppose a new Internet company Mumble.com was to require all employees to take a drug test. Mumble.com can aord only the inexpensive drug test { the one with a 5% false positive and a 10% false negative rate. (That means that ve percent of those who are not using drugs will incorrectly test positive, and ten percent of those who are actually using drugs will test negative.) Suppose that 10% of those who work for Mumble.com are using the drugs for which Mumble.com is checking. An employee is chosen at random.
(a) (3 marks). What is the probability the employee both uses drugs and tests positive?
(b) (2 marks). What is the probability the employee does not use drugs but tests positive anyway?
(c) (2 marks). What is the probability the employee tests positive?
(d) (2 marks). If the employee has tested positive, what is the probability he or she uses drugs?

Analogy between quotient groups and quotient topology
I am trying to see "analogy" between the two concepts in the title, although I am familiar with independent definitions of them.
(1) Let G be a group, N a normal subgroup. Define an equivalence relation on G by $g_1\sim <!--\; \sim \; -->g_2$ if $g_1{g}_2^{-<!--\; -\; -->1}\in <!--\; \in \; -->N$. Let $[g_1]$ denote equivalence class of g, and G/N the set of equivalence classes. Then we have a canonical map $\mathrm{\phi <!--\; \phi \; -->}:<!--\; :\; -->G\to <!--\; \to \; -->G/N$, where domain is group, and codomain is "set" right now.
Then the quotient structure on G/N is a "group structure" such that the map φ becomes a homomorphism. Such a structure is unique. We call G/N quotient group.
Step 2
(2) Let X be a topological space, ∼ an equivalence relation on X, and [x] denote equivalence class. Let X/∼ denote the set of equivalence classes. Then there is canonical map $\mathrm{\phi <!--\; \phi \; -->}:<!--\; :\; -->X\to <!--\; \to \; -->X/\sim <!--\; \sim \; -->$, where domain is topological space and codomain is just set right now.
The quotient topology on $X/\sim <!--\; \sim \; -->$ is a topology such that the map $\mathrm{\phi <!--\; \phi \; -->}$ is continuous.
But, I was feeling that there could be more than one topological structures on X/∼ which make the canonical map continuous. For example, take nice example of quotient topology on one hand, and structure of indiscrete topology on X/∼ on the other hand.
Question: What is exact definition of quotient topology which makes it unique in some sense? (compare with quotient structure in (1))

A potted plant rests on a window frame above the ground. The potted plant is off the frame and falls from rest to the grass below 1.93 s later.
a) How fast is the pot traveling just before impact with the grass?
b) How far did the pot fall?

How do you find the exact value of $\ln <!--\; \; -->e^{4.5}$?

If 𝐴 is indefinite and 𝐵 is positive definite, then 𝐴+𝐵 is indefinite.

You are the assistant manager at Extra Credit Union. Your boss, Robin Banks, has sent you the following email:
"Assistant Manager, as you know, we go the extra mile in our service to our clients. Part of that is knowing our clients well. Thus, we are wanting to solicit phone-call feedback from some of our clients. I know we don't have time to talk to all 10,000 clients, so I am wanting a smaller representative sample. Perhaps a random sample of 25 or 50 would be good. Across all 10,000 clients, the average account balance is $12,000 and the standard deviation is $5000. I want the sample we take to closely match our clients as a whole. Specifically, I want the probability to be at least 80% that the average of our sample is within $1000 of the average for all clients. Would a random sample of 25 be fine or do we need to take a random sample of 50? Use the magical statistics skills you learned back in college and get back to me ASAP! Thanks, Robin Banks"
In your response indicate whether 25 is sufficient or whether a sample of 50 would be necessary. (You will want to calculate the probability of the sample mean being between $11,000 and $13,000 for both a sample size of 25 and 50.)

Why "bother" with a null hypothesis at all?
whenever i am seeking to get into data (once more), i am constantly lost at speculation trying out.
My simple question is - why can we form a null hypothesis as a negation of what we need to show within the first region, and most effective then can we show or disprove the null speculation?
Why do we do it at all, in preference to just proving the authentic speculation?

Explain Strengths of the Experimental Method.

True or False: If $x\notin <!--\; \notin \; -->\mathbb{Q}$ then $\underset{m\ge <!--\; \ge \; -->0}{\sum <!--\; \sum \; -->}m{x}^{m-<!--\; -\; -->1}\notin <!--\; \notin \; -->\mathbb{Q},$, where $|x|<1.$
Considered the contra-positive of the above statement:
If $\underset{m\ge <!--\; \ge \; -->0}{\sum <!--\; \sum \; -->}m{x}^{m-<!--\; -\; -->1}\in <!--\; \in \; -->\mathbb{Q}$ then $x\in <!--\; \in \; -->\mathbb{Q}.$
Now if the contra-positive is true/false then the statement is true/false.
Thus, if
$$\underset{m\ge <!--\; \ge \; -->0}{\sum <!--\; \sum \; -->}m{x}^{m-<!--\; -\; -->1}\in <!--\; \in \; -->\mathbb{Q}$$
$$⟹<!--\; ⟹\; -->1+2x+3x^2+...\in <!--\; \in \; -->\mathbb{Q}$$
$$⟹<!--\; ⟹\; -->1^{\prime }+x^{\prime }+(x^2{)}^{\prime }+(x^3{)}^{\prime }+...\in <!--\; \in \; -->\mathbb{Q}$$
$$⟹<!--\; ⟹\; -->(1+x+x^2+x^3+...{)}^{\prime }\in <!--\; \in \; -->\mathbb{Q}$$
$$⟹<!--\; ⟹\; -->(\frac{1}{1-<!--\; -\; -->x}{)}^{\prime }\in <!--\; \in \; -->\mathbb{Q}$$
$$⟹<!--\; ⟹\; -->\frac{-<!--\; -\; -->1}{(1-<!--\; -\; -->x{)}^2}\in <!--\; \in \; -->\mathbb{Q}$$
Now from the last step we can conclude that $x\in <!--\; \in \; -->\mathbb{Q}$, since $|x|<1.$ Therefore the contrapositive is true, so the given statement is also true.
Is my analysis correct?

What is the exact value of $\sec -<!--\; -\; -->330$?

How do you find the exact value of $\cos <!--\; \; -->({\tan }^{-<!--\; -\; -->1}<!--\; \; -->2)$?

Given $X_1,X_2,\dots <!--\; \dots \; -->,X_n$ are i.i.d with $f(x)=\frac{1}{\mathrm{\pi <!--\; \pi \; -->}}\frac{1}{1+x^2}$. Find the minimal sufficient statistic for $\frac{1}{{\mathrm{\pi <!--\; \pi \; -->}}^n(\underset{i=1}{\overset{n}{\prod <!--\; \prod \; -->}}1+(x_i-<!--\; -\; -->\mathrm{\theta <!--\; \theta \; -->}{)}^2)}$.
If we are given the entire sample set, we could easily obtain the order statistic of that sample, so how could $T(X)=(x_1,\dots <!--\; \dots \; -->,x_n)$ is wrong?

Confidence Interval Probabilty
1) We have discovered a new method to produce a liquid containing Zinc. We measure the concentration $\mathrm{\mu <!--\; \mu \; -->}$ four times independently (from the same solution) and find the values
0.3, 0.33, 0.3, 0.27
a) Assuming that the standard deviation of the measurement error is 0.02 (when we make one measurement) estimate μ and give a 90%-confidence interval.
Answer:
We assume the measurement errors to be normal with zero expectation. The estimate for $\mathrm{\mu <!--\; \mu \; -->}$ is
${\mu }_X=(X_1+X_2+X_3+X_4)/4=(0.3+0.33+0.3+0.27)/4=\mathrm{0.3.}$
The same procedure as usual leads to a confidence interval
$[0.3-<!--\; -\; -->b_{\text{standard deviation}}/\sqrt{n},0.3+b_{\text{standard deviation}}/\sqrt{n}]$
where the standard deviation equals 0.02 and $\sqrt{n}=2$, whilst b is defined by the equation
$P(-b\le <!--\; \le \; -->N(0,1)\le <!--\; \le \; -->b)=90\mathrm{\%<!--\; \%\; -->}.$
The last equation above is equivalent to $\mathrm{\lambda <!--\; \lambda \; -->}(b)=0.95$ which we can solve with the help of a table and find $b=1.645$. Hence the 90% confidence interval is equal to $[0.3-0.0164,0.3+0.0164]=[0.2836,0.3164]$. The true concentration is thus between 0.2836 and 0.3164 with at least 95% probability.
My question is how were they able to get 95% probability and b?

A virus has been spread around a population. The prevalence of this virus is 84%. A diagnostic test, with a specificity of 94% and sensitivity of 15%, has been introduced. If a patient is drawn randomly from the population, what is the probability that:
a) a person has the virus, given that they tested positive?
b) a person has the virus, given that they tested negative?

How do you find the exact value of $\sin <!--\; \; -->(\arccos <!--\; \; -->(-<!--\; -\; -->\frac{2}{3}))$?

Suppose that you are constructing a survey of average student loan debt. You decide that instead of sampling 50 students, you are going to sample 75 students. When you do this, the
a. standard error of the mean decreases O
b. population mean increases O
c. standard deviation of the population decreases
d. standard error of the mean increases