Improve Your Understanding of College Level Statistics Problems

Suppose that the sample mean for a random sample of 110 cars is 28.3 miles and assume the standard deviation is 3.2 miles. Is 26.9 against the alternative hypothesis that it is not 26.9? What is the test statistic?

A chi square test is conducted to check whether a person's ability in Mathematics has an impact on his/her interest in Statistic.
The test statistic is 13.277 under the tested null hypothesis. write a recommended null hypothesis and an alternative hypothesis. Briefly describe your conclusion on this test at the 0.01 significance level.

Say that given an estimator $\stackrel{^<!--\; ^\; -->}{T}$, of a statistic $T$, we have that $\sqrt{n}(\stackrel{^<!--\; ^\; -->}{T}-<!--\; -\; -->T)\stackrel{\text{D}}{\to <!--\; \to \; -->}\mathcal{N}(0,Var)$, where $n$ is the sample size. Consider now $-<!--\; -\; -->\stackrel{^<!--\; ^\; -->}{T}$, does the asymptotic result hold also in this case? If yes/no, under which conditions?

Doomsday Prediction
I have a calculus problem I can't seem to figure out. Any help would be appreciated!
Doomsday prediction. In 1960, three electrical engineers at the University of Illinois published a paper in Science titled "Doomsday." Based on world population growth data from 1000 AD to 1960 AD, the engineers found that population growth was faster than proportional to the population size. Using the data, they modeled the growth of the population as
$dP/dt=0.4873P^2$
where P is the population size in billions and t is centuries after 1000 AD.
1) Solve this differential equation.

True and False?
Let $A$ and $B$ be nonempty sets and $1-<!--\; -\; -->1$ be a $1-<!--\; -\; -->1$ function. Then $f(X\cap <!--\; \cap \; -->Y)=f(X)\cap <!--\; \cap \; -->f(Y)$ for all nonempty subsets $X$ and $Y$ of $A$

The difference between a cluster sample and a stratified sample is:
a) cluster samples use oversampling; stratified random samples are undersampling
b) there is no difference between them
c) cluster samples study all possible clusters; stratified random samples randomly select strata
d) cluster samples use randomly selected clusters; stratified random samples use predetermined strata

The exercise statement (roughly): Assume there is a terrorist prevention system that has a 99% chance of correctly identifying a future terrorist and 99.9% chance of correctly identifying someone that is not a future terrorist. If there are 1000 future terrorists among the 300 million people population, and one individual is chosen randomly from the population, then processed by the system and deemed a terrorist. What is the chance that the individual is a future terrorist?
Attempted exercise solution:
I use the following event labels:
A -> The person is a future terrorist
B -> The person is identified as a terrorist
Then, some other data:
$$P(A)=\frac{{10}^3}{3\cdot <!--\; \cdot \; -->{10}^8}=\frac{1}{3\cdot <!--\; \cdot \; -->{10}^5}$$
$$P(\stackrel{¯<!--\; ¯\; -->}{A})=1-<!--\; -\; -->P(A)$$
$$P(B\mid <!--\; \mid \; -->A)=0.99$$
$$P(\stackrel{¯<!--\; ¯\; -->}{B}\mid <!--\; \mid \; -->A)=1-<!--\; -\; -->P(B\mid <!--\; \mid \; -->A)$$
$$P(\stackrel{¯<!--\; ¯\; -->}{B}\mid <!--\; \mid \; -->\stackrel{¯<!--\; ¯\; -->}{A})=0.999$$
$$P(B\mid <!--\; \mid \; -->\stackrel{¯<!--\; ¯\; -->}{A})=1-<!--\; -\; -->P(\stackrel{¯<!--\; ¯\; -->}{B}\mid <!--\; \mid \; -->\stackrel{¯<!--\; ¯\; -->}{A})$$
What I need to find is the chance that someone identified as a terrorist, is actually a terrorist. I express that through P(A | B) and use Bayes Theorem to find its value.
$$P(A\mid <!--\; \mid \; -->B)=\frac{P(A\cap <!--\; \cap \; -->B)}{P(B)}=\frac{P(B\mid <!--\; \mid \; -->A)\cdot <!--\; \cdot \; -->P(A)}{P(B\mid <!--\; \mid \; -->A)\cdot <!--\; \cdot \; -->P(A)+P(B\mid <!--\; \mid \; -->\stackrel{¯<!--\; ¯\; -->}{A})\cdot <!--\; \cdot \; -->P(\stackrel{¯<!--\; ¯\; -->}{A})}$$
The answer I get after plugging-in all the values is: $3.29\cdot <!--\; \cdot \; -->{10}^{-<!--\; -\; -->3}$, the book's answer is $3.29\cdot <!--\; \cdot \; -->{10}^{-<!--\; -\; -->4}$.
Can someone help me identify what I'm doing wrong? Also, in either case, I find that it is very unintuitive that the probability of success is so small. If someone could explain it to me in more intuitive terms I'd be very grateful.

Suppose X and Y are two independent random variables such that $$E{X}^4=2,E{Y}^2=1,E{X}^2=1$$, and EY = 0. Compute $$Var(X^{2}Y)$$

Normalizing sample data to unit variance
Suppose you have data points $X_1,...,X_n$ drawn from some distribution P (regard them as iid random variables) and let the distribution be "well-behaved" so that the mean is $\mathrm{\mu <!--\; \mu \; -->}$ and variance is ${\mathrm{\sigma <!--\; \sigma \; -->}}^2$. Now suppose that we want to normalize our data, i.e., find a mapping $X_i\mapsto <!--\; \mapsto \; -->Y_i$ such that $Y_i$ are iid drawn from a distribution with mean $=0$ and variance 1. Suppose that we don't know the true values $\mathrm{\mu <!--\; \mu \; -->},{\mathrm{\sigma <!--\; \sigma \; -->}}^2$, then it's not too hard to make the mean zero by using the map
$$X_i\mapsto <!--\; \mapsto \; -->Y_i\equiv <!--\; \equiv \; -->X_i-<!--\; -\; -->\stackrel{^<!--\; ^\; -->}{\mathrm{\mu <!--\; \mu \; -->}},\stackrel{^<!--\; ^\; -->}{\mathrm{\mu <!--\; \mu \; -->}}\equiv <!--\; \equiv \; -->\frac{1}{n}\underset{1}{\overset{n}{\sum <!--\; \sum \; -->}}{X}_i$$
Indeed, $\stackrel{^<!--\; ^\; -->}{\mathrm{\mu <!--\; \mu \; -->}}$ is the sample mean and it's quite easy to check that
$$\mathbb{E}{Y}_i=0$$
Now if I would want to do the same for the variance, the natural method would be something like
$$X_i\mapsto <!--\; \mapsto \; -->Y_i\equiv <!--\; \equiv \; -->\frac{X_i-<!--\; -\; -->\stackrel{^<!--\; ^\; -->}{\mathrm{\mu <!--\; \mu \; -->}}}{\stackrel{^<!--\; ^\; -->}{\mathrm{\sigma <!--\; \sigma \; -->}}},{\stackrel{^<!--\; ^\; -->}{\mathrm{\sigma <!--\; \sigma \; -->}}}^2\equiv <!--\; \equiv \; -->\frac{1}{n-<!--\; -\; -->1}\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}(X_i-<!--\; -\; -->\stackrel{^<!--\; ^\; -->}{\mathrm{\mu <!--\; \mu \; -->}}{)}^2$$
Now we know by the convergence in distribution, that when $n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}$, we see that $Y_i\to <!--\; \to \; -->(X_i-<!--\; -\; -->\mathrm{\mu <!--\; \mu \; -->})/\mathrm{\sigma <!--\; \sigma \; -->}$ in distribution, but the variance doesn't seem to be normalized for arbitrary n.
Question. Is is actually possible to normalize the variance of a sample data set without knowing the true mean and variance?

check the null hypothesis that the mean is extra than or identical to 50. Use a one tail and a importance level of .05.
A sample of 25 answers has an average=forty five and a variance=25.

Let ${\mathrm{\theta <!--\; \theta \; -->}}^{\prime }$, $\mathrm{\theta <!--\; \theta \; -->}\in <!--\; \in \; -->\mathrm{\Theta <!--\; \Theta \; -->}$ such that ${\mathrm{\theta <!--\; \theta \; -->}}^{\prime }\ne <!--\; \ne \; -->\mathrm{\theta <!--\; \theta \; -->}$. I want to prove that $T$ is a sufficient statistic if and only if
$$\frac{f(x,{\mathrm{\theta <!--\; \theta \; -->}}^{\prime })}{f(x,\mathrm{\theta <!--\; \theta \; -->})}$$
is a function dependent only on $T(x)$.

If the sample mean = 16, population mean = 8, sample standard deviation = 4, and sample size = 150, what is the one sided t-score of the sample?

A random sample of 10 motorists buying petrol are found to spend an average of £58.30 with estimated standard error £5.25. Calculate a 95% confidence interval for the expected spending of motorists at this petrol station.
I got 10 of these questions so if someone can help me do this one i think i can do the rest.

How do you find the exact values of $\tan <!--\; \; -->{15}^{\circ <!--\; \circ \; -->}$ using the half angle formula?

Showing that $\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{X}_i^2$ is a sufficient statistic for $\mathrm{\theta <!--\; \theta \; -->}$ from an $N(0,\mathrm{\theta <!--\; \theta \; -->})$ population

How does the risk of having an identical twin with schizophrenia compare to the risk of having a fraternal twin with schizophrenia?
A.The risk is the same.
B. The risk is twice as great
C. The risk is twice as great.
D The risk is four times as great

A sample of 50 values has a mean of 8.33 while the population standard deviation is 0.67. According to the Central Limit Theorem, the distribution of sample means is what?

For a relative risk of 1.53, what is the percent increase in risk? For a percent increase in risk of 140%, what is the relative risk?

Why is an “unbiased estimator” called as such?

Classify the following random variables as continuous or discrete: the number of hairs on a cat.