How to know when to use which test statistic? I'm studying for my stats exam and I seem to always get stuck on which test statistic to use. This is even after many examples that I have done.
For example: 
With this question:

I attempted to do this:
$\frac {0.2 - 0.1}{\sqrt{(0.2 \times 0.8) \div 200}}$
My thought process was to use: . When I should have used  and converted µ = np and $σ^2$ = np(1 − p).
Or for example, in this question:

I attempted to use the confidence interval and state that 0 is not in the confidence interval thus there is a difference in the rates. But the solutions indicate to use hypothesis testing.
I seem to understand how to use all these formulas when I'm told what formula to use, but this won't do me good on the final. How can I know what to use and when?
Thank you for any help!
 A: For the first question about traveling salesmen:
It is given that a firm is checking to see if a report is correct or incorrect.  In other words, they are conducting an experiment in which the results are going to be either "success" or "failure".  This should set off a flag in your mind that tells you the results are going to be of binomial distribution.
In the problem, they are conducting $n=200$ trials of the experiment with probability $p=.1$ of success (here, success is travel expenses are incorrectly documented).  If you let $X=$ the number of incorrectly documented expense reports, then $X\sim Bin(n=200,p=.1).$  Now, assuming you know about the normal approximation to the binomial distribution, then the Central Limit Theorem says for large $n$, typically $>20$ when $p$ is sufficiently far away from $0$ or $1$, one can use this approximation.  This example works for the approximation.  Therefore, we have that $X\sim N(\mu=np,\sigma^2=np(1-p))$ which yields $X\sim N(\mu=20, \sigma^2=18)$.  The question then asks you to find $$\Pr(X>40)=1-\Pr(X\le 40)=1-\Pr\bigg(\frac{X-\mu}{\sigma}\le\frac{40-\mu}{\sigma}\bigg)=1-\Pr\bigg(Z\le\frac{40-20}{\sqrt{18}}\bigg)=1-\Pr\bigg(Z\le\frac{20}{\sqrt{18}}\bigg)=1-\Pr(Z\le 4.71)$$
Using normal distribution table for standard normal r.v. we find that this is equal to $0$ (it is a very small number, although nonzero).
For the second question:
We want to test to see if there is a difference between the means (you were correct in that regard).  Usually, it will stated in the question if you are to construct a confidence interval.  This question is asking you to check to see if there is a difference between the means of the given data at a level of significance.  This is your hint on what to do with the problem.  You need to find determine your null hypothesis and alternative hypothesis.  Since we are testing to see if there is only a difference, i.e, not testing to see if one is larger than the other, then this is a two tailed test.  Then we have the following hypothesis:
$$\begin{cases}H_0:& \mu_1=\mu_2\\H_1:& \mu_1\ne\mu_2\end{cases}$$
Let $\bar X_1=$ the average nightly rate of Marriot and $\bar X_2=$ the average nightly rate of Radisson.  Then we have $\bar X_1=170, \bar X_2=140, n_1=n_2=50, s_1=15, s_2=10$.  I assume that they took $50$ billing statements from each hotel.  Since the population variances are unknown and we are using sample variances, we need to use t-test.  Moreover, since each $n$ is large enough, we don't have to pool the variances.  Now, we can proceed with the t-test.  Therefore, we have
$$t_0=\frac{\bar X_1-\bar X_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}=\frac{170-145}{\sqrt{\frac{15^2}{50}+\frac{10^2}{50}}}=\frac{25}{\sqrt{\frac{325}{50}}}=9.81$$
For $\alpha=.01$ with $99$ degrees of freedom, we have that $t=2.326$.  Since $t_0>t$, this means that we reject the null hypothesis, $H_0$.  That is, there is a difference between the means of the two hotels at the $\alpha=.01$ level of significance.
