# Tag Info

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Broad question! All I can offer is my own experiences regarding the above. How closely related are math and statistics? First of all, you need to make the distinction between applied math/stats and theoretical math/stats. I've found the applied/theoretical divide to be more substantial then the subject-level division per se. For example, take a look at a ...

3

Once you know the correct numbers it should take you at most 3 more guesses to find their positions (since any number in wrong position can be corrected in at most 3 guesses). Also you can find the four correct numbers by guessing 1234; 5678; 9012. So this algorithm is perhaps not the most efficient, but always gives the answer within 6 tries.

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A hypothesis test in the frequentist sense is a procedure by which one arrives at a decision about whether the data contains sufficient evidence to accept the alternative hypothesis. In other words, there are two choices: either you reject the null $H_0$, or the test is inconclusive. The reason why you cannot ever "accept the null" under such a test is ...

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Yes, using $\widehat p=\frac{Y}{n}$ instead of $p_0$ is valid (but your treacher's reason I think is incorrect). The reason that it is valid to use $\widehat p$ instead of $p_0$ is because the asymptotics are the same i.e. the hypothesis test is based on $$\sqrt{n}(\widehat p - p_0) \stackrel{d}{\rightarrow}N(0,p_0 (1-p_0))$$ By the continuous mapping ...

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For hypothesis testing purposes, you always use the actual null distribution, which requires no estimation. Therefore, I am suspicious of your teacher's response. However, your teacher may have been referring to how confidence intervals are formed, in which case, you do use estimates. Your case appears to look like the beginnings of the basic "textbook" CI ...

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When one is constructing a hypothesis test, you have to find a way to balance Type I error and Type II error the best. The problem is that Type I error and Type II error are usually inversely related (as in the case of this problem) which means that if you want decrease probability of Type I error in test you are going to increase Type II error. Thus the way ...

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Aren't you also given either the population or sample deviation? If any of these are given, you assume the data is distributed a certain way, e.g., normally, with $\mu=100, \sigma=\sigma_0$. Knowing this, i.e., the (assumed) parameters of the population you are sampling from, then allows you to compute the probability of obtaining the value of 103 under ...

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There are several algorithms and measures that can be used to quantify the difference/similarity between sequences. For short strings - as those reported in the question - using the Levenshtein distance could be a good choice. This is a simple measure of difference between two strings that computes the minimum number of single-character edits necessary to ...

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We use a Poisson model. If a random variable $X$ has Poisson distribution with parameter (mean) $\lambda$, then $\Pr(X=n)=e^{-\lambda}\frac{\lambda^n}{n!}$. a) The number of cyclones in a one month in-season period has Poisson distribution with parameter (mean)$\frac{12}{6}$. So the probability of exactly $5$ cyclones in a month is $e^{-2}\frac{2^5}{5!}$. ...

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statistics tell us the Z-score (i.e., $\frac{Actual\;Lifespan - Mean\;Lifespan}{SD}$) for the elephant is -0.7 while the Z-score for the lion is 2.0. we're assuming a normal distribution and a Z score of 2.0 means that it lived longer than 97.5% of the other lions while the elephant lived only ~ 40% longer than the other elephants (look up 68-95-99.7 rule) ...

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You can formulate this as a binomial test of a proportion and use either the Wilson score interval or the Clopper-Pearson interval. Specifically: let our sample space be all hostels under study and assume random sampling. Let X be a random variable that takes the value 1 if a selected hostel has at most 30 rooms and 0 otherwise. For a sample of N rooms the ...

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Below algorithm guarantee that no matter which code given, you can hack the security camera in maximum of 5 moves: It requires somewhat intelligent planning and roughly goes like this: Check your guess. If it's solution stop, otherwise continue. If there are amber (misplaced) digits, then try to find (if possible) precisely these digits' places in ...

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The constraints actually give six attempts, so while the following algorithm may fail to produce the answer in just five tries, it is still guaranteed to give it in at most six. (So far I could only find two examples of codes requiring six guesses, and I think these might be the only two.) For each digit define its status as $u,r,a,g$ meaning untested, ...

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For example, $f_{X\mid T}(t\mid t)=1/(1+\mathrm e^{-2t\mu})$ if $t\ne0$ hence $f_{X\mid T}(\ \mid t)$ does depend on $\mu$ except if $t=0$.

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OK, you found the ratio correctly. Now you need to construct the critical region in the form $\Omega = \{x:\Lambda(x) > C_{\alpha}\}$, where $C_{\alpha}$ corresponds to your significance level $\alpha$. Let's vary $C$ and look what we have. If $C=0$, then our critical region consists of all possible $x$, i.e. $\Omega_1 = [0, 1.1]$. In this case the ...

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In the example, there are three protocols. In the Bayesian viewpoint, the idea is that the number of events of interest is a binomial random variable whose probability of "success" is itself a random variable $p_i$. Specific realizations of the data may be drawn from realizations of the underlying random variables in the model, but that does not mean that ...

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In general, yes, your interpretation is correct. Bayesian models support a generative interpretation, unlike classical frequentist models (a grey area is mixed-effects modeling).

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How about this: If old_score*delta<0 then damp = 1 Otherwise If old_score < -90 then damp = 2+(old_score/50) If -90 < old_score < -10 then damp = 1.1+(old_score/100) If -10 <= old_score < 10 then damp = 1 If 10 <= old_score < 90 then damp = 1.1-(old_score/100) If 90 <= old_score then damp = 2-(old_score/50) ...

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You are performing a multinomial experiment. The wikipedia article has formulas for the distribution of the frequencies for a given (M,N). I will not reproduce them here. Specifically, you are looking at a multinomial distribution over M trials with N objects, with $p_i = \frac{1}{N}\;\; i\in \{1....N\}$. Note that when you generalize to more than one ...

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