Recent content by Siron

I think that you're going in the right direction. However, your calculation is not entirely correct. Suppose that we have given observations $x_1,\ldots,x_n$ from the given distribution. The likelihood is then given by $$\mathcal{L}(x_1,\ldots,x_n,\alpha,\beta) = \prod_{i=1}^{n} \frac{1}{\alpha...

Actually, I think the claim is true. First, I tried some elementary examples and they appeared to be true so my intuition was that the claim should be true. Suppose that he shots $n$ times during the game, then each time he shots we can compute his succes rate. The game starts and he shots ...

I was typing an answer but then I saw Jameson's post and he basically gave the answer. Your calculation is fine! Now, I want to add a comment to your final question: can we use a binomial distribution here? The answer is no. The binomial distribution can only be used in experiments where you...

First step in hypothesis testing is to formulate the zero hypothesis $H_0$ and the alternative hypothesis. Once you have defined those I can further explain the core principle of the test and how to formulate a conclusion.

In general, the sum of $n$ independent Bernoulli trials where the succes probability is $r$, follows a binomial distribution with parameters $n$ and $r$. Hence, $X \sim \mbox{Binomial}(5,r)$. Further, note that $X$ and $R$ are both discrete random variables. The (marginal) distribution of $R$ is...

Your reasoning for question $(1)$ looks good. Generally, a $95\%$ confidence interval (thus significance level $\alpha = 5\%$) is given by (assuming normality of the data) $$\left[\overline{x} - t_{\alpha/2}\frac{s}{\sqrt{n}}, \overline{x} + t_{\alpha/2}\frac{s}{\sqrt{n}}\right].$$ Your...

I will suppose you play one time. Let $X$ be your gain after one play. Since there are $16$ different spaces you have the following distribution for $X$ $$X = \left \{ \begin{array}{lllll} 4000, \quad p = \frac{1}{16} \\ 1500, \quad p = \frac{1}{16} \\ 1000, \quad p = \frac{2}{16} \\ -50, \quad...

Your work is fine so far! However, the question states that we only consider a single observation. Therefore, you just have to consider the quotient of the likelihoods for one observation $x$: The critical region $C$ is given by $$\frac{L(\theta_0 \ | \ x)}{L(\theta_1 \ | \ x)} \geq k.$$ An easy...

In addition to I Like Serena's post, making use of Bayes rule: $$P(B \ | \ A) = \frac{P(A \ | \ B) P(B)}{P(A)} = P(A \ | \ B) > P(A) = P(B).$$ Hence, first answer is indeed correct.

Where does $p = 0.63$ comes from? I would say it should be $p=0.6$ according to the question.

Thanks for the answer! The expressions are indeed quite messy to work so its difficult to conclude if a solution is indeed real. I did some small tests with some easy examples. Apparently, adding 'Reals' or 'Complexes' does not matter to Mathematica, it displays the same solutions. So it looks...

Hello! I let Mathematica run the following command: Solve[16 x^4 - 40 a x^3 + (15 a^2 + 24 b) x^2 - 18 a b x + 3 b^2 == 0 && 5 a x - 4 x^2 - b > 0 && 15 a x - 20 x^2 - 3 b < 0 && 4 x^3 - 8 c x^2 + 5 a c x - c b < 0 && a < 0 && x < 0 && c < 0, x] It displays a solutions in function of Roots...

UPDATE: If for instance $b>1.08148a^2$ then I replace $w$ by $w'$ such that $b(v,s,w') = 1.08148a^2$ and then I extract $\epsilon>0$ to guarantee that the new value $b'(v,s,w')<1.0848a^2$ as required. On the other hand I also need to guarantee that $q(a,b)<c<0$. Since I adjusted $b$, $q(a,b)$...

Given three parameters: $$a= \frac{(k-3)^2 \sqrt{v}}{s}, \ \ b = \frac{v}{s}(w-10s), \ \ c = s \sqrt{v}.$$ which exact values I know (that is, I know $v,s,k$ and $w$ exactly). I need to guarantee that $a<0$ (this is always satisfied in my calculations!) and $$0<b<1.08148a^2$$ For instance, if...

Your calculation of the expected value (=mean) is correct. Have you already figured out how to compute $\mathbb{P}(X \leq 2)$? Hint: use the link between a probability and the area under a density function