Therefore, the Factorization Theorem tells us that \(Y = \bar{X}\) is a sufficient statistic for \(\mu\). Now, \(Y = \bar{X}^3\) is also sufficient for \(\mu\), because if we are given the value of \( …

Factorization. Let \(X_1, X_2, \ldots, X_n\) denote random variables with joint probability density function or joint probability mass function \(f(x_1, x_2, \ldots, x_n; \theta)\), which depends on …

Let \(X_1, X_2, \ldots, X_n\) denote a random sample from a normal distribution \(N(\theta_1, \theta_2\). That is, \(\theta_1\) denotes the mean \(\mu\) and \(\theta_2\) denotes the variance …

To learn how to apply the Factorization Theorem to identify a sufficient statistic. To learn how to apply the Exponential Criterion to identify a sufficient statistic. To extend the definition of …

To learn how to apply the Factorization Theorem to identify a sufficient statistic. To learn how to apply the Exponential Criterion to identify a sufficient statistic. To extend the definition of …

To learn how to apply the Factorization Theorem to identify a sufficient statistic. To learn how to apply the Exponential Criterion to identify a sufficient statistic. To extend the definition of …

24.2 - Factorization Theorem; 24.3 - Exponential Form; 24.4 - Two or More Parameters; Lesson 25: Power of a Statistical Test. 25.1 - Definition of Power; 25.2 - Power Functions; 25.3 - …

Proof See Hogg and Tanis, pages 400-401 (8th edition pages 513-14). Well, okay, so perhaps the proof isn't all that particularly enlightening, but perhaps if we take a look at a simple example, …

Therefore, the Factorization Theorem tells us that \(Y=\sum_{i=1}^{n}K(X_i)\) is a sufficient statistic for \(\theta\). Let's try the Exponential Criterion out on an example. Example 24-5 Section

Proof. A theorem we learned (way) back in Stat 414 tells us that if the two conditions stated in the theorem hold, then: \(\dfrac{(n_i-1)W^2_i}{\sigma^2}\) follows a chi-square distribution with …

Proof. A theorem we learned (way) back in Stat 414 tells us that if the two conditions stated in the theorem hold, then: \(\dfrac{(n_i-1)W^2_i}{\sigma^2}\) follows a chi-square distribution with …

Proof. In order to derive the formulas for the intercept \(a\) and slope \(b\), we need to minimize: \(Q=\sum\limits_{i=1}^n (y_i-(a+b(x_i-\bar{x})))^2\) Time to put on your calculus cap, as …

Proof. A theorem we learned (way) back in Stat 414 tells us that if the two conditions stated in the theorem hold, then: \(\dfrac{(n_i-1)W^2_i}{\sigma^2}\) follows a chi-square distribution with …

In some cases, the levels of the factors are selected at random from a larger population. In this case, the inference made on the significance of the factor can be extended to the whole …

We'll work through five theorems in all, in each case first stating the theorem and then proving it. Then, once we've added the five theorems to our probability tool box, we'll close this lesson by …

Proof. The result follows directly from the previous theorem. All we need to do is recognize that the sample mean: \(\bar{X}=\dfrac{X_1+X_2+\cdots+X_n}{n}\) is a linear combination of …

Factor Analysis is a method for modeling observed variables, and their covariance structure, in terms of a smaller number of underlying unobservable (latent) “factors.” The factors typically …

Proof. Because \(Y\) is a continuous random variable, we need to use the definition of the conditional variance of \(Y\) given \(X=x\) for continuous random variables. That is: …

Proof. We'll again follow the strategy of first finding the cumulative distribution function \(G_r(y)\) of the \(r^{th}\) order statistic, and then differentiating it with respect to \(y\) to get the probability …

Proof. We select d so that: \[P(D_n \ge d)=\alpha\] Therefore, using the definition of \(D_n\), and the probability rule of complementary events, we get: \[P(sup_x|F_n(x) -F(x)| \le d) =1-\alpha\]