Question
Verify that, under the assumptions of the single factor ANOVA model, the random variable$$\hat{\theta}=\sum_i c_i \bar{Y}_i$$is normally distributed with mean and variance given by Equations 12.15 and 12.16.
Step 1
We assume that the observations \( Y_{ij} \) (where \( i \) indexes the groups and \( j \) indexes the observations within each group) are independent and normally distributed with a common mean \( \mu \) and a common variance \( \sigma^2 \). Specifically, we have Show more…
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Suppose again, as in Exercise 9, that a random sample of 10 observations is taken from the normal distribution with unknown mean ̄́ and unknown variance ̃̀^2, but suppose now that the following hypotheses are to be tested at the level of significance 0.05: H0: ̃̀^2 = 4, H1: ̃̀^2 ≠ 4. Suppose that the null hypothesis H0 is to be rejected if either S_n^2 ≤ c1 or S_n^2 ≥ c2, where the constants c1 and c2 are to be chosen so that, when the hypothesis H0 is true, Pr(S_n^2 ≤ c1) = Pr(S_n^2 ≥ c2) = 0.025. Determine the values of c1 and c2. Suppose that a random sample of 10 observations X1, . . . , X10 is taken from the normal distribution with unknown mean ̄́ and unknown variance ̃̀^2, and it is desired to test the following hypotheses: H0: ̃̀^2 ≤ 4, H1: ̃̀^2 > 4. Suppose that a test of the form described in Exercise 8 is to be carried out at the level of significance ̄́0 = 0.05. If the observed value of S_n^2 is 60, should the hypothesis H0 be rejected or not?
Suppose that $Y_{1}, Y_{2}, \ldots, Y_{n}$ is a random sample from a normal distribution with mean $\mu$ and variance $\sigma^{2}$. The independence of $\sum_{i=1}^{n}\left(Y_{i}-\bar{Y}\right)^{2}$ and $Y$ can be shown as follows. Define an $n \times n$ matrix A by and notice that $\mathbf{A}^{\prime} \mathbf{A}=\mathbf{I}$, the identity matrix. Then, $$\sum_{i=1}^{n} Y_{i}^{2}=\mathbf{Y}^{\prime} \mathbf{Y}=\mathbf{Y}^{\prime} \mathbf{A}^{\prime} \mathbf{A} \mathbf{Y}$$ where $Y$ is the vector of $Y_{i}$ values. a. Show that $$\mathbf{A Y}=\left[\begin{array}{c} \bar{Y} \sqrt{n} \\ U_{1} \\ U_{2} \\ \vdots \\ U_{n-1} \end{array}\right]$$ where $U_{1}, U_{2}, \ldots, U_{\mathrm{n}-1}$ are linear functions of $Y_{1}, Y_{2}, \ldots, Y_{n} .$ Thus, $$\sum_{i=1}^{n} Y_{i}^{2}=n \bar{Y}^{2}+\sum_{i=1}^{n-1} U_{i}^{2}$$ b. Show that the linear functions $Y \sqrt{n}, U_{1}, U_{2}, \ldots, U_{n-1}$ are pairwise orthogonal and hence independent under the normality assumption. (See Exercise $5.130 .$ ) c. Show that $$\sum_{i=1}^{n}\left(Y_{i}-\bar{Y}\right)^{2}=\sum_{i=1}^{n-1} U_{i}^{2}$$ and conclude that this quantity is independent of $Y$. d. Using the results of part (c), show that $$\frac{\sum_{i=1}^{n}\left(Y_{i}-Y\right)^{2}}{\sigma^{2}}=\frac{(n-1) S^{2}}{\sigma^{2}}$$ has a $\chi^{2}$ distribution with $(n-1)$ df.
The Analysis of Variance
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Transcript
For exercise 93, part A, we have that, let's see, sigma, I equals 1 to N of Y, sub I, is equal to Y prime, A, prime, A, Y equals AY, equals AY prime times AY. This equals N y bar squared plus sigma, I equals 1 to N minus 1, U squared, sub, I, and then for part B we have sigma j equals 1 2 n a sub j b sub j equal to i times 1 over root n times 1 over root i open parentheses i plus 1 close parentheses minus i divided by root i times i plus one close parentheses times one over root n equal to zero then we get that n from j equals 1 to n have a sub j b sub j equal to i times 1 over root k times k plus 1 close parentheses times 1 over the root i times i plus 1 close parentheses minus 1 divided by root k k plus 1 close parentheses times i divided by the root i times i plus 1 close parentheses equal to 0 and then for part c we have okay sigma i equals 1 to 1 to 8000 to 8 n of y sub i minus y bar squared and this is equal to sigma i equals one to n y sub i squared plus y bar squared minus two y bar Y sub I, close parentheses, equals sigma I equals 1 to N, Y squared, sub I plus sigma, I equals 1 to n, Y bar…
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