Suppose that independent samples of sizes n1, n2, …, nk are taken from each of k normally distri

Name: Problem 6, Chapter 13: Mathematical Statistics with Applications
Uploaded: 2022-10-31T19:50:26-08:00
Channel: Victor Salazar

step 1 For exercise 6A, I'm going to head up both. I understand you both love each other a lot. But I'm

Question

Suppose that independent samples of sizes $n_{1}, n_{2}, \ldots, n_{k}$ are taken from each of $k$ normally distributed populations with means $\mu_{1}, \mu_{2}, \ldots, \mu_{k}$ and common variances, all equal to $\sigma^{2} .$ Let $Y_{i j}$ denote the $j$ th observation from population $i,$ for $j=1,2, \ldots, n_{i}$ and $i=1,2, \ldots, k,$ and let $n=n_{1}+n_{2}+\cdots+n_{k}$ a. Recall that $$\mathrm{SSE}=\sum_{i=1}^{k}\left(n_{i}-1\right) S_{i}^{2} \quad \text { where } S_{i}^{2}=\frac{1}{n_{i}-1} \sum_{j=1}^{n_{i}}\left(Y_{i j}-\bar{Y}_{i \bullet}\right)^{2}$$ Argue that $\operatorname{SSE} / \sigma^{2}$ has a $x^{2}$ distribution with $\left(n_{1}-1\right)+\left(n_{2}-1\right)+\cdots+\left(n_{k}-1\right)=n-k$ df. b. Argue that under the null hypothesis, $H_{0}: \mu_{1}=\mu_{2}=\cdots=\mu_{k}$ all the $Y_{i j}$ 's are independent, normally distributed random variables with the same mean and variance. Use Theorem 7.3 to argue further that, under the null hypothesis, $$\text { Total } \mathrm{SS}=\sum_{i=1}^{k} \sum_{j=1}^{n_{i}}\left(Y_{i j}-\bar{Y}\right)^{2}$$ c. In Section 13.3 . we argued that SST is a function of only the sample means and that SSE is a function of only the sample variances. Hence, SST and SSE are independent. Recall that Total SS = SST + SSE. Use the results of Exercise 13.5 and parts (a) and (b) to show that, under the hypothesis $H_{0}: \mu_{1}=\mu_{2}=\cdots=\mu_{k}, \mathrm{SST} / \sigma^{2}$ has a $\chi^{2}$ distribution with $k-1 \mathrm{df}$ d. Use the results of parts (a), (b), and (c) to argue that, under the hypothesis $H_{0}$ : $\mu_{1}=\mu_{2}=\cdots=\mu_{k}, F=\mathrm{MST} / \mathrm{MSE}$ has an $F$ distribution with $k-1$ and $n-k$ numerator and denominator degrees of freedom, respectively.

   Suppose that independent samples of sizes $n_{1}, n_{2}, \ldots, n_{k}$ are taken from each of $k$ normally distributed populations with means $\mu_{1}, \mu_{2}, \ldots, \mu_{k}$ and common variances, all equal to $\sigma^{2} .$ Let $Y_{i j}$ denote the $j$ th observation from population $i,$ for $j=1,2, \ldots, n_{i}$ and $i=1,2, \ldots, k,$ and let $n=n_{1}+n_{2}+\cdots+n_{k}$
a. Recall that
$$\mathrm{SSE}=\sum_{i=1}^{k}\left(n_{i}-1\right) S_{i}^{2} \quad \text { where } S_{i}^{2}=\frac{1}{n_{i}-1} \sum_{j=1}^{n_{i}}\left(Y_{i j}-\bar{Y}_{i \bullet}\right)^{2}$$
Argue that $\operatorname{SSE} / \sigma^{2}$ has a $x^{2}$ distribution with $\left(n_{1}-1\right)+\left(n_{2}-1\right)+\cdots+\left(n_{k}-1\right)=n-k$ df.
b. Argue that under the null hypothesis, $H_{0}: \mu_{1}=\mu_{2}=\cdots=\mu_{k}$ all the $Y_{i j}$ 's are independent, normally distributed random variables with the same mean and variance. Use Theorem 7.3 to argue further that, under the null hypothesis,
$$\text { Total } \mathrm{SS}=\sum_{i=1}^{k} \sum_{j=1}^{n_{i}}\left(Y_{i j}-\bar{Y}\right)^{2}$$
c. In Section 13.3 . we argued that SST is a function of only the sample means and that SSE is a function of only the sample variances. Hence, SST and SSE are independent. Recall that Total SS = SST + SSE. Use the results of Exercise 13.5 and parts (a) and (b) to show that, under the hypothesis $H_{0}: \mu_{1}=\mu_{2}=\cdots=\mu_{k}, \mathrm{SST} / \sigma^{2}$ has a $\chi^{2}$ distribution with $k-1 \mathrm{df}$
d. Use the results of parts (a), (b), and (c) to argue that, under the hypothesis $H_{0}$ :
$\mu_{1}=\mu_{2}=\cdots=\mu_{k}, F=\mathrm{MST} / \mathrm{MSE}$ has an $F$ distribution with $k-1$ and $n-k$ numerator and denominator degrees of freedom, respectively.

Mathematical Statistics with Applications

Dennis D. Wackerly,… 7th Edition

Chapter 13, Problem 6

Instant Answer

Solved by Expert Victor Salazar

10/31/2022

Step 1

Therefore, we can write $\frac{SSE}{\sigma^2}$ as the sum of $k$ chi-squared distributed variables, each with $n_i - 1$ degrees of freedom. This gives us: $$\frac{SSE}{\sigma^2} = \sum_{i=1}^{k} \frac{(n_i - 1)S_i^2}{\sigma^2}$$ Each term in the sum is a Show more…

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Suppose that independent samples of sizes $n_{1}, n_{2}, \ldots, n_{k}$ are taken from each of $k$ normally distributed populations with means $\mu_{1}, \mu_{2}, \ldots, \mu_{k}$ and common variances, all equal to $\sigma^{2} .$ Let $Y_{i j}$ denote the $j$ th observation from population $i,$ for $j=1,2, \ldots, n_{i}$ and $i=1,2, \ldots, k,$ and let $n=n_{1}+n_{2}+\cdots+n_{k}$ a. Recall that $$\mathrm{SSE}=\sum_{i=1}^{k}\left(n_{i}-1\right) S_{i}^{2} \quad \text { where } S_{i}^{2}=\frac{1}{n_{i}-1} \sum_{j=1}^{n_{i}}\left(Y_{i j}-\bar{Y}_{i \bullet}\right)^{2}$$ Argue that $\operatorname{SSE} / \sigma^{2}$ has a $x^{2}$ distribution with $\left(n_{1}-1\right)+\left(n_{2}-1\right)+\cdots+\left(n_{k}-1\right)=n-k$ df. b. Argue that under the null hypothesis, $H_{0}: \mu_{1}=\mu_{2}=\cdots=\mu_{k}$ all the $Y_{i j}$ 's are independent, normally distributed random variables with the same mean and variance. Use Theorem 7.3 to argue further that, under the null hypothesis, $$\text { Total } \mathrm{SS}=\sum_{i=1}^{k} \sum_{j=1}^{n_{i}}\left(Y_{i j}-\bar{Y}\right)^{2}$$ c. In Section 13.3 . we argued that SST is a function of only the sample means and that SSE is a function of only the sample variances. Hence, SST and SSE are independent. Recall that Total SS = SST + SSE. Use the results of Exercise 13.5 and parts (a) and (b) to show that, under the hypothesis $H_{0}: \mu_{1}=\mu_{2}=\cdots=\mu_{k}, \mathrm{SST} / \sigma^{2}$ has a $\chi^{2}$ distribution with $k-1 \mathrm{df}$ d. Use the results of parts (a), (b), and (c) to argue that, under the hypothesis $H_{0}$ : $\mu_{1}=\mu_{2}=\cdots=\mu_{k}, F=\mathrm{MST} / \mathrm{MSE}$ has an $F$ distribution with $k-1$ and $n-k$ numerator and denominator degrees of freedom, respectively.

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Key Concepts

Analysis of Variance (ANOVA)

ANOVA is a statistical method used to compare means across multiple groups or populations. It partitions the overall variability found within the data into components attributable to differences among group means and variations within groups. This technique forms the basis for testing the null hypothesis that all group means are equal, using the ratios of these variance estimates.

Sum of Squares Decomposition

In the context of ANOVA, the total variability in the data is decomposed into the sum of squares due to treatments (SST), which reflects the variability between the group means, and the sum of squares error (SSE), which captures the variability within groups. This partitioning of the total sum of squares allows one to isolate the variability explained by the differences among the group means from the inherent random variability in the data.

Chi-Squared Distribution

The chi-squared distribution arises naturally in the analysis of variance when dealing with sums of squared independent standard normal variables. In ANOVA, scaled sums of squares, such as SSE/?² and SST/?² under certain conditions, follow a chi-squared distribution. The degrees of freedom of these distributions reflect the number of independent components contributing to the sums of squares.

Independence of Orthogonal Sum of Squares

A key property in ANOVA is that the sum of squares due to treatments (SST) and the sum of squares error (SSE) are independent. This independence stems from the orthogonality of the components of variability: SST is determined solely by the group sample means while SSE is determined exclusively by within-group variations. This separation is critical for developing valid statistical tests in ANOVA.

F Distribution

The F distribution is used in ANOVA to test the null hypothesis that all group means are equal. It is obtained as the ratio of two independent chi-squared variables (SST/?² and SSE/?²) divided by their respective degrees of freedom. Specifically, the test statistic F = MST/MSE, where MST and MSE are the mean squares for treatments and error respectively, follows an F distribution with degrees of freedom corresponding to the number of groups minus one and the total number of observations minus the number of groups.

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