Mathematical Statistics with Applications
Mathematical Statistics with Applications
7th Edition
ISBN: 9780495110811
Author: Dennis Wackerly, William Mendenhall, Richard L. Scheaffer
Publisher: Cengage Learning
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Chapter 13.4, Problem 6E

Suppose that independent samples of sizes n1, n2,…, nk are taken from each of k normally distributed populations with means μ1, μ2,…,μk and common variances, all equal to σ2. Let Yij denote the jth observation from population i, for j = 1, 2, …, ni and i = 1, 2,…, k, and let n = n1 + n2 + …+ nk

  1. a. Recall that

    S S E = i = 1 k ( n i 1 ) s i 2     where  s i 2 = 1 n i 1 j = 1 n i ( Y i j Y ¯ i · ) 2

    Argue that SSE/σ2 has a χ2distribution with (n1 − 1) + (n2 − l) + … + (nk − 1) = nk df.

    1. Exercises proceed by an asterisk are optional.

  2. b. Argue that under the null hypothesis. H0:μ1 = μ2 = …= μk all the Yij’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.

    T o t a l   SS = i = 1 k j = 1 n i ( Y i j Y ¯ ) 2

    is such that (Total SS)/σ2 has a χ2 distribution with n – 1 df.

  3. 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 H01 = μ2 = …= μk SST/σ2: has χ2 distribution with k − 1 df.
  4. d. Use the results of part(a)–(c) to argue that, under the hypothesis H0: μ1 = μ2 = …= μkF = MST/MSE has an F distribution with k − 1 and nk numerator and denominator degrees of freedom, respectively.

a.

Expert Solution
Check Mark
To determine

Show that SSEσ2 has a chi-square distribution with the (nk) degrees of freedom.

Explanation of Solution

Result:

  • The function (ni1)Si2σ2 follows a chi-square distribution with (ni1).
  • If X and Y are an independent random variable and have a chi-square distribution with ‘a’ and ‘b’ the degrees of freedom; then, X+Y follows a chi-square distribution with a+b the degrees of freedom.

The formula for the sum of squares for error is as follows:

SSE = i=1k(ni1)Si2Where Si2=1(ni1)i=1k(YijY¯i.)2

Thus, the SSEσ2 can be expressed as follows:

SSEσ2=i=1k(ni1)Si2σ2=i=1k(ni1)Si2σ2=(n11)S12σ2+(n21)S22σ2+....+(nk1)Sk2σ2

That is,

SSEσ2 is a sum of k chi-square distributions. Therefore, SSEσ2 has a chi-square distribution.

The degree of freedom is calculated as follows:

Degrees of freedom=(n11)+(n21)+...+(nk1)=i=1knik=nk

Thus, SSEσ2 has a chi-square distribution with the (nk) degrees of freedom.

b.

Expert Solution
Check Mark
To determine

Prove that Total SSσ2 has a chi-square distribution with the (n1) degrees of freedom.

Explanation of Solution

The formula for the sum of squares for total is as follows:

Total SS = i=1kj=1ni(YijY¯)2

If the null hypothesis is true; then, Yij is independent and normally distributed with the same mean and same variance.

Thus, the Total SSσ2 can be expressed as follows:

Total SS σ21σ2i=1kj=1ni(YijY¯)2=i=1kj=1ni(YijY¯)2σ2

Use Theorem7.3 Total SSσ2 that has a chi-square distribution with the (n1) degrees of freedom.

c.

Expert Solution
Check Mark
To determine

Show that SSTσ2 has a chi-square distribution with the (k1) degrees of freedom.

Explanation of Solution

From Part (a), SSEσ2 has a chi-square distribution with the (nk) degrees of freedom.

From Part (b), Total SSσ2 has a chi-square distribution with the (n1) degrees of freedom.

Formula for the total sum of square is as follows:

Total SS=SST+SSETotal SSσ2=SSTσ2+SSEσ2SSTσ2=Total SSσ2SSEσ2

From Exercise 13.5, it is clear that SSTσ2 has a chi-square distribution.

The degrees of freedom associated with SSTσ2 is calculated as follows:

Degrees of freedom=(n1)(nk)=(k1)

Thus, SSTσ2 has a chi-square distribution with the (k1) degrees of freedom.

d.

Expert Solution
Check Mark
To determine

Prove that F=MSTMSE has an F distribution with the (k1) and (nk) numerator and denominator degrees of freedom.

Explanation of Solution

From Part (a), SSEσ2 has a chi-square distribution with the (nk) degrees of freedom.

From Part (c), SSTσ2 has a chi-square distribution with the (k1) degrees of freedom.

Formula for the F statistic is as follows:

F=MSTMSE=SST(k1)SSE(nk)

That is, F statistic is a ratio of two chi-square distributions divided by their respective degrees of freedom. Thus, F=MSTMSE has an F distribution with the (k1) and (nk) numerator and denominator degrees of freedom, respectively.

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Chapter 13 Solutions

Mathematical Statistics with Applications

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