Q12E Suppose that in the experiment d... [FREE SOLUTION]

91影视

Probability And Statistics For Engineering And Sciences

Jay L. Devore

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 714 pages

ISBN: 9781305251809

Chapter 11: Q12E (page 450)

Suppose that in the experiment described in Exercise 6 the five houses had actually been selected at random from among those of a certain age and size, so that factor B is random rather than fixed. Test ${H_0}:\sigma _B^2 = 0$ versus ${H_a}:\sigma _B^2 > 0$using a level .01 test.

Short Answer

Expert verified

The five houses had actually been selected at random from among those of a certain age and size, so that factor B is random rather than fixed. So, reject null hypothesis.

Step by step solution

Derive the equation using mean and variance

Let

And where ${E_{ij}}$are independent normally distributed random variable with mean 0 and variance${\sigma ^2}$. The hypothesis of interest is

${H_{0B}}:\sigma _B^2 = 0\quad {\rm{\;versus\;}}{H_{aB}}:\sigma _B^2 > 0$

With given information

$\begin{aligned}{*{20}{c}}{SSA = 11.7}\\{SSB = 113.5}\\{SSE = 25.6}\end{aligned}$

It is easier to compute corresponding mean squares are the F statistic value.

Fundamental Identity:

$SST = SSA + SSB + SSE$

By the fundamental identity, the $SSE$can be computed as$SST = SSA + SSB + SSE = 11.7 + 113.5 + 25.6 = 150.8$

Use hypotheses and test the static values

When testing hypotheses${H_{0B}}{\rm{\;versus\;}}{H_{aB}}$, the test statistic value is

${f_B} = \frac{{MSA}}{{MSE}}$

And the P-value is the area under the ${F_{I - 1,(I - 1)(J - 1)}}$curve to the right of the test statistic value ${f_A}$

The degrees of freedom are

$\begin{aligned}{*{20}{c}}{d{f_T} = IJ - 1 = 3 \cdot 5 - 1 = 14}\\{d{f_A} = I - 1 = 3 - 1 = 2}\\{d{f_B} = J - 1 = 5 - 1 = 4}\\{d{f_E} = (I - 1)(J - 1) = (3 - 1) \cdot (5 - 1) = 8.}\end{aligned}$

The mean squares are

$\begin{aligned}{*{20}{c}}{MSB = \frac{1}{{J - 1}} \cdot SSB = \frac{1}{4} \cdot 113.5 = 28.38}\\{MSE = \frac{1}{{(I - 1)(J - 1)}} \cdot SSE = \frac{1}{8} \cdot 25.6 = 3.2.}\end{aligned}$

${f_B} = \frac{{MSB}}{{MSE}} = \frac{{28.38}}{{3.2}} = 8.87$

There are two ways to make conclusion. Using the table in the appendix or compute P value using software.

Use hypotheses and test the given values

The${P_B}$value when testing hypotheses${H_{0B}}{\rm{\;versus\;}}{H_{aB}}$is

${P_A} = P\left( {F > {f_B}} \right) = P(F > 8.87) = 0.005$

${P_B} = 0.005 < 0.01 = \alpha $

Reject hypothesis ${H_{0B}}$

At given significance level. There is sufficient evidence to make conclusion that there is significance difference in the variance of three assessors.

Using the fact that

${F_{\alpha ,J - 1,(I - 1)(J - 1)}} = {F_{0.01,4,8}} = 7.01$

Which was obtained from the table in the appendix, and the fact that

${F_{0.01,4,8}} = 7.01 < 8.87 = {f_B}$

Reject hypothesis ${H_{0B}}$

At given significance level.