Reading 5

Complete this two-way ANOVA table by matching each labeled blank, (a) through (d), to its value.

(a) matches Choice, 2 2 (b) matches Choice, 4 4 (c) matches Choice, 9 9 (d) matches Choice, 46 46

A BLANK in a factorial experiment is an independent variable. Any one of the values it can take is called a BLANK.

Blank 1: factor Blank 2: level or treatment

The error mean square is defined by MSE = ______.

SSE/N-I

The treatment mean square is defined by MSTr = ______.

SSTr/I-1

Match each quantity in a two-factor analysis of variance on the left with its description on the right.

Xijk = μ + αi + βj + γij + εijk matches Choice, Two-way ANOVA model Xijk = μ + αi + βj + εijk matches Choice, Additive model μijk = μ + αi + βj + γij matches Choice, True mean for treatment ij under full two-way ANOVA model μijk = μ + αi + βj matches Choice, True mean for treatment ij under additive model

The quantity εijk = Xijk - μij, the difference between the kth observation and the ij treatment (population) mean is called an BLANK.

error

Match each quantity in a two-factor analysis of variance on the left with its description on the right.

μ matches Choice, Population grand mean μμ.j matches Choice, jth column mean βj = μμ.j - μ matches Choice, jth column effect, the difference between the jth column mean and the population grand mean K matches Choice, Number of replicates per treatment J matches Choice, Number of columns

A one-factor experiment tests the strength of rope splices across five different types of rope by breaking three spliced ropes of each type. To test H0: μi - μj = 0 at the 5% level for all C = 5(5-1)/2 = 10 pairs of means, the table value used is qI,N-I,α = ______.

4.65

In which case is it reasonable to suppose the assumptions of one-way ANOVA are met?

A normal probability plot of residuals shows points nearly on a line and a plot of residuals versus fitted values shows reasonably uniform spreads.

Match two-way ANOVA sum of squares on the left to its degrees of freedom on the right.

Choice, I- 1 I - 1 SSB = IK∑i=1Iβˆ2j∑Ii=1β̂j2 matches Choice, J- 1 SSAB = K∑i=1I∑j=1Jγˆ2ij∑Ii=1∑Jj=1γ̂ij2 matches SSE = ∑i=1I∑j=1J∑k=1K(Xijk−Xij.)∑Ii=1∑Jj=1∑Kk=1Xijk-Xij.2 matches Choice, IJ(K- 1) SST = ∑i=1I∑j=1J∑k=1K(Xijk−X...)∑Ii=1∑Jj=1∑Kk=1Xijk-X...2 matches Choice, IJK- 1

In a two-way ANOVA, which null hypothesis must be evaluated before it makes sense to test the others?

H0: γ11 = ... = γIJ = 0

A canoe maker weaves six cane seats, two each using three types of cane, to find out which type is most durable. He labels the seats from cane type A as A1 and A2, those from type B as B1 and B2, and those from type C as C1 and C2. He will install them in three rental canoes labeled X, Y, and Z. Which installation plan corresponds to a completely randomized experiment?

Put paper labels A1, A2, B1, B2, C1, and C2 in a hat. For each of the six canoe seat positions, draw a label and install the selected seat.

When observations are taken on every possible treatment in a factorial experiment, the design is said to be ______.

complete full factorial

A factorial experiment in which experimental units are assigned to treatments at random, with all possible assignments being equally likely, is called a completely BLANK experiment.

randomized

Calculations for Tukey-Kramer simultaneous 100(1 - α)% confidence intervals and simultaneous level α tests for all C = I(I - 1)/2 differences μi - μj are like those for Fisher's least significant difference method except that ______.

√MSE(1Ji+1Jj)MSE1Ji+1Jj is replaced by √MSE2(1Ji+1Jj)MSE21Ji+1Jj tN-I,α/2 is replaced by qI,N-I,α

Match the two-way ANOVA sum of squares abbreviation on the left to its value on the right. Rows (SSA)

Choice, JK∑Ii=1α̂i2 JK∑i=1Iαˆ2i∑Ii=1α̂i2 Columns (SSB) matches Choice, IK∑Ii=1β̂j2 Interactions (SSAB) matches Choice, K∑Ii=1∑Jj=1γ̂ij2 Error (SSE) matches Choice, ∑Ii=1∑Jj=1∑Kk=1Xijk-Xij.2 Total (SST) matches Choice, ∑Ii=1∑Jj=1∑Kk=1Xijk-X...2

Match each null hypothesis used in two-way ANOVA on the left with its interpretation on the right.

H0: γ11 = ... = γIJ = 0 matches Choice, All interactions are zero. That is, the additive model holds. H0: α1 = ... = αI = 0 matches Choice, All row effects are zero. That is, the row factor does not affect the outcome. H0: β1 = ... = βJ = 0 matches Choice, All column effects are zero. That is, the column factor does not influence the outcome.

Match each mean square on the left with its purpose in two-way ANOVA on the right.

MSA = SSAI-1SSAI-1 matches Choice, Nonzero row effects make this mean square large. MSB = SSBJ-1SSBJ-1 matches Choice, Nonzero column effects make this mean square large. MSAB = SSAB(I-1)(J-1)SSAB(I-1)(J-1) matches Choice, Nonzero interactions make this mean square large. MSE = SSEIJ(K-1)SSEIJ(K-1) matches Choice, This is the mean square to which others are compared.

A confidence interval for the ith treatment mean μi in a one-way ANOVA requires the standard deviation of the point estimate XXi.. It is estimated as σXi.σXi. ≈ ______.

MSE/Ji√ where MSE can be described as a pooled estimate of the variance σ2.

One-way ANOVA requires that treatment populations are normal with the same variance. Which methods are appropriate for checking these assumptions?

Make and evaluate a normal probability plot of the residuals Xij - XXi.. Make and evaluate a normal probability plot of the residuals Xij - XXi. versus fitted values XXi..

Match the factorial experiment term on the left with its description on the right.

Response variable matches Choice, The dependent variable Factor matches Choice, An independent variable Level matches Choice, One of several values taken by a factor Treatment matches Choice, A combination of factor levels Experimental unit matches Choice, Object of measurement Replicate matches Choice, Units assigned to a treatment

Match the term or phrase from a two-factor experiment on the left with its description on the right.

Row and column factors matches Choice, Explanatory variables in the experiment Treatment matches Choice, A combination of factor values Complete experiment matches Choice, One in which observations are taken on every possible treatment Balanced experiment matches Choice, One in which the same number K of replicates is used for each treatment

Match each quantity from a one-way ANOVA on the left to its value from the ANOVA table, below, on the right. Source DF SS MS F P Treatment 2 42 21 21 0.002 Error 6 6 1 Total 8 48

SSTr matches Choice, 42 MSE matches Choice, 1 F matches Choice, 21 P-value matches Choice, 0.002

Match the two-way ANOVA assumption on the left with a way to check it on the right.

The design is complete. matches Choice, Check that there are observations for each treatment. Check that there are observations for each treatment. The design is balanced. matches Choice, Check that the same number K of replicates is used for each treatment. Check that the same number K of replicates is used for each treatment. K ≥ 2. matches Choice, Check that the number of replicates K for each treatment is at least two. Check that the number of replicates K for each treatment is at least two. Observations in any treatment are a simple random sample from a normal population. matches Choice, Check that, in a normal probability plot of residuals, points roughly line up. Check that, in a normal probability plot of residuals, points roughly line up. All treatment populations have the same variance σ2. matches Choice, Check a plot of residuals Xijk-Xij. versus fitted values Xij. for reasonably uniform vertical spread. Check a plot of residuals Xijk - XXij. versus fitted values XXij. for reasonably uniform vertical spread.

Which checks of plots would be useful for deciding whether the assumptions for two-way ANOVA are met?

When there are many replicates per treatment, for each sample, make a normal probability plot of observations and check that points roughly line up. Check that a plot of residuals versus fitted values shows reasonably uniform vertical spread. Check that points roughly line up in a normal probability plot of residuals from all samples combined.

A confidence interval for the ith treatment mean μi in a one-way ANOVA is given by XXi ± tN-I,α/2√MSEJiMSEJi. Match each quantity from this interval on the left with its description on the right.

XXi matches Choice, Point estimate for μi tN-I,α/2 matches Choice, Table value for the confidence level √MSEJiMSEJi matches Choice, Estimated standard deviation of point estimate

Match the notation used in a one-factor experiment on the left with its description on the right.

XXi. = 1Ji1Ji∑i=1JiXij∑Jii=1Xij matches Choice, ith sample mean XX.. = 1N1N∑i=1I∑j=1JiXij∑Ii=1∑Jij=1Xij matches Choice, Sample grand mean Ji matches Choice, ith sample size I matches Choice, Number of samples (or levels)

Match each quantity from a one-way ANOVA on the left to its value from the ANOVA table, below, on the right. Source DF SS MS F P Treatment 2 42 21 21 0.002 Error 6 6 1 Total 8 48

treatment mean square matches Choice, 21 error sum of squares matches Choice, 6 degrees of freedom for SSE matches Choice, 6 degrees of freedom for SSTr matches Choice, 2

Match the notation used in a one-factor experiment on the left with its description on the right.

μ1, ..., μI matches Choice, treatment (population) means H0: μ1 = ...= μI versus H1: at least two of the μi are different matches Choice, Hypotheses for analysis of variance test SSTr = ∑i=1IJi(Xi.-X..)2∑Ii=1Ji(Xi.-X..)2 matches Choice, Treatment sum of squares, which measures variation across samples Xij - XXi. matches Choice, Residual of a point about its sample mean SSE = ∑i=1I∑j=1Ji(Xij∑Ii=1∑Jij=1(Xij - XXi.)2 matches Choice, Error sum of squares, which measures variation within samples

Match each quantity in a two-factor analysis of variance on the left with its description on the right.

μij matches Choice, treatment mean for row i and column j μμi. matches Choice, ith row mean αi = μμi. - μ matches Choice, ith row effect, the difference between the ith row mean and the population grand mean γij = μij - (μ + αi + βj) = μij - μμi. - μμ.j + μ matches Choice, ij interaction, the difference between the treatment mean for row i and column j and what is predicted by the additive model I matches Choice, number of rows