Unit XX - California State University, East Baysfan/SubPages/CSUteach/st3503/minita… · Web view(In a word processor, we have formatted the output below for a compact display

Stat 6305 — Unit 7: Two Factor ANOVA Model With Interaction Term — Partial Solutions

7.1.1. For your convenience the 30 obser-vations are given below (reading down columns for Condition in the table). Cut and paste them into c1 of a Minitab worksheet; label it Time. Then use the patterned data feature to make two columns of subscripts in c2 and c3 for Disp and Cond, respectively. Finally, print out the columns or look at the worksheet to verify that your data and subscripts are as shown below. (In a word processor, we have formatted the output below for a compact display in your browser.)

A printout of the stacked data with appropriate subscripts for levels of both treatments is as follows:

MTB > print c1-c3

Data Display

Row Time Cond Disp

1 18 1 1 2 16 1 1 3 13 1 2 4 15 1 2 5 24 1 3 6 28 1 3

7 31 2 1 8 35 2 1 9 33 2 2 10 30 2 2 11 42 2 3 12 46 2 3

13 22 3 1 14 27 3 1 15 24 3 2 16 21 3 2 17 40 3 3 18 37 3 3

19 39 4 1 20 36 4 1 21 35 4 2 22 38 4 2 23 52 4 3 24 57 4 3

25 15 5 1 26 12 5 1 27 10 5 2 28 16 5 2 29 28 5 3 30 24 5 3

7.1.2. Use Minitab to make tables as follows:(a) A data table similar to the one shown in this section. Use the menu path or commands below. What happens if you reverse the order in which you list the classification (subscript) variables?"

MTB > table 'Disp' 'Cond';SUBC> data 'Time'.

Tabulated statistics: Disp, Cond

Rows: Disp Columns: Cond

1 2 3 4 5

1 18 31 22 39 15 16 35 27 36 12

2 13 33 24 35 10 15 30 21 38 16

3 24 42 40 52 28 28 46 37 57 24

Cell Contents: Time : DATA

If the order of the factors reversed, Cond will be printed as the rows and Disp will be printed as the columns.

(b) A data table that shows means in the 15 cells of the table, and also for each row and column.

MTB > table 'Disp' 'Cond';SUBC> means 'Time'.

Tabulated statistics: Disp, Cond

Rows: Disp Columns: Cond

1 2 3 4 5 All

1 17.00 33.00 24.50 37.50 13.50 25.102 14.00 31.50 22.50 36.50 13.00 23.503 26.00 44.00 38.50 54.50 26.00 37.80All 19.00 36.17 28.50 42.83 17.50 28.80

Cell Contents: Time : Mean

Based on notes by Elizabeth Ellinger, Spring 2004, as expanded and modified by Bruce E. Trumbo, 2005-08. Copyright © 2005, 2008 by Bruce E. Trumbo. All rights reserved.

Stat 6305 — Unit 7: Partial Solutions 2 of 15

7.1.3. Make box plots of the data: first broken out by Display, and then broken out by Condition. (Use either character or pixel graphics.) Does either of these two comparisons of box plots give a clue as to the possible importance of the factors, Display or Condition, in influencing the time to stabilize?

Character boxplots of Time broken out by Display are as follows:

MTB > gstdMTB > boxp 'Time';SUBC> by 'Disp'.

Boxplot

Disp

--------------------1 ----I + I---- --------------------

-------------------2 -----I + I----- -------------------

-------------------3 ----I + I----------- ------------------- ----+---------+---------+---------+---------+---------+--Time 10 20 30 40 50 60

Professional boxplots of Time broken out by Condition is as follows:

MTB > gproMTB > boxp Time * Cond

The boxplots by Display suggest that the time to stabilize Display 3 might be greater than for the other two displays. The boxplots by Condition suggest possible differences among Conditions, perhaps with Conditions 1 and 5 easiest to stabilize, and Condition 4 the most difficult.



7.1.4. Use R to repeat problem 7.1.3.

time = c(18, 16, 13, 15, 24, 28, 31, 35, 33, 30, 42, 46, 22, 27, 24, 21, 40, 37, 39, 36, 35, 38, 52, 57, 15, 12, 10, 16, 28, 24)cond = as.factor(rep(1:5, each=6)); disp = as.factor(rep(1:3, each=2, times=5))par(mfrow=c(1,2)); boxplot(time ~ cond); boxplot(time ~ disp); par(mfrow=c(1,1))

7.2.1. Use the menu path STAT ANOVA Balanced Interaction plot, response: Time, factors: Cond, Disp to make profile plots. Make two figures: (i) a profile plot with Display on the horizontal axis (that is, reverse the order of the factors to get a different plot than the one shown in this section), and (ii) a full matrix plot, which shows two plots (check the appropriate box)…. Which plot do you find easier to interpret in terms of the Display main effect–the one with the three Displays on the horizontal axis, or the one with the three Displays represented as three broken-line segments? Is it clear from this plot which Display is best (has smallest Time values)? Do you see evidence of significant interaction in either plot?


Stat 6305 — Unit 7: Partial Solutions 4 of 15Plot (i):

Plot (ii):

From the plot with Cond on the x-axis, it is clear that Display 3 has consistently higher stabilization times than the other Displays across all five levels of Condition. However, the performance of Displays 1 and 2 seems indistinguishable. There is no significant interaction inferred from either of the two graphs – the line segments are nearly parallel. That is, remembering that the vertices (cell means) are random variables, we should probably not expect them to be more nearly parallel. In any case there is no disorderly interaction because the line segments do not cross one another.

If a report has room for only one of these interaction plots, then the one with five Conditions on the horizontal axis and three broken-line paths corresponding to Displays would probably be preferable. As a general guideline, it is usually easier to look at the smallest possible number of broken line paths, so the factor with the larger number of levels should be put on the horizontal axis. (Usually, it is best to look at both.)



7.2.2. Profile plots can be thought of as two-dimensional projections of a three-dimensional response surface. Use the menu path GRAPH 3D Surface Wireframe, Z=Time, Y=Cond, X=Disp to make a three-dimensional plot. Try to imagine projections of the surface onto the two front faces of the plotting "box." (a) Rotate the image so that the Condition axis is perpendicular to you, then so that the Display axis is perpendicular to you. What profile plot is generated by each of these rotations.

One set of edges is hidden. (Thus the question in part b)



(b) How could you renumber the levels of one or both factors so that no part of the surface is hidden from view (before rotation)?

Put the levels of Condition in increasing order of means.

MTB > desc c1;SUBC> by c2;SUBC> means. Descriptive Statistics: Time

Variable Cond MeanTime 1 19.00 2 36.17 3 28.50 4 42.83 5 17.50

That is change Condition codes 5 to 1, 1 to 2, 3 to 3, 2 to 4, and 4 to 5.

MTB > name c4 'OrdCond'MTB > code (5)11 (1)12 (3)13 (2)14 (4)15 c2 c4MTB > let c4 = c4 - 10

Also, for best results, put Display codes in decreasing order of means.

MTB > name c5 'OrdDisp'MTB > code (3)11 (2)13 (1)12 c3 c5MTB > let c5 = c5 - 10

Then make the wireframe plot using the suitably ordered codes for levels of the factors.

(c) When there is no (significant) interaction, what is the (approximate) shape of the 8 segments of this surface?

When there is no significant interactions, the 8 segments will be approximately parallelograms.



7.2.3. In a two-way ANOVA, show (by example with contrived cell means) that it is possible for one of the two possible profile plots to show markedly crossing line segments while the other does not. The interaction is either significant or it is not. But if it is significant, it can be disorderly with respect to one effect and not the other.

As a hypothetical example, consider treatments A and B with two levels each, and response R (two observations per cell). The data in the following table produce orderly interactions in the R vs. B graph, but disorderly interactions in the R vs. A graph:

A ——————————————— B 1 2————————————————————— 1 4 6 10 10 2 24 26 22 18

Tabulated statistics: A, B

Rows: A Columns: B

1 2

1 5 10

2 25 20

Cell Contents: R : Mean

The interaction plot matrix is as follows:

Note: For some reason, few introductory texts on experimental design mention the possibility that one profile plot may indicate disorderly interaction and the other may not. (Anyhow, we cannot immediately find such a reference.) In a subsequent unit, you will see a real example of this. Maybe you can find a way to alter the data of the Condition/Display data to give an example.



7.3.1. Perform the ANOVA of this section again so that you can see the EMS table. Also, try an alternate notation for the model: instead of Cond Disp Cond*Disp use Cond | Disp. On standard US keyboards, the "vert" symbol (|) is located just below the Backspace key along with the backslash (\); use Shift to get vert. Whether you use commands or menus, make a printout from the Session window to show what you specified. Translate from Minitab's notation with parentheses and brackets to write the EMS for each row of the ANOVA table in terms of parameters of the model given at the beginning of Section 2. Interpret the EMSs to say which MS is in the denominator of the F-statistic for each of the three tests of hypothesis (Condition, Display, and Interaction).

The commands and resulting ANOVA are as follows:

MTB > name c6 'RESI2' [If using commands. Automatically named in menus.]MTB > ANOVA 'Time' = Cond| Disp;SUBC> Restrict;SUBC> EMS;SUBC> Residuals 'RESI2'. ANOVA: Time versus Cond, Disp

Factor Type Levels ValuesCond fixed 5 1, 2, 3, 4, 5Disp fixed 3 1, 2, 3

Analysis of Variance for Time

Source DF SS MS F PCond 4 2850.13 712.53 100.83 0.000Disp 2 1227.80 613.90 86.87 0.000Cond*Disp 8 44.87 5.61 0.79 0.617Error 15 106.00 7.07Total 29 4228.80

S = 2.65832 R-Sq = 97.49% R-Sq(adj) = 95.15%

Expected Mean Square for Each Term (using Variance Error restricted Source component term model)1 Cond 4 (4) + 6 Q[1]2 Disp 4 (4) + 10 Q[2]3 Cond*Disp 4 (4) + 2 Q[3]4 Error 7.067 (4)

The EMS table may be translated as follows:

E[MS(Cond)] = + 6C

E[MS(Disp)] = + 10D

E[MS(Interaction)] = + 2CD

E[MS(Error)] =

In general, Q[ ] indicates a fixed factor with effect subscripted by the name of the row whose number is in [ ],and ( ) indicates a random factor with variance component 2 subscripted by the name of the column whose number is in ( ).

In all three cases, MSE is in the denominator of the F ratio. When one of the 's is 0, then the corresponding MS has expected value

Note: In subsequent units, we will see examples where MS(Error) is not in the denominator of F.


Stat 6305 — Unit 7: Partial Solutions 9 of 157.3.2. Make a dotplot and a normal probability plot of the residuals. Notice that the distribution is markedly bimodal. Is this enough of a departure from normality to cause the Anderson-Darling test to reject? Look at the original data table. It seems suspicious that, while variability within cells is rather small and observations are rounded to the nearest integer, there are no cells among the 15 in which the two replications agree exactly or almost exactly to give a residual of 0 or 1. This peculiarity explains the bimodality of the residuals. One must wonder if the successive observations in each cell are truly independent.

The distribution of the residuals in the dotplot is bimodal:

Dotplot: RESI1

: : : : : : . : : : : : : : : . -------+---------+---------+---------+---------+---------RESI1 -2.4 -1.2 0.0 1.2 2.4 3.6

The dotplot is precisely symmetrical, since, with n = 2, the mean is equal to (x1 + x2)/2 and the residual for x1 is equal to x1 – (x1 + x2)/2 = (x1 – x2)/2 and the residual for x2 is equal to x2 – (x1 + x2)/ = –(x1 - x2)/2. An Anderson-Darling normality test confirms that the null hypothesis of normality of the residuals must be rejected, with a P-value of < .005.

This may be as much due to the granularity of the data as to the bimodality. The normal distribution is continuous and the data here take very few values because of the limited range and rounding of reported values to integers. (Many nonparametric tests assume that data come from a continuous distribution, even if maybe not a normal one. Nonparametric tests are not at all robust to granularity of data. In contrast, normal-based tests, such as the F-tests in ANOVAs are not usually adversely affected by granularity unless it is quite severe.

Ignorant of the circumstances that gave rise to the data, one is entitled to speculate how such strongly bimodal data may have arisen. The peculiar thing here is that there are several cells in the data table where the two observations differ by 2, 3, or 4 seconds, but none that differ by 0 or 1 second. This is almost impossible if the data are independent and (rounded) normal. [One possible explanation is that the original data had only one observation per cell and that, in order to illustrate an ANOVA in which interaction could be discussed, someone fabricated two observations per cell having the same cell means as the original data. The corresponding dataset in O/L 6e (Table 14.27, p918) are expressed to one decimal place, and so they have residuals with fewer ties and perhaps a more realistic distribution.]


Stat 6305 — Unit 7: Partial Solutions 10 of 157.3.3. By hand, use Tukey's HSD method to find the pattern of differences among Displays. Then use HSD to find the pattern of differences among Conditions. In each case you need to use MS(Error) as the estimate of the error variance. To summarize your findings, provide "underline" diagrams for each of the main effects. (See, for example, page 920 of Ott/ Longnecker for the formula.) You can use Minitab to verify your answer: STAT ANOVA glm; Comparisons, Tukey.

For Displays, Tukey's HSD = 3.67[7.07 / 10]1/2 = 3.086, where 3.67 is the 95th percentile of the Studentized range for 3 levels and df(Error) = 15, where 7.07 is MS(Error), and where 10 = 2(5) is the number of observations averaged into the sample mean for each level of the Display factor. Below we see that the length of each of the CIs for differences in Display means is 2(3.086) = 6.17. For example, the CI for comparing Displays 1 and 2 is(-4.685, 1.485), which has length 6.17.

Similarly, for Conditions, HSD = 4.37[7.07 / 6]1/2 = 4.744. Twice this is 9.49, which is also the length of each of the CIs for differences in Condition means. For example, the length of the CI comparing Conditions 4 and 5 is |–30.08 – (–20.59)| = 9.49.

From Minitab's glm procedure (upon selecting Tukey CIs, but not tests):

MTB > GLM 'Time' = Cond | Disp;SUBC> Brief 2 ;SUBC> Pairwise Disp Cond;SUBC> Tukey;SUBC> NoTest. General Linear Model: Time versus Cond, Disp


Analysis of Variance for Time, using Adjusted SS for Tests

Source DF Seq SS Adj SS Adj MS F PCond 4 2850.13 2850.13 712.53 100.83 0.000Disp 2 1227.80 1227.80 613.90 86.87 0.000Cond*Disp 8 44.87 44.87 5.61 0.79 0.617Error 15 106.00 106.00 7.07Total 29 4228.80

S = 2.65832 R-Sq = 97.49% R-Sq(adj) = 95.15%

Tukey 95.0% Simultaneous Confidence IntervalsResponse Variable TimeAll Pairwise Comparisons among Levels of DispDisp = 1 subtracted from:

Disp Lower Center Upper --------+---------+---------+--------2 -4.685 -1.600 1.485 (----*----)3 9.615 12.700 15.785 (----*----) --------+---------+---------+-------- 0.0 6.0 12.0

Disp = 2 subtracted from:

Disp Lower Center Upper --------+---------+---------+--------3 11.21 14.30 17.39 (----*----) --------+---------+---------+-------- 0.0 6.0 12.0



Tukey 95.0% Simultaneous Confidence IntervalsResponse Variable TimeAll Pairwise Comparisons among Levels of CondCond = 1 subtracted from:

Cond Lower Center Upper ---------+---------+---------+-------2 12.424 17.167 21.909 (--*--)3 4.757 9.500 14.243 (--*--)4 19.091 23.833 28.576 (--*--)5 -6.243 -1.500 3.243 (--*--) ---------+---------+---------+------- -16 0 16

Cond = 2 subtracted from:

Cond Lower Center Upper ---------+---------+---------+-------3 -12.41 -7.67 -2.92 (--*--)4 1.92 6.67 11.41 (--*--)5 -23.41 -18.67 -13.92 (--*--) ---------+---------+---------+------- -16 0 16


Cond Lower Center Upper ---------+---------+---------+-------4 9.59 14.33 19.076 (--*--)5 -15.74 -11.00 -6.257 (--*--) ---------+---------+---------+------- -16 0 16


Cond Lower Center Upper ---------+---------+---------+-------5 -30.08 -25.33 -20.59 (--*--) ---------+---------+---------+------- -16 0 16

The Display means are:

Variable Disp MeanTime 1 25.10 2 23.50 3 37.80

And the underline diagram for Displays is: 2 1 3, which agrees with out speculation in Problem 7.1.3.(The only CI that covers 0 is the one for the difference between Displays 1 and 2.)

The Condition means are:

Variable Cond MeanTime 1 19.00 2 36.17 3 28.50 4 42.83 5 17.50

And the underline diagram for Conditions is: 5 1 3 2 4, which is consistent with our speculation in Problem 7.1.3.(The only CI that covers 0 is the one for the difference between Conditions 1 and 5.)

Note: The restricted model is not an option in the glm command. For this analysis, restricted and unrestricted models have the same F-tests.



7.3.4. Use R to make an interaction plot and ANOVA tables with and without interaction.

# assumes earlier code, providing the data, is repeated--or still present in Rinteraction.plot(cond, disp, time)tapply(time, list(disp, cond), mean)anova(lm(time ~ cond * disp))anova(lm(time ~ cond + disp))

> tapply(time, list(disp, cond), mean) 1 2 3 4 51 17 33.0 24.5 37.5 13.52 14 31.5 22.5 36.5 13.03 26 44.0 38.5 54.5 26.0

> anova(lm(time ~ cond * disp))Analysis of Variance Table

Response: time Df Sum Sq Mean Sq F value Pr(>F) cond 4 2850.13 712.53 100.8302 1.188e-10 ***disp 2 1227.80 613.90 86.8726 5.644e-09 ***cond:disp 8 44.87 5.61 0.7936 0.6167 Residuals 15 106.00 7.07 ---Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

> anova(lm(time ~ cond + disp))Analysis of Variance Table

Response: time Df Sum Sq Mean Sq F value Pr(>F) cond 4 2850.13 712.53 108.63 1.386e-14 ***disp 2 1227.80 613.90 93.59 8.913e-12 ***Residuals 23 150.87 6.56 ---Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1



7.4.1. Write the reduced model, using the same symbols as in the full model shown in Section 2, where appropriate. Repeat the ANOVA procedure of this section so that you can see the EMSs. Translate Minitab's EMS table into the notation of your model.

The reduced model is:

Yijk = + i + j + eijk; where i = 1, 2, 3; j = 1, 2,…5; k = 1, 2. Restrictions are that i i = 0 and j j = 0. Also eijk are iid N(0, 2).

The Minitab output for the reduced model is as follows:

ANOVA: Time versus Cond, Disp


Analysis of Variance for Time

Source DF SS MS F PCond 4 2850.13 712.53 108.63 0.000Disp 2 1227.80 613.90 93.59 0.000Error 23 150.87 6.56Total 29 4228.80

S = 2.56114 R-Sq = 96.43% R-Sq(adj) = 95.50%

Expected Mean Square for Each Term (using Variance Error restricted Source component term model)1 Cond 3 (3) + 6 Q[1]2 Disp 3 (3) + 10 Q[2]3 Error 6.559 (3)

The EMS table may be translated as follows:

E[MS(Cond)] = + 6C

E[MS(Disp)] = + 10D

E[MS(Error)] =

Thus the denominator for both F-tests is MS(Error).

7.4.2. Show how you could obtain the entire ANOVA table for the reduced model from that of the full model with a few simple arithmetic operations. In particular, how are DF(Error) and SS(Error) of the table for the reduced model obtained from entries in the table for the full model?

SS(Error)red = SS(Error)full + SS(Cond*Disp)full = 106.0 + 44.87 = 150.87

DF(Error)red = DF(Error)full + DF(Cond*Disp)full = 15 + 8 = 23

MS(Error)red = SS(Error)red /DF(Error)red = 150.87 / 23 = 6.56



7.4.3. Make a dotplot of the residuals from the reduced model. What is the P-value of the Anderson-Darling normality test on these residuals? Comment.

The dotplot and Anderson-Darling normality test of the residuals are as follows:

Dotplot: RESI2 . . : ...: . . ....... : ... : :. . . -----+---------+---------+---------+---------+---------+-RESI2 -4.0 -2.0 0.0 2.0 4.0 6.0

The residuals in the boxplot for the reduced model do not have a bimodal distribution (as they did in the full model). The Anderson-Darling P-value of .765 for the residuals in the reduced model is consistent with their normality. The bimodality disappears because the we are essentially dealing with the 15 cell means, not the strangely distributed individual values mentioned previously. The granularity disappears because the residuals from the reduced model are computed from a slightly more complex formula.

For the full model, the residuals are rijk = Yijk – Y–

ij . .

For the reduced model, the residuals are rijk = Yijk – Y–

i.. – Y–.j. + Y–

... . For example, r111 = 18 – 25.1 – 19.0 + 28.8 = 2.7.

7.4.4. Compare the 1% critical value of F for testing the Condition main effect in the reduced model with the critical value for the full model. Do the same for the Display effect. Notice that the F-statistics for Condition and Display are changed little in moving from the full model to the reduced model. What has been gained by pooling?

Condition: Freduced = F(.01, 4, 23) = 4.26, Ffull = F(.01, 4, 15) = 4.89

Display: Freduced = F(.01, 2, 23) = 5.66, Ffull = F(.01, 2, 15) = 6.36

By pooling the interactions with the error term, we increased the degrees of freedom of the error term from 15 to 23. As a result the critical value of F decreases from 4.89 to 4.26 for Condition and from 6.36 to 5.66 for Display. The tests for both effects are somewhat more powerful under the reduced model.

7.4.5. Without doing the computations, say whether Tukey's HSD for differences between Conditions is larger or smaller for the reduced model than for the full model. (Give two reasons.)

Tukey's HSD will be smaller for the reduced model because (i) MSE is smaller and (ii) the degrees of freedom of the error term are greater making the critical value of the Studentized range smaller.

To illustrate, we show the computation of HSD for Display with full and reduced models: Above for the full model, HSD = 3.67[7.07 / 10]1/2 = 3.086 based on df = 15. By contrast, here for the reduced model HSD 3.54[6.56 / 10]1/2 = 2.87 based on df = 23.



7.4.6. Error degrees of freedom are obtained at the expense of additional experimentation. Compute the entries in the DF column of the ANOVA table for the full model with 3 Displays, 5 Conditions, and 3 (instead of 2) replications per cell. The effect of pooling on DF(Error) is not as great as the effect of doing half again as many replications, but pooling doesn't cost anything.

The DF column of the ANOVA table would be as follows:

Source DFCond 4Disp 2Cond*Disp 8Error 30Total 44

Taking 3 observations per cell instead of 2 increases df(Error) from 15 to 30, which costs money; removing interaction from the model (after testing to see interaction is not significant) increases df(Error) from 15 to 23at no monetary cost.

7.4.7. Look at the table of cell means in Problem 7.1.2(b). Suppose you did not have the complete dataset available and had to draw conclusions only from the information in this table. What kind of model is this? Compare the SS and DF columns from this analysis with those in the ANOVA of the complete dataset. How is variability (error) estimated in the ANOVA from the table? From this point of view, what important information is lost when one deals only with the summary table of means?

If we had only the table of means, then n = 1 and so the third subscript is missing. There are only two subscripts (i for Cond and j for Disp). This is a two-factor ANOVA with one observation per cell. (You might also call it a block design, except that both factors are fixed.)

DF(Cond) and DF(Disp) are the same as for the actual dataset (with two observations per cell). DF(Error) here is the same as DF(Cond*Disp) for the actual dataset. SS(Cond) and SS(Disp) are half as large here as are their counterparts for the actual data. SS(Error) here is half as large as SS(Cond*Disp) for the actual dataset.

It is not possible to include an interaction term in the model because it would have the same subscripts as the error term. (One says that "interaction is confounded with error.") The main information lost is that it is not possible to test whether there is significant interaction. If interaction were present, it would inflate MS(Error) in the ANOVA based on means, and possibly mask significant main effects.

ANOVA: Time2 versus Cond2, Disp2

Factor Type Levels ValuesCond2 fixed 5 1, 2, 3, 4, 5Disp2 fixed 3 1, 2, 3

Analysis of Variance for Time2

Source DF SS MS F PCond2 4 1425.07 356.27 127.05 0.000Disp2 2 613.90 306.95 109.46 0.000Error 8 22.43 2.80Total 14 2061.40

S = 1.67456 R-Sq = 98.91% R-Sq(adj) = 98.10%

Expected Mean Square for Each Term (using Variance Error restricted Source component term model)1 Cond2 3 (3) + 3 Q[1]2 Disp2 3 (3) + 5 Q[2]3 Error 2.804 (3)


Documents

Unit XX - California State University, East Baysfan/SubPages/CSUteach/st3503/minita… · Web view(In a word processor, we have formatted the output below for a compact display