**STAT 330: 95-3
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Final Examination: 9 December 1995 Partial Solutions
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Instructor: Richard Lockhart
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INSTRUCTIONS: This is an open book exam. You may use notes, text and
other books as well as a calculator. You MUST be clear about what you are doing;
if not I will assume you don't know what you are doing and mark accordingly.
The exam is out of 77.
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**The attached SAS output, A, shows parts of the input and output for the analysis of the results of the following experiment. A total of 48 animals were each given one of 3 poisons, labelled I, II and III, so that 16 animals were given each poison. Each animal was then treated with 1 of 4 treatments, labelled A, B, C and D. Animals were assigned at random to one of the 12 possible poison/treatment combinations. For each animal, the response measured was time to death. The output analyzes the rate of dying defined to be 1 over the time to death.**- From the incomplete output produce a complete ANOVA table for the
analysis of this data set. [5 marks]
**Solution**: You fill in the ANOVA table for a two way layout with replications.Sum of MS F Source DF Squares POISON 2 34.88 17.44 72.77 TREATMNT 3 20.41 6.80 28.33 POISON*TREATMNT 6 1.57 0.26 1.08 Error 36 8.65 0.24 Corrected Total 47 65.51

You cannot produce*P*values. - State and test relevant hypotheses, describing conclusions in
real world terms in so far as possible. You will have to do fixed
level testing in this problem. [5 marks]
**Solution**: The hypothesis of no interaction between the poison and the treatment is accepted. Those of no main effect of poison and no main effect for treatment are rejected even at the 1% level. - Which treatment is most effective at
*reducing*the death rate? [2 marks]**Solution**: Tukey's intervals show that treatment's B and D give definitely lower rates than A or C. The comparison of D to B is not significant but the estimated rate is lowest for B. - In SAS output B you will see the full ANOVA table for the original
response variable, time to death. Why did I transform to do the analysis? [1 marks]
**Solution**: The answer, which was hard for students to come up with, is that there is less evidence of interactions in the transformed data. - What is the fitted residual corresponding to the first response
in the cell Poison=I and Treatment=A (i.e. the value 0.31 at the top
of the data table) for the model fitted in (d) above. [2 marks]
**Solution**:

- From the incomplete output produce a complete ANOVA table for the
analysis of this data set. [5 marks]
**Consider the following experiment. From each of 11 litters two rats are selected. One of the two is picked at random and raised in isolation while the other is raised in a cage with other rats and ``enriched'' surroundings. The rats are sacrificed on reaching adulthood and the cortexes are weighed. The data are analyzed two ways on SAS outputs C and D.**- Does isolation reduce cortex weight? [5 marks]
**Solution**: This is paired comparisons; output D gives a two sided*P*value of 0.0088; the test is one sided so*P*=0.0044. In any case there is clear evidence of a reduction in cortex weight. - About how much effect on cortex weight does isolation have? Give an
approximate 80% confidence interval. [5 marks]
**Solution**: from output D

- Does isolation reduce cortex weight? [5 marks]
**A popular model in fisheries predicts that, except for error, the number***y*of fish returning to spawn is proportional to some power of*x*the number of young salmon (smolts) ``escaping'' 4 years previously. In symbols . After taking logarithms this can be written asThis model is fitted to the data shown in SAS output E.

- Give a 95% confidence interval for [5 marks]
**Solution**: - Is less than 1? (This is an important scientific question.) [5 marks]
**Solution**:Get

*P*values from*t*tables with 18 degrees of freedom.

- Give a 95% confidence interval for [5 marks]
**Four variations, labelled A, B, C, and D of a process for making penicillin are investigated. One of the basic ingredients is corn liquor. This liquor must be blended in batches. The batches are big enough to split into 4 and so the experiment is run as follows. Five batches of the liquor are made. Each is split into 4 parts and each of the four parts is used with one of the 4 variations of the process. Each variation is tried on exactly one of the 4 parts of each batch of liquor. The yield of penicillin is then recorded for each of the resulting 20 runs. The data and output of 2 SAS models (both done in one run) are in SAS output F.**- Is there a difference between the four variations of the process
in terms of average yield? [4 marks]
**Solution**: This is a randomized complete blocks design. The first model fitted is a one way layout model which is wrong. The two way additive model shows an*F*statistic of 3.96 for treatment; the corresponding*P*value is 0.0229. There is relatively good evidence for a difference. - Which variant produces the highest yield? [2 marks]
**Solution**: Use the Tukey intervals! The intervals show that while treatment C had the highest mean you cannot be sure that C is the highest yield.

- Is there a difference between the four variations of the process
in terms of average yield? [4 marks]
**A new process for measuring the concentration of a chemical in water is being investigated. A total of***n*samples are prepared in which the concentrations are the known numbers for ; the new process is used to measure the concentrations for these samples. It is thought likely that the concentrations measured by the new process, which we denote , will be related to the true concentrations viawhere the are independent, have mean 0 and all have the same variance which is unknown.

- If this model is fitted by least squares, (that is by minimizing
) show that the least squares estimate of
is
[5 marks]

**Solution**:Set this equal to 0 to get

Solve for to get the desired answer.

- Show that the estimator in part (a) is unbiased. [4 marks]
**Solution**: - Compute (give a formula for) the standard error of . [4 marks]
**Solution**:So the standard error of is the square root of the variance or

- The error sum of squares for this model is which may be shown to have
*n*-1 degrees of freedom. If the are the numbers 1, 2, 3 and 4, and the error sum of squares is 0.12 find a 95% confidence interval for and explain what further assumptions you must make to do so. [4 marks]**Solution**: The estimate of is . The estimated standard error of is which is so the confidence interval is

- If this model is fitted by least squares, (that is by minimizing
) show that the least squares estimate of
is
**A simple random sample is drawn from a large population and***p*is the proportion of that population favouring sovereignty for Québec.- If the true value of
*p*is 0.496 then how big does the sample need to be to give an 80% chance of rejecting the hypothesis*p*>=0.5 at the level 0.05. [5 marks]**Solution**: Use the formula for*N*on the bottom of page 329 for a one-tailed test with , , , and . - If the true value of
*p*is 0.496 what is the probability that a survey of 1600 people will have ? [5 marks]NOTE: You may assume the population is very large so that there is no harm in pretending the sampling is carried out with replacement.

**Solution**:*X*is Binomial(1600,0.496) andwhich is the area to the right of 0.32 under a standard normal curve or 0.374 approximately.

- If the true value of
**A sample of 50 people is divided into 5 groups of 10 people each. Four of the groups are put on special diets; the fifth serves as control. After 13 weeks the blood concentration of high density lipoprotein (HDL) is measured for each person. Data (hypothetical), sas commands and output are in SAS output G.**- If the average concentration of HDL in the population at large is
reported in the literature to be 1.24 could sampling fluctuation for
simple random sampling explain the difference between the control average
and this reported value? [4 marks]
**Solution**: You are supposed to test the hypothesis that the control mean is 1.24 based on a sample of size 10 with and*s*=0.116. This leads to with 9 degrees of freedom. - Are there any differences between the diets (including control)? [4 marks]
**Solution**: This is a 1 way layout problem. The*P*value for the*F*test is 0.0077 which is quite decidedly significant so, yes, there are differences. - Which diets lead to the lowest level of HDL? [2 marks]
**Solution**: The Tukey intervals show that Diets A and B are clearly lower than the control level. They do not distinguish clearly between the other 4 diets.

- If the average concentration of HDL in the population at large is
reported in the literature to be 1.24 could sampling fluctuation for
simple random sampling explain the difference between the control average
and this reported value? [4 marks]

**DATA**

0.31 I A 0.45 I A 0.46 I A 0.43 I A 0.82 I B 1.10 I B 0.88 I B 0.72 I B 0.43 I C 0.45 I C 0.63 I C 0.76 I C 0.45 I D 0.71 I D 0.66 I D 0.62 I D 0.36 II A 0.29 II A 0.40 II A 0.23 II A 0.92 II B 0.61 II B 0.49 II B 1.24 II B 0.44 II C 0.35 II C 0.31 II C 0.40 II C 0.56 II D 1.02 II D 0.71 II D 0.38 II D 0.22 III A 0.21 III A 0.18 III A 0.23 III A 0.30 III B 0.37 III B 0.38 III B 0.29 III B 0.23 III C 0.25 III C 0.24 III C 0.22 III C 0.30 III D 0.36 III D 0.31 III D 0.33 III D

options pagesize=60 linesize=80; data poison; infile 'poison.dat'; input time poison $ treatmnt $ ; rate=1/time; proc glm data=poison; class poison treatmnt; model rate = poison|treatmnt; means treatmnt / tukey cldiff alpha=0.05; means treatmnt / tukey ; run;

General Linear Models Procedure Dependent Variable: RATE Sum of Source DF Squares POISON 34.88 TREATMNT 20.41 POISON*TREATMNT 1.57 Corrected Total 47 65.51 R-Square C.V. Root MSE RATE Mean 0.868055 18.68478 0.4899853 2.6223763 Tukey's Studentized Range (HSD) Test for variable: RATE Alpha= 0.05 Confidence= 0.95 df= 36 MSE= 0.240086 Critical Value of Studentized Range= 3.809 Minimum Significant Difference= 0.5387 Comparisons significant at the 0.05 level are indicated by '***'. Simultaneous Simultaneous Lower Difference Upper TREATMNT Confidence Between Confidence Comparison Limit Means Limit A - C 0.0334 0.5721 1.1109 *** A - D 0.8196 1.3583 1.8971 *** A - B 1.1187 1.6574 2.1961 *** C - A -1.1109 -0.5721 -0.0334 *** C - D 0.2475 0.7862 1.3249 *** C - B 0.5465 1.0853 1.6240 *** D - A -1.8971 -1.3583 -0.8196 *** D - C -1.3249 -0.7862 -0.2475 *** D - B -0.2397 0.2991 0.8378 B - A -2.1961 -1.6574 -1.1187 *** B - C -1.6240 -1.0853 -0.5465 *** B - D -0.8378 -0.2991 0.2397 Tukey's Studentized Range (HSD) Test for variable: RATE Alpha= 0.05 df= 36 MSE= 0.240086 Critical Value of Studentized Range= 3.809 Minimum Significant Difference= 0.5387 Means with the same letter are not significantly different. Tukey Grouping Mean N TREATMNT A 3.5193 12 A B 2.9472 12 C C 2.1610 12 D C C 1.8619 12 B

**CODE**

options pagesize=60 linesize=80; data poison; infile 'poison.dat'; input time poison $ treatmnt $ ; proc glm data=poison; class poison treatmnt; model time = poison|treatmnt; means treatmnt / tukey cldiff alpha=0.05; means treatmnt / tukey ; run;

General Linear Models Procedure Dependent Variable: TIME Sum of Mean Source DF Squares Square F Value Pr > F Model 11 2.20435625 0.20039602 9.01 0.0001 Error 36 0.80072500 0.02224236 Corrected Total 47 3.00508125 R-Square C.V. Root MSE TIME Mean 0.733543 31.11108 0.1491387 0.4793750 Source DF SS Mean Square F Value Pr > F POISON 2 1.03301250 0.51650625 23.22 0.0001 TREATMNT 3 0.92120625 0.30706875 13.81 0.0001 POISON*TREATMNT 6 0.25013750 0.04168958 1.87 0.1123 Tukey's Studentized Range (HSD) Test for variable: TIME Alpha= 0.05 Confidence= 0.95 df= 36 MSE= 0.022242 Critical Value of Studentized Range= 3.809 Minimum Significant Difference= 0.164 Simultaneous Simultaneous Lower Difference Upper TREATMNT Confidence Between Confidence Comparison Limit Means Limit B - D -0.02148 0.14250 0.30648 B - C 0.12019 0.28417 0.44815 *** B - A 0.19852 0.36250 0.52648 *** D - B -0.30648 -0.14250 0.02148 D - C -0.02231 0.14167 0.30565 D - A 0.05602 0.22000 0.38398 *** C - B -0.44815 -0.28417 -0.12019 *** C - D -0.30565 -0.14167 0.02231 C - A -0.08565 0.07833 0.24231 A - B -0.52648 -0.36250 -0.19852 *** A - D -0.38398 -0.22000 -0.05602 *** A - C -0.24231 -0.07833 0.08565 Tukey's Studentized Range (HSD) Test for variable: TIME Alpha= 0.05 df= 36 MSE= 0.022242 Critical Value of Studentized Range= 3.809 Minimum Significant Difference= 0.164 Means with the same letter are not significantly different. Tukey Grouping Mean N TREATMNT A 0.67667 12 B A B A 0.53417 12 D B B C 0.39250 12 C C C 0.31417 12 A

**DATA**

Enriched 689 Isolated 657 Enriched 656 Isolated 623 Enriched 668 Isolated 652 Enriched 660 Isolated 654 Enriched 679 Isolated 658 Enriched 663 Isolated 646 Enriched 664 Isolated 600 Enriched 647 Isolated 640 Enriched 694 Isolated 605 Enriched 633 Isolated 635 Enriched 653 Isolated 642

The SAS code

options pagesize=60 linesize=80; data cortex; infile 'cortex.dat'; input treatmnt $ weight ; proc sort data=cortex; by treatmnt; proc ttest cochran; class treatmnt; run;

The output

TTEST PROCEDURE Variable: WEIGHT TREATMNT N Mean Std Dev Std Error ------------------------------------------------------------------------------- Enriched 11 664.18181818 17.93777122 5.40844152 Isolated 11 637.45454545 20.15124630 6.07582937 Variances T Method DF Prob>|T| -------------------------------------------------------- Unequal 3.2857 Satterthwaite 19.7 0.0037 Cochran 10.0 0.0082 Equal 3.2857 20.0 0.0037 For H0: Variances are equal, F' = 1.26 DF = (10,10) Prob>F' = 0.7200

**DATA**

689 657 656 623 668 652 660 654 679 658 663 646 664 600 647 640 694 605 633 635 653 642

The SAS code

options pagesize=60 linesize=80; data cortexpr; infile 'cortexpr.dat'; input enriched isolated; diff=enriched-isolated; proc means mean std stderr t prt maxdec=2; run;

The output

Variable Mean Std Dev Std Error T Prob>|T| -------------------------------------------------------------------------- ENRICHED 664.18 17.94 5.41 122.80 0.0001 ISOLATED 637.45 20.15 6.08 104.92 0.0001 DIFF 26.73 27.33 8.24 3.24 0.0088 --------------------------------------------------------------------------

The data follow. The fourth column is the logarithm of the second and the fifth is the log of the third. The second column is the number of returning adults in millions and the third is the number of smolts escaping in millions.

1960 1.154 14.4 0.1432342 2.667228 1961 1.159 25.0 0.1475576 3.218876 1962 1.422 24.6 0.3520643 3.202746 1963 2.156 32.7 0.7682547 3.487375 1964 2.455 28.7 0.8981268 3.356897 1965 2.576 31.9 0.9462378 3.462606 1966 3.160 30.2 1.1505720 3.407842 1967 2.429 35.2 0.8874797 3.561046 1968 1.818 31.6 0.5977370 3.453157 1969 3.005 41.5 1.1002776 3.725693 1970 3.974 33.6 1.3797731 3.514526 1971 2.162 40.7 0.7710337 3.706228 1972 2.164 47.7 0.7719584 3.864931 1973 3.256 42.0 1.1804994 3.737670 1974 1.916 39.6 0.6502397 3.678829 1975 4.272 41.8 1.4520821 3.732896 1976 1.190 42.7 0.1739533 3.754199 1977 1.859 39.7 0.6200387 3.681351 1978 1.778 53.1 0.5754891 3.972177 1979 1.422 46.9 0.3520643 3.848018The SAS code

data fish; infile 'fish.dat'; input year adults smolts logretrn logescap ; proc reg; model logretrn = logescap; run;The output

Model: MODEL1 Dependent Variable: LOGRETRN Analysis of Variance Sum of Mean Source DF Squares Square F Value Prob>F Model 1 0.30002 0.30002 2.078 0.1666 Error 18 2.59839 0.14436 C Total 19 2.89841 Root MSE 0.37994 R-square 0.1035 Dep Mean 0.74593 Adj R-sq 0.0537 C.V. 50.93494 Parameter Estimates Parameter Standard T for H0: Variable DF Estimate Error Parameter=0 Prob > |T| INTERCEP 1 -0.766287 1.05238687 -0.728 0.4759 LOGESCAP 1 0.425772 0.29533681 1.442 0.1666

**DATA**

1 A 86 1 B 86 1 C 97 1 D 96 2 A 81 2 B 75 2 C 92 2 D 81 3 A 78 3 B 85 3 C 87 3 D 87 4 A 84 4 B 90 4 C 89 4 D 86 5 A 76 5 B 79 5 C 80 5 D 90The code

options pagesize=60 linesize=80; data pen; infile 'pen.dat'; input blend treatmnt $ yield; proc glm data=pen; class blend treatmnt ; model yield = treatmnt; means treatmnt / tukey cldiff alpha=0.05; means treatmnt / tukey ; proc glm data=pen; class blend treatmnt ; model yield = treatmnt blend; means treatmnt / tukey cldiff alpha=0.05; means treatmnt / tukey ; run; end{verbatim} The output \begin{verbatim} General Linear Models Procedure Dependent Variable: YIELD Sum of Mean Source DF Squares Square F Value Pr > F Model 3 223.75000000 74.58333333 2.44 0.1025 Error 16 490.00000000 30.62500000 Corrected Total 19 713.75000000 R-Square C.V. Root MSE YIELD Mean 0.313485 6.491479 5.5339859 85.250000 Source DF Type I SS Mean Square F Value Pr > F TREATMNT 3 223.75000000 74.58333333 2.44 0.1025 Tukey's Studentized Range (HSD) Test for variable: YIELD NOTE: This test controls the type I experimentwise error rate. Alpha= 0.05 Confidence= 0.95 df= 16 MSE= 30.625 Critical Value of Studentized Range= 4.046 Minimum Significant Difference= 10.014 Comparisons significant at the 0.05 level are indicated by '***'. Simultaneous Simultaneous Lower Difference Upper TREATMNT Confidence Between Confidence Comparison Limit Means Limit C - D -9.014 1.000 11.014 C - B -4.014 6.000 16.014 C - A -2.014 8.000 18.014 D - C -11.014 -1.000 9.014 D - B -5.014 5.000 15.014 D - A -3.014 7.000 17.014 B - C -16.014 -6.000 4.014 B - D -15.014 -5.000 5.014 B - A -8.014 2.000 12.014 A - C -18.014 -8.000 2.014 A - D -17.014 -7.000 3.014 A - B -12.014 -2.000 8.014 General Linear Models Procedure Tukey's Studentized Range (HSD) Test for variable: YIELD NOTE: This test controls the type I experimentwise error rate, but generally has a higher type II error rate than REGWQ. Alpha= 0.05 df= 16 MSE= 30.625 Critical Value of Studentized Range= 4.046 Minimum Significant Difference= 10.014 Means with the same letter are not significantly different. Tukey Grouping Mean N TREATMNT A 89.000 5 C A A 88.000 5 D A A 83.000 5 B A A 81.000 5 A General Linear Models Procedure Dependent Variable: YIELD Sum of Mean Source DF Squares Square F Value Pr > F Model 7 487.75000000 69.67857143 3.70 0.0229 Error 12 226.00000000 18.83333333 Corrected Total 19 713.75000000 R-Square C.V. Root MSE YIELD Mean 0.683363 5.090603 4.3397389 85.250000 Source DF Type I SS Mean Square F Value Pr > F TREATMNT 3 223.75000000 74.58333333 3.96 0.0356 BLEND 4 264.00000000 66.00000000 3.50 0.0407 Tukey's Studentized Range (HSD) Test for variable: YIELD NOTE: This test controls the type I experimentwise error rate. Alpha= 0.05 Confidence= 0.95 df= 12 MSE= 18.83333 Critical Value of Studentized Range= 4.199 Minimum Significant Difference= 8.1485 Comparisons significant at the 0.05 level are indicated by '***'. Simultaneous Simultaneous Lower Difference Upper TREATMNT Confidence Between Confidence Comparison Limit Means Limit C - D -7.149 1.000 9.149 C - B -2.149 6.000 14.149 C - A -0.149 8.000 16.149 D - C -9.149 -1.000 7.149 D - B -3.149 5.000 13.149 D - A -1.149 7.000 15.149 B - C -14.149 -6.000 2.149 B - D -13.149 -5.000 3.149 B - A -6.149 2.000 10.149 A - C -16.149 -8.000 0.149 A - D -15.149 -7.000 1.149 A - B -10.149 -2.000 6.149 General Linear Models Procedure Tukey's Studentized Range (HSD) Test for variable: YIELD NOTE: This test controls the type I experimentwise error rate, but generally has a higher type II error rate than REGWQ. Alpha= 0.05 df= 12 MSE= 18.83333 Critical Value of Studentized Range= 4.199 Minimum Significant Difference= 8.1485 Means with the same letter are not significantly different. Tukey Grouping Mean N TREATMNT A 89.000 5 C A A 88.000 5 D A A 83.000 5 B A A 81.000 5 A

**DATA**

1.064 Control 1.221 Control 1.053 Control 1.123 Control 0.989 Control 1.142 Control 1.110 Control 1.247 Control 1.132 Control 1.399 Control 0.997 Diet_A 0.998 Diet_A 0.920 Diet_A 1.240 Diet_A 0.778 Diet_A 0.970 Diet_A 0.909 Diet_A 1.046 Diet_A 0.865 Diet_A 0.845 Diet_A 0.810 Diet_B 0.897 Diet_B 1.088 Diet_B 1.006 Diet_B 1.121 Diet_B 1.054 Diet_B 0.822 Diet_B 1.039 Diet_B 0.756 Diet_B 1.125 Diet_B 0.983 Diet_C 1.041 Diet_C 0.834 Diet_C 0.947 Diet_C 1.105 Diet_C 1.218 Diet_C 0.980 Diet_C 1.003 Diet_C 0.984 Diet_C 1.104 Diet_C 0.993 Diet_D 0.855 Diet_D 1.004 Diet_D 1.069 Diet_D 0.941 Diet_D 1.103 Diet_D 0.941 Diet_D 1.001 Diet_D 1.240 Diet_D 1.074 Diet_D

options pagesize=60 linesize=80; data hdl; infile 'hdl.dat'; input hdl diet $ ; proc means; by diet; proc glm data=hdl; class diet ; model hdl = diet; means diet / tukey cldiff alpha=0.05; run;

Analysis Variable : HDL ---------------------------------- DIET=Control -------------------------------- N Mean Std Dev Minimum Maximum ---------------------------------------------------------- 10 1.1480000 0.1163586 0.9890000 1.3990000 ---------------------------------------------------------- ---------------------------------- DIET=Diet_A --------------------------------- N Mean Std Dev Minimum Maximum ---------------------------------------------------------- 10 0.9568000 0.1283145 0.7780000 1.2400000 ---------------------------------------------------------- ---------------------------------- DIET=Diet_B --------------------------------- N Mean Std Dev Minimum Maximum ---------------------------------------------------------- 10 0.9718000 0.1384275 0.7560000 1.1250000 ---------------------------------------------------------- ---------------------------------- DIET=Diet_C --------------------------------- N Mean Std Dev Minimum Maximum ---------------------------------------------------------- 10 1.0199000 0.1045652 0.8340000 1.2180000 ---------------------------------------------------------- ---------------------------------- DIET=Diet_D --------------------------------- N Mean Std Dev Minimum Maximum ---------------------------------------------------------- 10 1.0221000 0.1062570 0.8550000 1.2400000 ---------------------------------------------------------- General Linear Models Procedure Dependent Variable: HDL Sum of Mean Source DF Squares Square F Value Pr > F Model 4 0.22636708 0.05659177 3.96 0.0077 Error 45 0.64251500 0.01427811 Corrected Total 49 0.86888208 R-Square C.V. Root MSE HDL Mean 0.260527 11.67224 0.1194911 1.0237200 Source DF Sum of Squares Mean Square F Value Pr > F DIET 4 0.22636708 0.05659177 3.96 0.0077 General Linear Models Procedure Tukey's Studentized Range (HSD) Test for variable: HDL NOTE: This test controls the type I experimentwise error rate. Alpha= 0.05 Confidence= 0.95 df= 45 MSE= 0.014278 Critical Value of Studentized Range= 4.018 Minimum Significant Difference= 0.1518 Comparisons significant at the 0.05 level are indicated by '***'. Simultaneous Simultaneous Lower Difference Upper DIET Confidence Between Confidence Comparison Limit Means Limit Control - Diet_D -0.02594 0.12590 0.27774 Control - Diet_C -0.02374 0.12810 0.27994 Control - Diet_B 0.02436 0.17620 0.32804 *** Control - Diet_A 0.03936 0.19120 0.34304 *** Diet_D - Control -0.27774 -0.12590 0.02594 Diet_D - Diet_C -0.14964 0.00220 0.15404 Diet_D - Diet_B -0.10154 0.05030 0.20214 Diet_D - Diet_A -0.08654 0.06530 0.21714 Diet_C - Control -0.27994 -0.12810 0.02374 Diet_C - Diet_D -0.15404 -0.00220 0.14964 Diet_C - Diet_B -0.10374 0.04810 0.19994 Diet_C - Diet_A -0.08874 0.06310 0.21494 Diet_B - Control -0.32804 -0.17620 -0.02436 *** Diet_B - Diet_D -0.20214 -0.05030 0.10154 Diet_B - Diet_C -0.19994 -0.04810 0.10374 Diet_B - Diet_A -0.13684 0.01500 0.16684 Diet_A - Control -0.34304 -0.19120 -0.03936 *** Diet_A - Diet_D -0.21714 -0.06530 0.08654 Diet_A - Diet_C -0.21494 -0.06310 0.08874 Diet_A - Diet_B -0.16684 -0.01500 0.13684

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Thu Apr 2 16:08:41 PST 1998