ST7003-1

TRINITY COLLEGE DUBLIN THE UNIVERSITY OF DUBLIN

Faculty of Engineering, Mathematics and Science

School of Computer Science and Statistics

Postgraduate Certificate in Statistics Hilary Term 2015

DESIGN AND ANALYSIS OF EXPERIMENTS

Wednesday 29 April 2015 Sports Centre 14.00 – 17.00

Professor Stuart, Professor Parnell

______________________________________________________________________

Instructions to Candidates: Answer all 3 questions. Questions 1 and 2 carry 30 marks each. Question 3 carries 40 marks. Answer each question in a separate answer book. Appendix 1, pages 12 - 14, gives tables of critical values of the t distribution and selected critical values of the F distribution.

Materials permitted for this examination:

Non-programmable calculators are permitted for this examination; please indicate the make and model of your calculator on each answer book used.

2

1 Chemicals are used to increase the water retention capacity of meats (or "preserve the integrity of the moisture content in meats"). An experiment was conducted using two such chemicals, identified as A and B. Each chemical was used at three levels in a 3x3 factorial design, in duplicate. Water retention was measured in millilitres. The results are shown in Table 1.

Table 1: Water retention capacity (ml H2O) using three levels of Factor A and three levels of Factor B, in duplicate.

Factor A

Factor B 1 2 3

1

1.14 2.23 0.74

1.05 2.30 0.50

2

1.87 3.13 1.43

1.60 3.00 1.00

3

1.70 2.80 0.10

1.80 1.95 0.05

An analysis of variance produced the following results.

Analysis of Variance for Water Retention Capacity

Source DF SS MS F P

A 2 1.6631 0.8315 13.86 0.002

B 2 11.2170 5.6085 93.45 0.000

A*B 4 0.9487 0.2372 3.95 0.040

Error 9 0.5401 0.0600

Total 17 14.3690

S = 0.244983

(a) Provide a brief report on the statistical significance of the results (3 marks)

(b) Find the 5% critical value for the F ratio for interaction and explain how it

relates to the p-value. (3 marks)

(c) Why are there 9 degrees of freedom for Error? (2 marks)

Table 2 on page 3 shows summary data.

(d) Draw an interaction plot showing profiles of levels of Factor B with levels

of Factor A on the horizontal axis. (4 marks)

Table 2: Mean water retention capacity (ml H2O) using three levels of Factor A and three levels of Factor B

3

Factor A Factor B

Means 1 2 3

1 1.095 2.265 0.620 1.327

2 1.735 3.065 1.215 2.005

3 1.750 2.375 0.075 1.400

Means 1.527 2.568 0.637 1.577

(e) Provide a brief interpretation of the interaction plot of part (d). (4 marks)

(f) Explain why interpretation of main effects is not recommended in this

case. (2 marks)

(g) Identify the optimum combination of factor levels (assuming high water

retention capacity is desirable) and calculate a 95% confidence interval for

the mean water retention capacity when using that combination. (5 marks)

The following diagnostic plot was produced along with the analysis of variance.

The cases with "deleted" residuals approximating +4 and –4 correspond to the

duplicate design points with Factor A at level 3 and Factor B at level 2.

3.02.52.01.51.00.50.0

5

4

3

2

1

0

-1

-2

-3

-4

Fitted Value

Dele

ted

Resid

ual

4

(h) Explain the advantage of using "deleted" residuals as distinct from ordinary (raw)

residuals in plots such as this. (2 marks)

(i) Provide a brief interpretive comment on the plot. (2 marks)

(j) What action(s) would you recommend based on your interpretation of this plot? (3 marks)

5

2. In an experimental incineration plant, three versions of the basic burner were

evaluated with a view to identifying the most efficient version. The measure of

efficiency used in this case was the residual amount of a particular toxic

chemical, smaller is better. A complete burning cycle took approximately two

hours so that at most three burns could be completed in a single working day.

To allow for the possibility that burner efficiency might be subject to variation

depending on changing conditions from day to day, all three burners were used,

in random order, on each of four successive days. The results were as follows;

efficiency is recorded per cent multiplied by 100. (Thus, the efficiency recorded

using Burner 1 on Day 1 was 0.21%).

Burner

Day (Block)

B1 B2 B3 Mean

1 21 23 23 22.33

2 18 19 22 19.67

3 18 21 20 19.67

4 17 20 21 19.33

Mean 18.50 20.75 21.50 20.25

An analysis of variance was calculated using Minitab, with the following results.

Two-way ANOVA: Efficiency versus Burner, Day (Block)

Source DF SS MS F P

Burner 2 19.5000 9.75000 11.32 0.009

Day (Block) 3 17.5833 5.86111 6.81 0.023

Error 6 5.1667 0.86111

Total 11 42.2500

Tukey 95.0% Simultaneous Confidence Intervals

All Pairwise Comparisons among Levels of Burner

Burner = 1 subtracted from:

Burner Lower Center Upper ------+---------+---------+---------+

2 0.2363 2.250 4.264 (---------*---------)

3 0.9863 3.000 5.014 (---------*---------)

------+---------+---------+---------+

0.0 2.0 4.0 6.0

Burner = 2 subtracted from:

Burner Lower Center Upper ------+---------+---------+---------+

3 -1.264 0.7500 2.764 (---------*---------)

------+---------+---------+---------+

0.0 2.0 4.0 6.0

6

(a) Report on the statistical significance of the results shown in the Analysis

of Variance table, in terms of F ratios and p-values. (6 marks)

(b) Discuss the F test for Burner effect:

what is measured / estimated by the Error Mean Square (MS)?

what is measured / estimated by the Burner Mean square?

what is measured by the Burner F-ratio?

what hypothesis is tested by the Burner F-ratio? (6 marks)

(c) Summarise the results of the Tukey pairwise comparisons.

(3 marks)

(d) Briefly explain how and why the simultaneous confidence intervals shown

above differ from confidence intervals for differences between individual

pairs of means. (3 marks)

A one-way analysis of variance, ignoring blocking, resulted as follows.

Source DF SS MS F P

Burner 2 19.50 9.75 3.86 0.062

Error 9 22.75 2.53

Total 11 42.25

Tukey 95% Simultaneous Confidence Intervals

All Pairwise Comparisons among Levels of Burner

Burner = 1 subtracted from:

Burner Lower Center Upper ---+---------+---------+---------+---

2 -0.890 2.250 5.390 (----------*---------)

3 -0.140 3.000 6.140 (---------*---------)

---+---------+---------+---------+---

-3.0 0.0 3.0 6.0

Burner = 2 subtracted from:

Burner Lower Center Upper ---+---------+---------+---------+---

3 -2.390 0.750 3.890 (----------*---------)

---+---------+---------+---------+---

-3.0 0.0 3.0 6.0

7

(e) Compare the results of the two-way analysis with those of the one-way

analysis, referring to both analysis of variance and Tukey pairwise

comparisons. Discuss the benefits of blocking in the light of these

comparisons. (4 marks)

(f) Explain why randomization might be used in experiments such as this and

how it achieves its goal.

(4 marks)

(g) Describe how a spreadsheet might be used to implement the

randomization in this case. (4 marks)

8

3 When weeds occur in fields where food crops are being grown, there is

competition between the weeds and the food crops for nutrients supplied via the

soil in which the crops are planted and any added fertilisers. Evidence suggests

that different weed species may have different competitive effects. Experiments

may be carried out in which a standard wheat variety is grown in combination

with different weed species in different plots and the wheat yields from the

different plots are compared with a view to estimating the differential weed

species effects.

A complicating factor is that irrigation has an effect on wheat yield and this effect

may vary, depending on the competing weed species.

The matter is further complicated by the fact that, whereas wheat seed may be

sown combined with various weed seed combinations in relatively small plots of

land, the water piping arrangements required for irrigation mean that irrigated

areas will necessarily be larger. This means that plots treated with the same

level of irrigation (Irrigation or No irrigation) will be made up of a number of the

smaller plots treated with the different weed species.

An experiment was conducted in which a single variety of wheat was sown in

combination with three weed species and none in two sets of four neighbouring

plots, with one set being irrigated while the other was not irrigated, and the whole

arrangement was replicated four times, resulting in four blocks of eight plots

each. The weed species were black-grass (Bg), cleavers (Cl) and chickweed

(Cw), with no weed being designated as Nw. Irrigation was applied to one half of

each block, selected at random, while the other half was not irrigated. Weed

species (and none) were applied randomly within each set of four plots. The

yield (Y) of grain from each plot at 85% dry matter in tonnes per hectare was

measured. The results of the experiment are shown in Table 3 that follows.

9

Table 3 Wheat yields corresponding to different weed species for both irrigated

and non irrigated areas in four blocks.

Block I

Block II

Block III

Block IV

Irrigation Y N

Y N

Y N

Y N

Weed

Species

Bg 3.62 4.12

2.52 3.19

1.97 2.92

2.73 3.71

Cl 4.49 7.59

4.70 5.72

2.20 6.90

4.91 6.51

Cw 5.70 6.77

5.91 6.32

4.91 6.64

5.78 6.65

Nw 7.92 9.11

7.05 8.02

5.54 8.18

8.22 7.16

(a) Sketch a layout for this experiment showing a plausible assignment of

irrigation levels in the four blocks and a plausible assignment of weed

species in one of the blocks. (5 marks)

(b) Identify the whole plots and the whole plot treatments, the sub plots and

the subplot treatments. (4 marks)

(c) Show the plot and treatment structure diagram for these data. (5 marks)

(d) Write down the components of a Minitab style model for these data,

separating the terms in accordance with the plot structure and identifying

random term(s). (3 marks)

(e) Indicate how a split plot design facilitates assessing how the effect of

irrigation on wheat yield may vary, depending on the competing weed

species. (2 marks)

The data are illustrated in Figure 1 that follows.

10

Irrigation

Figure 1 Wheat yields corresponding to different weed species for both irrigated

and non irrigated areas in four blocks.

(f) Ignoring statistical significance, provide a commentary on the results with

respect to effects of all factors, including the blocking factor, and key

interactions. Refer to evidence in the graphs to support your commentary.

(6 marks)

The Analysis of Variance produced by Minitab for these data resulted as follows, where

B, I, W represent Block, Irrigation, Weed species, respectively

NwCwClBg

9

8

7

6

5

4

3

Weed Species

Wh

eat

Yie

ldBlock I

NwCwClBg

8

7

6

5

4

3

2

Weed Species

Wh

eat

Yie

ld

Block II

NwCwClBg

9

8

7

6

5

4

3

2

Weed Species

Wh

eat

Yie

ld

Block III

NwCwClBg

9

8

7

6

5

4

3

2

Weed Species

Wh

eat

Yie

ld

Block IV

11

Source DF Adj SS Adj MS F-Value P-Value

B 3 6.647 2.2158 1.48 0.378

I 1 14.231 14.2311 9.48 0.054

B*I 3 4.504 1.5012 5.75 0.006

W 3 85.926 28.6419 109.73 0.000

I*W 3 4.371 1.4571 5.58 0.007

Error 18 4.699 0.2610

Total 31 120.378

(Note: The B*W interaction was not at all significant and so was omitted

from the analysis.)

(g) Identify the errors terms corresponding to each of the other terms in the

model underlying the analysis. Confirm the values of the relevant F ratios;

show the relevant calculations. With reference to your answer to part (f),

comment on the validity of the whole plots error term. (5 marks)

(h) Discuss the statistical significance of the results. Make cross references

to your answer to part (f). (5 marks)

(i) Comment on the effectiveness of blocking in this case. (1 marks)

(j) Calculate a new Whole Plot variation by combining the B and B*I sources

of variation. Use this to recalculate the F ratio for Irrigation. Assess its

statistical significance by reference to the tables of the F distribution at the

end of this paper. Comment. (4 marks)

12

Appendix 1 Statistical Tables

Selected critical values for the t-distribution

is the proportion of values in a t distribution with degrees of freedom

which exceed in magnitude the tabled value. For example, 25% of the

values in a t distribution with 1 degree of freedom are outside ±2.41.

.25 .10 .05 .02 .01 .002 .001

= 1 2.41 6.31 12.71 31.82 63.66 318.32 636.61

2 1.60 2.92 4.30 6.96 9.92 22.33 31.60 3 1.42 2.35 3.18 4.54 5.84 10.22 12.92 4 1.34 2.13 2.78 3.75 4.60 7.17 8.61 5 1.30 2.02 2.57 3.36 4.03 5.89 6.87 6 1.27 1.94 2.45 3.14 3.71 5.21 5.96 7 1.25 1.89 2.36 3.00 3.50 4.79 5.41 8 1.24 1.86 2.31 2.90 3.36 4.50 5.04 9 1.23 1.83 2.26 2.82 3.25 4.30 4.78 10 1.22 1.81 2.23 2.76 3.17 4.14 4.59 12 1.21 1.78 2.18 2.68 3.05 3.93 4.32 15 1.20 1.75 2.13 2.60 2.95 3.73 4.07 20 1.18 1.72 2.09 2.53 2.85 3.55 3.85 24 1.18 1.71 2.06 2.49 2.80 3.47 3.75 30 1.17 1.70 2.04 2.46 2.75 3.39 3.65 40 1.17 1.68 2.02 2.42 2.70 3.31 3.55 60 1.16 1.67 2.00 2.39 2.66 3.23 3.46 120 1.16 1.66 1.98 2.36 2.62 3.16 3.37 ∞ 1.15 1.64 1.96 2.33 2.58 3.09 3.29

13

Selected critical values for the F distribution

with 1 numerator and 2 denominator degrees of freedom

For example, 10% of the values in an F distribution with 1 numerator and 2 denominator degrees of freedom exceed 8.5.

10% critical values for the F distribution

1 2 3 4 5 6 7 8 10 12 24 ∞

1 39.9 49.5 53.6 55.8 57.2 58.2 58.9 59.4 60.2 60.7 62.0 63.3

2 8.5 9.0 9.2 9.2 9.3 9.3 9.3 9.4 9.4 9.4 9.4 9.5

3 5.5 5.5 5.4 5.3 5.3 5.3 5.3 5.3 5.2 5.2 5.2 5.1

4 4.5 4.3 4.2 4.1 4.1 4.0 4.0 4.0 3.9 3.9 3.8 3.8

5 4.1 3.8 3.6 3.5 3.5 3.4 3.4 3.3 3.3 3.3 3.2 3.1

6 3.8 3.5 3.3 3.2 3.1 3.1 3.0 3.0 2.9 2.9 2.8 2.7

7 3.6 3.3 3.1 3.0 2.9 2.8 2.8 2.8 2.7 2.7 2.6 2.5

8 3.5 3.1 2.9 2.8 2.7 2.7 2.6 2.6 2.5 2.5 2.4 2.3

9 3.4 3.0 2.8 2.7 2.6 2.6 2.5 2.5 2.4 2.4 2.3 2.2

10 3.3 2.9 2.7 2.6 2.5 2.5 2.4 2.4 2.3 2.3 2.2 2.1

12 3.2 2.8 2.6 2.5 2.4 2.3 2.3 2.2 2.2 2.1 2.0 1.9

15 3.1 2.7 2.5 2.4 2.3 2.2 2.2 2.1 2.1 2.0 1.9 1.8

20 3.0 2.6 2.4 2.2 2.2 2.1 2.0 2.0 1.9 1.9 1.8 1.6

40 2.8 2.4 2.2 2.1 2.0 1.9 1.9 1.8 1.8 1.7 1.6 1.4

120 2.7 2.3 2.1 2.0 1.9 1.8 1.8 1.7 1.7 1.6 1.4 1.2

∞ 2.7 2.3 2.1 1.9 1.8 1.8 1.7 1.7 1.6 1.5 1.4 1.0

5% critical values for the F distribution

1 2 3 4 5 6 7 8 10 12 24 ∞

1 161.4 199.5 215.7 224.6 230.2 234.0 236.8 238.9 241.9 243.9 249.1 254.3

2 18.5 19.0 19.2 19.2 19.3 19.3 19.4 19.4 19.4 19.4 19.5 19.5

3 10.1 9.6 9.3 9.1 9.0 8.9 8.9 8.8 8.8 8.7 8.6 8.5

4 7.7 6.9 6.6 6.4 6.3 6.2 6.1 6.0 6.0 5.9 5.8 5.6

5 6.6 5.8 5.4 5.2 5.1 5.0 4.9 4.8 4.7 4.7 4.5 4.4

6 6.0 5.1 4.8 4.5 4.4 4.3 4.2 4.1 4.1 4.0 3.8 3.7

7 5.6 4.7 4.3 4.1 4.0 3.9 3.8 3.7 3.6 3.6 3.4 3.2

8 5.3 4.5 4.1 3.8 3.7 3.6 3.5 3.4 3.3 3.3 3.1 2.9

9 5.1 4.3 3.9 3.6 3.5 3.4 3.3 3.2 3.1 3.1 2.9 2.7

10 5.0 4.1 3.7 3.5 3.3 3.2 3.1 3.1 3.0 2.9 2.7 2.5

12 4.7 3.9 3.5 3.3 3.1 3.0 2.9 2.8 2.8 2.7 2.5 2.3

15 4.5 3.7 3.3 3.1 2.9 2.8 2.7 2.6 2.5 2.5 2.3 2.1

20 4.4 3.5 3.1 2.9 2.7 2.6 2.5 2.4 2.3 2.3 2.1 1.8

30 4.2 3.3 2.9 2.7 2.5 2.4 2.3 2.3 2.2 2.1 1.9 1.6

40 4.1 3.2 2.8 2.6 2.4 2.3 2.2 2.2 2.1 2.0 1.8 1.5

120 3.9 3.1 2.7 2.4 2.3 2.2 2.1 2.0 1.9 1.8 1.6 1.3

∞ 3.8 3.0 2.6 2.4 2.2 2.1 2.0 1.9 1.8 1.8 1.5 1.0

14

2.5% critical values for the F distribution

1 1 2 3 4 5 6 7 8 10 12 24 ∞

1 647.8 799.5 864.2 899.6 921.8 937.1 948.2 956.6 968.6 976.7 997.3 1018.3

2 38.5 39.0 39.2 39.2 39.3 39.3 39.4 39.4 39.4 39.4 39.5 39.5

3 17.4 16.0 15.4 15.1 14.9 14.7 14.6 14.5 14.4 14.3 14.1 13.9

4 12.2 10.6 10.0 9.6 9.4 9.2 9.1 9.0 8.8 8.8 8.5 8.3

5 10.0 8.4 7.8 7.4 7.1 7.0 6.9 6.8 6.6 6.5 6.3 6.0

6 8.8 7.3 6.6 6.2 6.0 5.8 5.7 5.6 5.5 5.4 5.1 4.8

7 8.1 6.5 5.9 5.5 5.3 5.1 5.0 4.9 4.8 4.7 4.4 4.1

8 7.6 6.1 5.4 5.1 4.8 4.7 4.5 4.4 4.3 4.2 3.9 3.7

9 7.2 5.7 5.1 4.7 4.5 4.3 4.2 4.1 4.0 3.9 3.6 3.3

10 6.9 5.5 4.8 4.5 4.2 4.1 3.9 3.9 3.7 3.6 3.4 3.1

12 6.6 5.1 4.5 4.1 3.9 3.7 3.6 3.5 3.4 3.3 3.0 2.7

15 6.2 4.8 4.2 3.8 3.6 3.4 3.3 3.2 3.1 3.0 2.7 2.4

20 5.9 4.5 3.9 3.5 3.3 3.1 3.0 2.9 2.8 2.7 2.4 2.1

30 5.6 4.2 3.6 3.2 3.0 2.9 2.7 2.7 2.5 2.4 2.1 1.8

40 5.4 4.1 3.5 3.1 2.9 2.7 2.6 2.5 2.4 2.3 2.0 1.6

120 5.2 3.8 3.2 2.9 2.7 2.5 2.4 2.3 2.2 2.1 1.8 1.3

∞ 5.0 3.7 3.1 2.8 2.6 2.4 2.3 2.2 2.0 1.9 1.6 1.0

1% critical values for the F distribution

1 2 3 4 5 6 7 8 10 12 24 ∞

1 4052.2 4999.3 5403.5 5624.3 5764.0 5859.0 5928.3 5981.0 6055.9 6106.7 6234.3 6365.6

2 98.5 99.0 99.2 99.3 99.3 99.3 99.4 99.4 99.4 99.4 99.5 99.5

3 34.1 30.8 29.5 28.7 28.2 27.9 27.7 27.5 27.2 27.1 26.6 26.1

4 21.2 18.0 16.7 16.0 15.5 15.2 15.0 14.8 14.5 14.4 13.9 13.5

5 16.3 13.3 12.1 11.4 11.0 10.7 10.5 10.3 10.1 9.9 9.5 9.0

6 13.7 10.9 9.8 9.1 8.7 8.5 8.3 8.1 7.9 7.7 7.3 6.9

7 12.2 9.5 8.5 7.8 7.5 7.2 7.0 6.8 6.6 6.5 6.1 5.6

8 11.3 8.6 7.6 7.0 6.6 6.4 6.2 6.0 5.8 5.7 5.3 4.9

9 10.6 8.0 7.0 6.4 6.1 5.8 5.6 5.5 5.3 5.1 4.7 4.3

10 10.0 7.6 6.6 6.0 5.6 5.4 5.2 5.1 4.8 4.7 4.3 3.9

11 9.6 7.2 6.2 5.7 5.3 5.1 4.9 4.7 4.5 4.4 4.0 3.6

12 9.3 6.9 6.0 5.4 5.1 4.8 4.6 4.5 4.3 4.2 3.8 3.4

14 8.9 6.5 5.6 5.0 4.7 4.5 4.3 4.1 3.9 3.8 3.4 3.0

16 8.5 6.2 5.3 4.8 4.4 4.2 4.0 3.9 3.7 3.6 3.2 2.8

18 8.3 6.0 5.1 4.6 4.2 4.0 3.8 3.7 3.5 3.4 3.0 2.6

20 8.1 5.8 4.9 4.4 4.1 3.9 3.7 3.6 3.4 3.2 2.9 2.4

25 7.8 5.6 4.7 4.2 3.9 3.6 3.5 3.3 3.1 3.0 2.6 2.2

30 7.6 5.4 4.5 4.0 3.7 3.5 3.3 3.2 3.0 2.8 2.5 2.0

40 7.3 5.2 4.3 3.8 3.5 3.3 3.1 3.0 2.8 2.7 2.3 1.8

120 6.9 4.8 3.9 3.5 3.2 3.0 2.8 2.7 2.5 2.3 2.0 1.4

∞ 6.6 4.6 3.8 3.3 3.0 2.8 2.6 2.5 2.3 2.2 1.8 1.0

© UNIVERSITY OF DUBLIN 2015.