Units and allocation

Our study used 16 pens in four rooms, with six pigs per pen. Each pen received one diet and one housing treatment.

1. What was the experimental unit? Explain why.
2. What was the observational unit when we weighed every pig?
3. Where did we randomise the four treatment combinations?

Write one sentence for each answer before continuing.

Replication and evidence

1. How many independent pen replicates did each of the four treatment combinations have?
2. A layout has equal treatment counts in every room. Does that show that someone randomised it?
3. One pig leaves a pen. How does this differ from losing all usable observations from that pen?

For each answer, state what you would count or check.

Units: answers

The unit of observation depends on the response. Next class, we will work with one response per pen.

Replication: answers

• Four independent pens per combination: one in each of four rooms.
• Balance describes the result. To establish randomisation, check how the allocation was generated and recorded.
• One pig lost: the pen can still supply an outcome, but with fewer pig measurements.
• One pen outcome lost: one combination now has only three usable pen outcomes, and that room is incomplete.

Repeated measurements cannot replace a missing independent pen replicate.

The LEGO model

The allocation table

One possible room in the completed factorial design:

• Keep one allocation row per pen, with a unique pen identifier.
• Check that each of the other rooms also contains all four combinations.
• Preserve the method used to generate the allocation.

Why rooms matter

• Rooms may differ in conditions that influence growth.
• Comparing treatments within rooms helps separate treatment differences from room differences.
• Blocking is most useful when units within a block are relatively similar.

Design structure

The two structural decisions are compatible: a factorial experiment can use a randomised complete block design.

An interaction?

Hypothetical mean daily gain, in grams per day:

• Calculate high-fibre minus standard diet within each housing condition.
• Does one diet difference adequately describe both conditions?
• Explain the biological meaning in one sentence.

Interaction: answer

More factors

With 16 independent pens and two levels per treatment factor:

These counts assume equal allocation and set aside the four-room complete-block restriction.

With four factors, each combination occurs once. Six pigs per pen still give only one independent allocation per combination.

Targeted reading

In Ruxton and Colegrave, Experimental Design for the Life Sciences, fourth edition:

• Chapter 8, §§8.1–8.3: revisit treatment combinations and interactions. Explain why testing both diets under both housing conditions is useful.
• Chapter 9, §§9.1–9.2: revisit blocking. Explain why each room contains every treatment combination.

Write two short paragraphs using your group’s experiment. Revisit Chapter 5 if the distinction between pigs and pen replicates remains unclear.

Next class’s study

We will use a simpler experiment to concentrate on simulating responses.

Housing will not vary as an experimental treatment in the Class 5 exercise.

A response per pen

Possible values

For each response, describe its possible values. Do this before choosing a statistical distribution.

Pause and write your answers before continuing.

Values: answers

A probability of 0.2 can describe the chance that a pen needs treatment. The observed yes/no outcome is still 0 or 1.

Design before data

Illustrative rows for the new diet-only study:

• The allocation table can exist before any outcome is measured.
• A simulation supplies an outcome under stated assumptions.
• A fresh simulated dataset can use the same allocation table.

Assumed means

Suppose we assume these expected mean daily gains:

• The assumed diet difference is 20 g/day in both rooms.
• The assumed room difference is 30 g/day for both diets.
• Individual pen outcomes can vary around these means.

These numbers describe a hypothetical scenario, not results from the workshop.

What changes?

Suppose we retain the same allocation and the same assumed means, but generate another set of pen outcomes.

• Must the observed diet difference be exactly 20 g/day?
• Does a different set of outcomes mean that the treatment allocation changed?
• What additional information do we need beyond the expected means to simulate realistic observations?

Pause and explain each answer in words. No calculation or R code is needed.

Simulation: answers

• A simulated treatment estimate can differ from the assumed 20 g/day because outcomes vary among pens.
• The allocation can remain fixed while the simulated responses change.
• We need assumptions about the response distribution and variation, as well as expected means.

Next class, we will distinguish randomising treatment allocation from generating responses, and record the random seeds used for each operation.

Ready for Class 5

Bring your allocation table and a short response to these prompts:

• Which pen-level outcome would you like to investigate, and over what period?
• What values could one observation take?
• What diet difference would matter biologically?
• What evidence could inform your assumptions about room differences and variation among pens?

You may write “I need more information” where appropriate. State what information you would seek.

Sources

Ruxton, G. D., and Colegrave, N. (2016). Experimental Design for the Life Sciences. Fourth edition. Oxford University Press. Book and contents.

• Units and replication: Chapters 4–5.
• Randomisation: Chapter 7.
• Factorial designs and interactions: Chapter 8.
• Blocking: Chapter 9.

The simulation preparation follows the course’s Class 5 teaching plan. All numerical cases are hypothetical. Illustrations are reused course artwork.