Transcript Capacity

Lecture 5 – Short Term Memory
The Study of Memory
Part 1 – Short Term Memory
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Lecture 5 – Short Term Memory
Three questions:
1. Why did people originally believe in STM
independent of LTM?
2. What do we think of those reasons now?
3. Do we need the STM construct?
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Lecture 5 – Short Term Memory
1. Why did people originally believe in STM
independent of LTM?
Because of STM – LTM differences in:
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Patient data
Capacity
Duration
Type of code
Serial position effect
Mechanism of loss
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Lecture 5 – Short Term Memory
Patient data.
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Lecture 5 – Short Term Memory
Patient data
Issue: is there a patient with a selective inability
to add to LTM , with sparing of STM?
If so, that selective impairment could be used in
an argument for an independent STM.
The most famous of all memory patients is HM.
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Lecture 5 – Short Term Memory
HM (Scoville & Milner, 1957)
• Surgery to relieve severe epilepsy, in 1953, at
age 27.
• bilateral excision of medial temporal lobe
• after surgery, HM had profound anterograde
amnesia. Capable of little if any new learning.
• Some retrograde amnesia.
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In HM’s words:
"At this moment everything looks clear to me, but
what happened just before? That's what worries
me. It's like waking from a dream; I just don't
remember".
Lecture 5 – Short Term Memory
Psychological studies
by Brenda Milner.
HM has:
• Good vocabulary
and language; normal
IQ
• No attention
disorder.
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HM – does not know, for example:
where he lives
who cares for him
what he ate at his last meal
what year it is
who the President of the United States is
or how old he is.
In 1982, he failed to recognize a picture of himself
that had been taken on his 40th birthday in 1966.
Lecture 5 – Short Term Memory
Declarative tasks – asking HM what he knows:
HM cannot learn (and later recall) new
photographs of people
verbal material
sequences of digits
complex geometric designs
nonsense patterns.
Cannot expand his digit span.
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Lecture 5 – Short Term Memory
Procedural tasks – observing what HM can do.
Milner (1962) trained H. M. on a mirror-drawing
task.
• HM, like normal people, improves with
practice. But he denies having practice.
Cohen and Corkin (1981) showed a similar result
on the Tower of Hanoi puzzle.
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Lecture 5 – Short Term Memory
HM – Conclusion:
Though HM can learn procedures he cannot
acquire new declarative learning.
LTM impaired. But STM spared.
Argument in favour of view that STM and LTM
are independent.
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Lecture 5 – Short Term Memory
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Capacity
Lecture 5 – Short Term Memory
Capacity
Issue: is STM different from LTM in capacity?
If so, that supports view that LTM and STM are
independent.
Capacity of LTM is essentially infinite.
What is capacity of STM?
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Capacity
• Shepard & Tehgtsoonian (1961)
• Presented 200 3-digit numbers in a row.
• E.g. … 492, 865, 931, 758… 865, …
• Task: report when you hear a repeated number
Lecture 5 – Short Term Memory
Shepard & Teghtsoonian (1961)
I.V.: Interval between 1st and 2nd appearance
D.V.: Probability of noticing repetition
• Repetition can only be noticed if first
occurrence is still in memory.
• Question: Are there separate forgetting
functions for LTM and STM?
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Shepard & Teghtsoonian (1961)
Result:
• P(noticing repetition) fell dramatically at first
• Steep decline ended at interval = 7 items
• P(noticing) then fell more gradually,
asymptoting at 60%
Lecture 5 – Short Term Memory
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Shepard & Teghtsoonian (1961) – Interpretation:
• Initial steep decline in P (noticing) occurs
because response coming from STM.
• More gradual occurs when response depends
upon LTM.
• Two forgetting functions – two memory stores,
one large and one small.
Lecture 5 – Short Term Memory
Question:
Why should STM have so small a capacity?
Sensory memory has large capacity. LTM has
large capacity.
Why did we evolve a limited capacity store
between two large capacity stores?
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Answer:
If STM was any larger, it would take too long to
search through.
When we need information from STM, to choose
or guide a response, we need it fast.
Things have to be processed fast in STM…
Lecture 5 – Short Term Memory
Duration
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Duration.
Issue: how long do STM traces last?
LTM traces last a long time – possibly your whole
life.
If STM traces last less time, that supports the view
that STM and LTM are independent.
Lecture 5 – Short Term Memory
Duration – how long does stuff last in STM?
Brown (1958) and Peterson & Peterson (1959)
Task: subjects see a simple stimulus (e.g., BRG)
and have to report it back after an interval.
Rehearsal is prevented by having them count
backwards during retention interval.
I.V. = length of interval in seconds.
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% Correct as function of delay in Brown/Peterson task
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Brown / Peterson:
Result: after 18 seconds, subjects can no longer
report stimulus.
Interpretation: there is a memory system in which
things must be rehearsed, or they are lost.
But we don’t have to rehearse things in LTM – so
there must be a second memory system – STM.
Lecture 5 – Short Term Memory
Type of code
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Lecture 5 – Short Term Memory
Type of code.
Issue: every stimulus has multiple aspects – e.g.
colour
brightness
shape
category
name
All are found in LTM. Which are found in STM?
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Lecture 5 – Short Term Memory
Brown/Peterson task
• Many similar studies reported in the ’60s.
• Most errors phonological – e.g., P for T.
• Errors based on shape were rare – e.g., C
for O.
• No semantic errors observed (or possible).
• Conclusion: STM uses a phonological code.
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Serial Position Effect
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Lecture 5 – Short Term Memory
Serial Position Effect:
In ordered recall, subjects recall a list in the order
it was given.
Out-of-order responses are counted as errors.
Accuracy is higher for the beginning and end of
the list, lower for the middle of the list.
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%
Position in list
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Serial Position Effect
Better performance at beginning of list is called
Primacy Effect.
Better performance at end of list is called Recency
effect.
Theory: Primacy produced by LTM. Recency
produced by STM
Lecture 5 – Short Term Memory
Mechanism of Loss
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How are things lost from memory – if at all?
Decay?
Interference?
Retrieval failure?
Originally, LTM loss was blamed on interference
and STM loss was blamed on decay – as in Brown
/ Peterson paradgim.
Lecture 5 – Short Term Memory
2. What do we think of those reasons now?
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Lecture 5 – Short Term Memory
Differences between STM and LTM (taken in
slightly different order this time):
• Type of code
• Serial position effect
• Mechanism of loss
• Patient data
• Capacity
• Duration
Do these reasons survive?
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Type of code:
Original argument – any kind of code in LTM,
only phonological codes in STM.
We now know – that STM can contain any kind of
code.
• See, for example, Brooks (1968), and Wickens
Release from Proactive Inhibition studies.
Lecture 5 – Short Term Memory
Serial Position Effect:
Original argument – Primacy effect produced by
LTM, Recency effect produced by STM.
We now know – that both Primacy and Recency
effects can be found in pure LTM studies (e.g.,
recalling U.S. Presidents).
Thus, recency effect cannot be taken as
“empirical signature” of STM.
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Mechanism of loss:
Original argument – information lost from STM
through decay, from LTM through interference.
We now know – that information can be lost from
STM through interference.
Lecture 5 – Short Term Memory
Duration
Original argument:
• newly-acquired memories must be rehearsed
to survive
• but older memories do not need to be
• therefore, new and old memories must be in
separate stores.
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Lecture 5 – Short Term Memory
Duration:
Alternative account:
• traces in LTM are vulnerable until they have
been consolidated.
• new items are more vulnerable to loss than
‘established’ items.
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Lecture 5 – Short Term Memory
HM:
If traces in LTM are vulnerable until they have
been consolidated, then HM’s problem is that he
cannot consolidate.
He has normal digit span because new items can
be inserted in LTM.
But he has anterograde amnesia because new
items cannot be consolidated in LTM.
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That leaves only Capacity …
Capacity is capacity of the Articulatory Loop –
which is used for rehearsal of information and for
planning articulation.
AL is not a short-term memory.
• For example, you cannot search your articulatory
loop, the way you can search memory.
Lecture 5 – Short Term Memory
Articulatory loop.
Capacity is determined by rate of loss. You can
rehearse about 7 items.
If you try to rehearse more than 7 items, the first
ones will be lost before you finish one cycle
through the list and go back to the beginning.
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Articulatory loop in action:
Memory load = r l z t c j a
Articulatory loop rehearses:
r l z t c j a .. r l z t c j a .. r l z t c j a ..
‘r’ is still in loop when you finish ‘a.’
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Lecture 5 – Short Term Memory
Articulatory loop in action:
Memory load = r l z t c j a m k s c p y
Articulatory loop rehearses:
r l z t c j a m k s c p y ..
‘r’ is no longer in loop by the time you finish ‘y’
so cannot be rehearsed – is lost.
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3. Do we need the STM construct?
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No.
We can explain all memory phenomena in terms
of LTM and the articulatory loop.
All we need is two premises:
a. Limited capacity in articulatory loop.
b. Items in LTM are vulnerable to loss until they
have been consolidated.