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Introductory Psychology surveys biological bases of behaviour, sensation and perception, learning, memory, cognition, development, personality, psychological disorders, and social psychology. Exams are mostly multiple-choice, testing recognition of key theorists, experimental findings, and terminology. Practice tests reveal which concepts you can reliably distinguish under pressure.

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Topics typically tested

  • Neuroscience & biological bases of behaviour
  • Sensation & perception
  • Classical & operant conditioning
  • Memory models & forgetting
  • Psychological disorders (DSM categories)
  • Social influence & conformity

Sample Intro to Psychology exam questions

Three original questions in the style of a real Intro to Psychology exam, each with a worked solution. Your own practice exam can be built from your course's notes and syllabus.

Question 1Multiple choiceClassical & operant conditioning

Each time a four-year-old whines in the grocery store, his father eventually hands him a cookie and the whining stops. Over several weeks two things change: the child whines more often, and the father hands over the cookie sooner. In operant-conditioning terms, which statement correctly identifies the consequence that is strengthening each person's behaviour?

  1. A.

    The child's whining is positively reinforced; the father's cookie-giving is negatively reinforced.

  2. B.

    The child's whining is negatively reinforced; the father's cookie-giving is positively reinforced.

  3. C.

    The child's whining is positively reinforced; the father's cookie-giving is positively punished.

  4. D.

    Both behaviours are negatively reinforced, because each one ends an unpleasant state.

Worked solution
Answer:A

Work through each behaviour with the two questions that define any operant consequence: is a stimulus added or removed, and does the behaviour increase or decrease afterward?

The child. Behaviour: whining. Consequence: a cookie is added. Result: whining becomes more frequent. A stimulus is added and the behaviour increases, so the whining is positively reinforced.

The father. Behaviour: handing over a cookie. Consequence: the whining (an aversive stimulus) is removed. Result: he gives the cookie sooner each time, so the behaviour is strengthened. A stimulus is removed and the behaviour increases, so the cookie-giving is negatively reinforced.

Both behaviours increase, so neither is punished, which rules out option C. Option D is the most tempting mistake: the child is not escaping anything, a cookie is being added, so only the father's side involves removal. Option B swaps the two labels. The exchange is a classic reinforcement trap in which each person's behaviour maintains the other's.

Answer: A — the child's whining is positively reinforced and the father's cookie-giving is negatively reinforced.

Question 2Short answerMemory models & forgetting

Participants hear a list of 15 unrelated words read aloud at one word per second and then immediately write down as many as they can, in any order. Their recall, plotted against each word's position in the list, is U-shaped. A second group hears the same list but must count backward by threes from 300 for 30 seconds before recalling. Describe how the second group's recall curve differs from the first group's, and explain the difference using the multi-store (Atkinson–Shiffrin) model of memory.

Worked solution
Answer:

The recency effect disappears while the primacy effect remains: the 30-second counting task displaces the last few words from short-term memory before they can be reported, whereas the first few words were already rehearsed into long-term memory.

The U-shaped curve is the serial position effect. Its two ends have different explanations in the multi-store model.

Primacy effect (early words). The first words arrive when short-term memory is empty, so each gets more rehearsal than later words. Rehearsal transfers them into long-term memory, which holds them through a delay.

Recency effect (last words). The final few words are recalled well in immediate recall only because they are still sitting in short-term memory, which has a small capacity and holds unrehearsed information for only a matter of seconds.

The 30-second counting task. Counting backward is a demanding, attention-filling task. It prevents rehearsal and displaces the last few words from short-term memory before the participant can report them. The early words, already in long-term memory, are unaffected. So the second group's curve keeps its high left end, but its right end drops to roughly the level of the middle items.

This is the pattern Glanzer and Cunitz reported in 1966: a filled delay before recall abolished the recency effect but left the primacy effect intact, which is strong evidence that the two effects come from separate memory stores.

Answer: the recency effect is eliminated and the primacy effect is preserved, because the distractor task clears short-term memory while the rehearsed early items are already in long-term memory.

Question 3Worked problemSensation & perception

In a lifted-weight experiment, a participant compares a standard weight with a comparison weight and reports which is heavier. With a 100 g standard, the smallest difference the participant detects reliably (the just-noticeable difference, JND) is 2 g. Assume Weber's law holds over the whole range tested.

(a) Compute the Weber fraction for this participant.

(b) Predict the JND when the standard is 1,500 g.

(c) The participant is handed a 1,500 g standard and a 1,520 g comparison. Will the difference be detected reliably? State the lightest comparison weight that would be.

Worked solution
Answer:

(a) k = 0.02; (b) JND = 30 g; (c) no, 20 g is below the 30 g JND; the lightest reliably heavier comparison is 1,530 g.

Weber's law states that the just-noticeable difference is a constant proportion of the standard stimulus:

ΔII=k\frac{\Delta I}{I} = k

where II is the standard intensity, ΔI\Delta I the JND, and kk the Weber fraction for that sense.

(a) Weber fraction. Use the measured pair to find the constant: k=ΔI/I=2 g/100 g=0.02k = \Delta I / I = 2\text{ g} / 100\text{ g} = 0.02. The participant needs a 2% change in weight to notice a difference.

(b) JND at 1,500 g. Because the proportion is constant, scale it to the new standard: ΔI=k×I=0.02×1,500 g=30 g\Delta I = k \times I = 0.02 \times 1{,}500\text{ g} = 30\text{ g}. A heavier standard needs a larger absolute change, which is why a 2 g difference that is obvious at 100 g is invisible at 1,500 g.

(c) Testing 1,520 g. The difference offered is 1,520−1,500=20 g1{,}520 - 1{,}500 = 20\text{ g}, which is less than the 30 g JND, so the participant will not reliably tell the weights apart. The lightest comparison that reaches the threshold is 1,500+30=1,530 g1{,}500 + 30 = 1{,}530\text{ g}.

As a check, the same 2% rule applied to the original pair gives k×100=0.02×100=2k \times 100 = 0.02 \times 100 = 2 g, matching the measured JND, and a 30 g difference at 1,500 g is again exactly 2%.

Answer: (a) k=0.02k = 0.02; (b) JND = 30 g; (c) no, 20 g is below threshold; the comparison must be at least 1,530 g.

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