Research Methods and Statistics for Psychology: cram pack

Formula sheets, the key method steps for every topic and 16 practice problems, with full solutions at the back.

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1Formulas and methods

Descriptive statistics

Mean
x̄ = Σx/n
Sample standard deviation
s = √(Σ(x − x̄)²/(n − 1))
Median
middle value of the sorted data
z-score
z = (x − μ)/σ
Normal distribution
68% within ±1σ, 95% within ±2σ

Probability

Complement
P(not A) = 1 − P(A)
Addition rule
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Independent events
P(A ∩ B) = P(A)·P(B)
Conditional probability
P(A|B) = P(A ∩ B)/P(B)
Binomial
P(X = k) = C(n,k)·pk(1 − p)n−k

Normal distribution

z-score
z = (x − μ)/σ
Below x
P(X < x) = Φ(z)
Above x
P(X > x) = 1 − Φ(z)
Between
Φ(z2) − Φ(z1)
Percentile
x = μ + z·σ

Method

  1. Convert the value to a z-score: z = (x − μ)/σ.
  2. Look up Φ(z), the area to the left of z.
  3. For the area to the right use 1 − Φ(z); for an interval subtract the two left areas.
  4. For a percentile, work backwards: find z for the area, then x = μ + z·σ.

Confidence intervals

σ known
x̄ ± z·σ/√n
σ unknown
x̄ ± t·s/√n, df = n − 1
Critical z
90%: 1.645, 95%: 1.96, 99%: 2.576
Sample size
n = (z·σ/E)², rounded up

Method

  1. Decide between z (population σ known) and t (only the sample s, with n − 1 degrees of freedom).
  2. Find the critical value for the confidence level.
  3. Margin of error: E = critical value × standard deviation/√n.
  4. Interval: x̄ ± E.

Hypothesis tests & confidence intervals

z-test for a mean
z = (x̄ − μ0)/(σ/√n)
t-test for a mean
t = (x̄ − μ0)/(s/√n), df = n − 1
Two-sided at 5%
reject H0 if |z| > 1.96
Confidence interval
x̄ ± z·σ/√n

Method

  1. State H₀ and H₁ and decide whether the test is one- or two-sided.
  2. Calculate the test statistic: (x̄ − μ₀)/(σ/√n), or with s and t.
  3. Compare with the critical value, or find the p-value.
  4. Reject H₀ if the statistic is beyond the critical value (p below α), and state the conclusion in words.

Linear regression & correlation

Sums of squares
Sxx = Σ(x − x̄)² = Σx² − (Σx)²/n, Sxy = Σxy − ΣxΣy/n
Slope
b = Sxy/Sxx
Intercept
a = ȳ − b·x̄
Correlation
r = Sxy/√(SxxSyy)
Residual
e = y − ŷ

Chi-square tests

Test statistic
χ² = Σ(O − E)²/E
Goodness of fit
E = n·p, df = k − 1 − (estimated parameters)
Independence
E = row total × column total / grand total, df = (r − 1)(c − 1)
Critical values (α = 0.05)
df 1: 3.841, 2: 5.991, 3: 7.815, 4: 9.488
Decision
reject H0 if χ² > χ²crit

2Practice problems

1.Research methods in psychologyMedium

A researcher studies the effect of gender on memory scores. Why is this a quasi-experiment rather than a true experiment?

  1. AThe DV is not measured
  2. BThe IV is a pre-existing characteristic, so participants cannot be randomly allocated to it
  3. CThe study takes place in a laboratory
  4. DThere are only two conditions, and a true experiment needs at least three levels of the IV
2.Research methods in psychologyHard

The "replication crisis" in psychology refers to:

  1. AMany published findings failing to reproduce in new studies
  2. BCopying of questionnaires without permission
  3. CJournals being legally required to reject all replication studies submitted
  4. DA shortage of participants for experiments
3.Descriptive statisticsMedium
1618202224262830

Find the sample standard deviation s of the data: 19, 25, 16, 24, 31.

n = 5
s =
4.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 100 and standard deviation 8. Find the z-score z of a score of 84.

μ = 100σ = 8x = 84
z =
5.ProbabilityMedium
xp(x)00.310.3520.35μ = 1.05

For the distribution shown, find the standard deviation σ of X.

x: 0, 1, 2p(x): 0.3, 0.35, 0.35
σ =
6.ProbabilityHard
n = 7, p = 0.3: P(X ≤ 1) = ?k01234567

7 components are tested, and each one fails with probability 0.3, independently. What is the probability P that at most one fails?

n = 7p = 0.3
P =
7.Normal distributionMedium
200 kg202.8 kgT = X₁ + … + X₈?

The masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 8 of them is below 202.8 kg. Round z to two decimals.

μ = 25 kgσ = 0.5 kgn = 8< 202.8 kg
P =
8.Normal distributionHard
557085100115130145pointstop 30%

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. Drag the cut-off so the shaded top 30% is to the right of it. Find the cut-off x.

μ = 100 pointsσ = 15 points30%
x = points
9.Confidence intervalsMedium
μ₁ − μ₂0L = ?Ux̄₁ − x̄₂E99%

The battery life of two phone models is compared. Model 1: 40 phones, mean 84.5 h, known standard deviation 10 h. Model 2: 60 phones, mean 97 h, known standard deviation 14 h. Find the lower limit of the 99% confidence interval for μ₁ − μ₂.

n₁ = 40x̄₁ = 84.5 hσ₁ = 10 hn₂ = 60x̄₂ = 97 hσ₂ = 14 h99%
lower limit = h
10.Confidence intervalsHard
σ = 9, n = ?x̄ − Ex̄ + Ex̄E = 2.595%

The population standard deviation is 9. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 2.5?

σ = 9E = 2.595%
n =
11.Hypothesis tests & confidence intervalsMedium
−2.1312.131t = ?df = 15

16 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is 4.94 and the standard deviation of the differences is 7.6. Find the paired test statistic t for H₀: μd = 0.

n = 16d̄ = 4.94sd = 7.6
t =
12.Hypothesis tests & confidence intervalsHard
z −1.64501.645

A sample of 81 measurements has mean 122.2. The population standard deviation is 20. Find the upper limit U of the 90% confidence interval for the mean.

x̄ = 122.2σ = 20n = 8190%
U =
13.Linear regression & correlationMedium
xy18762192241002712730132xy20253080100120

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 120Σy = 527Σx² = 2970Σy² = 57793Σxy = 13089
r =
14.Linear regression & correlationHard
xy145717772090231022611929127xy15202530100x = 18.5

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. A new observation (18.5, 84) is made. Find its residual e = y − ŷ.

n = 6(18.5, 84)
e =
15.Chi-square testsMedium
yesnoTotalmen3466100women206080Total54126180

The contingency table shows observed counts. Compute the chi-square statistic χ² for the test of independence between the row and column variables. Do not use a continuity correction.

N = 1802 × 2
χ² =
16.Chi-square testsHard

A chi-square test of independence is carried out on a contingency table with 3 rows and 6 columns (not counting the totals). How many degrees of freedom df does the test have?

3 × 6
df =

3Solutions

1.Research methods in psychology

Answer: B · The IV is a pre-existing characteristic, so participants cannot be randomly allocated to it

Why this is the answer

In a quasi-experiment the IV is a naturally occurring difference between people. Without random allocation, other differences between groups may confound the results.

Answer

B · The IV is a pre-existing characteristic, so participants cannot be randomly allocated to it
2.Research methods in psychology

Answer: A · Many published findings failing to reproduce in new studies

Why this is the answer

Large replication projects found many effects were smaller or absent, linked to publication bias, small samples and flexible analysis. Pre-registration is one response.

Answer

A · Many published findings failing to reproduce in new studies
3.Descriptive statistics

Answer: s = 5.788

Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.

Mean

x̄ = 115/5 = 23

Squared deviations

Σ(x − x̄)² = (−4)² + 2² + (−7)² + 1² + 8² = 134

Standard deviation

s = √(134/4) = 5.788

Answer

s = 5.788
4.Descriptive statistics

Answer: z = −2

Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.

Standardise

z = (84 − 100)/8 = −2

Answer

z = −2
5.Probability

Answer: σ = 0.8047

Discrete random variables: a probability distribution lists every value x a variable can take with its probability p(x); the probabilities are between 0 and 1 and add up to 1. The expected value (mean) is E(X) = Σx·p(x), the long-run average. The variance is Var(X) = Σ(x − μ)²·p(x) = E(X²) − μ², and the standard deviation σ = √Var(X). For a linear change Y = aX + b: E(Y) = a·E(X) + b, Var(Y) = a²·Var(X) and σY = |a|·σX; adding a constant does not change the spread.

Mean

μ = 0·0.3 + 1·0.35 + 2·0.35 = 1.05

Mean of the squares

E(X²) = 0·0.3 + 1·0.35 + 4·0.35 = 1.75

Variance

Var(X) = 1.75 − 1.05² = 0.6475
σ = √0.6475 = 0.8047

Answer

σ = 0.8047
6.Probability

Answer: P = 0.3294

Probability: with equally likely outcomes, P(A) = favourable/total. For independent events multiply: P(A and B) = P(A)·P(B). The complement often helps: P(at least one) = 1 − P(none). Binomial: in n independent trials with success probability p, P(X = k) = C(n, k)·pᵏ·(1 − p)ⁿ⁻ᵏ.

Zero and one failure

P(X = 0) = 0.77 = 0.08235
P(X = 1) = 7·0.3·0.76 = 0.2471

Add them

P = 0.08235 + 0.2471 = 0.3294

Answer

P = 0.3294
7.Normal distribution

Answer: P = 0.9761

Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).

Distribution of the total

μT = 200 kg, σT = 0.5·√8 = 1.414 kg

Standardise

z = (202.8 − 200)/1.414 = 1.98

Probability

P = Φ(1.98) = 0.9761

Answer

P = 0.9761
8.Normal distribution

Answer: x = 107.8 points

From an area back to z: Φ(z) is the area to the left of z. For a left-tail area A, find z with Φ(z) = A in the table. A right-tail area A means Φ(z) = 1 − A. A central area A between −z and z leaves (1 − A)/2 in each tail, so Φ(z) = (1 + A)/2. For a value x, use x = μ + zσ.

Area to the left

Φ(z) = 0.7

z from the table

z ≈ 0.52

Back to the original units

x = 100 + 0.52·15 = 107.8 points

Answer

x = 107.8 points
9.Confidence intervals

Answer: lower limit = −18.69 h

Interval for a difference: estimate ± critical value × standard error. For two independent means with known σ: SE = √(σ₁²/n₁ + σ₂²/n₂) and z. With equal unknown variances: pooled s_p and t with n₁ + n₂ − 2 degrees of freedom. For paired data, work with the differences: d̄ ± t·s_d/√n.

Point estimate

x̄₁ − x̄₂ = 84.5 − 97 = −12.5

Standard error

SE = √(10²/40 + 14²/60) = 2.401

Margin of error

E = 2.576·2.401 = 6.186

Interval

−12.5 ± 6.186 → [−18.69, −6.314]

Answer

lower limit = −18.69 h

The interval does not contain 0, so the means differ significantly.

10.Confidence intervals

Answer: n = 50

Confidence interval for a mean: x̄ ± margin of error. With a known population σ the margin is z·σ/√n; with only the sample standard deviation s, use t with n − 1 degrees of freedom: t·s/√n. Critical values: z = 1.645 (90%), 1.96 (95%), 2.576 (99%). To get a margin E or smaller, you need n ≥ (z·σ/E)², rounded up.

Solve for n

n = (1.96·9/2.5)² = 49.79

Round up

A sample size is a whole number, and rounding down would make the margin too large.

n = 50

Answer

n = 50
11.Hypothesis tests & confidence intervals

Answer: t = 2.6

Testing a mean: the test statistic measures how many standard errors the sample mean lies from the value claimed in H₀: z = (x̄ − μ₀)/(σ/√n) when σ is known, t = (x̄ − μ₀)/(s/√n) when it is estimated from the sample. A confidence interval is x̄ ± (critical value)·(standard error).

Standard error of the mean difference

SE = 7.6/√16 = 1.9

Test statistic

t = 4.94/1.9 = 2.6

Answer

t = 2.6

With df = 15, the two-sided 5% critical value is 2.131; |t| is larger, so H₀ is rejected.

12.Hypothesis tests & confidence intervals

Answer: U = 125.9

Testing a mean: the test statistic measures how many standard errors the sample mean lies from the value claimed in H₀: z = (x̄ − μ₀)/(σ/√n) when σ is known, t = (x̄ − μ₀)/(s/√n) when it is estimated from the sample. A confidence interval is x̄ ± (critical value)·(standard error).

Margin of error

E = 1.645·20/√81 = 3.656

Interval limit

U = 122.2 + 3.656 = 125.9

Answer

U = 125.9
13.Linear regression & correlation

Answer: r = 0.9806

Least-squares regression: with x̄ and ȳ the means, Sxx = Σ(x − x̄)², Syy = Σ(y − ȳ)² and Sxy = Σ(x − x̄)(y − ȳ), the line ŷ = a + b·x has slope b = Sxy/Sxx and intercept a = ȳ − b·x̄, so it always passes through (x̄, ȳ). Pearson's correlation coefficient r = Sxy/√(Sxx·Syy) lies between −1 and 1; r² is the fraction of the variation in y explained by the line.

Centred sums

Sxy = 13089 − 120·527/5 = 441
Sxx = 2970 − 120²/5 = 90
Syy = 57793 − 527²/5 = 2247

Correlation coefficient

r = 441/√(90·2247) = 0.9806

Answer

r = 0.9806
14.Linear regression & correlation

Answer: e = 2.61

Least-squares regression: with x̄ and ȳ the means, Sxx = Σ(x − x̄)², Syy = Σ(y − ȳ)² and Sxy = Σ(x − x̄)(y − ȳ), the line ŷ = a + b·x has slope b = Sxy/Sxx and intercept a = ȳ − b·x̄, so it always passes through (x̄, ȳ). Pearson's correlation coefficient r = Sxy/√(Sxx·Syy) lies between −1 and 1; r² is the fraction of the variation in y explained by the line.

Means

x̄ = 129/6 = 21.5
ȳ = 572/6 = 95.33

Sums of squares and products

Sxx = Σ(x − x̄)² = 157.5
Sxy = Σ(x − x̄)(y − ȳ) = 732

Regression line

b = 732/157.5 = 4.648
a = 95.33 − 4.648·21.5 = −4.59
ŷ = −4.59 + 4.648·x

Predicted value

ŷ = −4.59 + 4.648·18.5 = 81.39

Residual

e = 84 − 81.39 = 2.61

Answer

e = 2.61

The point lies above the line.

15.Chi-square tests

Answer: χ² = 1.714

Chi-square tests: compare observed counts O with the counts E expected under H₀ using χ² = Σ(O − E)²/E. Goodness of fit: E = n·p for each category, df = k − 1 (minus one more for each parameter estimated from the data). Independence in an r × c table: E = row total × column total / grand total, df = (r − 1)(c − 1). Reject H₀ when χ² exceeds the critical value for the chosen α.

Expected counts

men: 100·54/180 = 30, 100·126/180 = 70
women: 80·54/180 = 24, 80·126/180 = 56

Chi-square statistic

χ² = (34 − 30)²/30 + (66 − 70)²/70 + (20 − 24)²/24 + (60 − 56)²/56
χ² = 1.714

Answer

χ² = 1.714

df = 1. At α = 0.05 the critical value is 3.841, so H₀ (independence) is not rejected.

16.Chi-square tests

Answer: df = 10

Chi-square tests: compare observed counts O with the counts E expected under H₀ using χ² = Σ(O − E)²/E. Goodness of fit: E = n·p for each category, df = k − 1 (minus one more for each parameter estimated from the data). Independence in an r × c table: E = row total × column total / grand total, df = (r − 1)(c − 1). Reject H₀ when χ² exceeds the critical value for the chosen α.

Degrees of freedom

df = (3 − 1)·(6 − 1) = 10

Answer

df = 10

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