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

Operationalisation means:

  1. ADefining variables in a precise, measurable way
  2. BGiving participants instructions
  3. CChoosing a statistical test
  4. DCarrying out the research in a real operational setting, such as a hospital
2.Research methods in psychologyHard

The "replication crisis" in psychology refers to:

  1. AMany published findings failing to reproduce in new studies
  2. BJournals being legally required to reject all replication studies submitted
  3. CCopying of questionnaires without permission
  4. DA shortage of participants for experiments
3.Descriptive statisticsMedium
152025303540455055

Find the sample standard deviation s of the data: 30, 12, 22, 42, 55, 46.

n = 6
s =
4.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 50 and standard deviation 5. Find the z-score z of a score of 42.5.

μ = 50σ = 5x = 42.5
z =
5.ProbabilityMedium
xp(x)00.3510.3520.3μ = 0.95

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

x: 0, 1, 2p(x): 0.35, 0.35, 0.3
σ =
6.ProbabilityHard
n = 9, p = 0.4: P(X = 1) = ?k0123456789

A test has 9 multiple-choice questions. A student who guesses gets each one right with probability 0.4, independently. What is the probability P of exactly 1 correct answers?

n = 9p = 0.4k = 1
P =
7.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.4581

Drag z so that the shaded area between −z and z under the standard normal curve is 0.4581. Find z to two decimals.

A = 0.4581
z =
8.Normal distributionHard
0.03 mm0D − d?

Hole diameters D are normal with mean 25.03 mm and standard deviation 0.02 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.01 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.03 mm, σ = 0.02 mmshaft: μ = 25 mm, σ = 0.01 mm
P =
9.Confidence intervalsMedium
n = 16, s = 12.5L = ?Ux̄ = 94E99%

A sample of 16 items has mean 94 and standard deviation s = 12.5. The population is roughly normal. Find the lower limit of the 99% confidence interval for the mean.

n = 16x̄ = 94s = 12.599%t(15) = 2.947
lower limit =
10.Confidence intervalsHard
σ = 25, n = ?x̄ − Ex̄ + Ex̄E = 0.599%

The population standard deviation is 25. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 0.5?

σ = 25E = 0.599%
n =
11.Hypothesis tests & confidence intervalsMedium
t −2.01102.011

A sample of 49 students has a mean score of 497.8 and a sample standard deviation of 5. Find the test statistic t for H₀: μ = 500.

μ₀ = 500x̄ = 497.8s = 5n = 49
t =
12.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 1α = 0.01βσ = 7, power 95%, n = ?

A one-sided z test at α = 0.01 must detect a shift of the mean of 1 units with a power of 95%. The standard deviation is 7. Find the minimum sample size n.

α = 0.01power = 95%δ = 1σ = 7zα = 2.326zβ = 1.645
n =
13.Linear regression & correlationMedium
xy166618722091229124105xy2080100

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

n = 5
r =
14.Linear regression & correlationHard
xy224861181010161216xy51051015ŷ = 8

The table gives 6 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 8?

n = 6ŷ = 8
x =
15.Chi-square testsMedium
coughno coughTotalsmokers152540non-smokers3466100Total4991140

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 = 1402 × 2
χ² =
16.Chi-square testsHard
dfα = 0.05α = 0.0113.8416.63525.9919.21037.81511.34549.48813.277511.07015.086612.59216.812714.06718.475815.50720.090916.91921.6661018.30723.209

A chi-square goodness-of-fit test compares observed counts in 6 categories with fully specified expected proportions (no parameters are estimated from the data). Using the table, find the critical value χ²crit at significance level α = 0.01.

α = 0.01
χ²crit =

3Solutions

1.Research methods in psychology

Answer: A · Defining variables in a precise, measurable way

Why this is the answer

Operationalising variables turns abstract concepts into specific measurements so that research can be replicated.

Answer

A · Defining variables in a precise, measurable way
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 = 16.07

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 = 207/6 = 34.5

Squared deviations

Σ(x − x̄)² = (−4.5)² + (−22.5)² + (−12.5)² + 7.5² + 20.5² + 11.5² = 1292

Standard deviation

s = √(1292/5) = 16.07

Answer

s = 16.07
4.Descriptive statistics

Answer: z = −1.5

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 = (42.5 − 50)/5 = −1.5

Answer

z = −1.5
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 = 0·0.35 + 1·0.35 + 2·0.3 = 0.95

Mean of the squares

E(X²) = 0·0.35 + 1·0.35 + 4·0.3 = 1.55

Variance

Var(X) = 1.55 − 0.95² = 0.6475
σ = √0.6475 = 0.8047

Answer

σ = 0.8047
6.Probability

Answer: P = 0.06047

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.

Binomial formula

P(X = 1) = 9·0.41·0.68 = 9·0.4·0.0168 = 0.06047

Answer

P = 0.06047
7.Normal distribution

Answer: z = 0.61

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 of z

Φ(z) = (1 + 0.4581)/2 = 0.7291

Read the z-table backwards

Find 0.7291 inside the table and read off z.

z = 0.61

Answer

z = 0.61
8.Normal distribution

Answer: P = 0.9099

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 difference

μ = 25.03 − 25 = 0.03 mm, σ = 0.02236 mm

Standardise

z = (0 − 0.03)/0.02236 = −1.34

Probability

P = 1 − Φ(−1.34) = 0.9099

Answer

P = 0.9099
9.Confidence intervals

Answer: lower limit = 84.79

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.

Critical value

df = n − 1 = 15, t = 2.947

Margin of error

E = 2.947·12.5/√16 = 9.209

Interval

94 ± 9.209 → [84.79, 103.2]

Answer

lower limit = 84.79
10.Confidence intervals

Answer: n = 16590

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 = (2.576·25/0.5)² = 16589

Round up

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

n = 16590

Answer

n = 16590
11.Hypothesis tests & confidence intervals

Answer: t = −3.08

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

SE = 5/√49 = 5/7 = 0.7143

Test statistic

t = (497.8 − 500)/0.7143 = −3.08

Answer

t = −3.08
t0.025; df = 48 = 2.011

|t| > 2.011, so H₀ is rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: n = 773

Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.

Solve for n

n = ((2.326 + 1.645)·7/1)² = 772.7

Round up

n = 773

Answer

n = 773
13.Linear regression & correlation

Answer: r = 0.969

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 = 100/5 = 20
ȳ = 425/5 = 85

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 194
Syy = Σ(y − ȳ)² = 1002

Correlation coefficient

r = 194/√(40·1002) = 0.969

Answer

r = 0.969
14.Linear regression & correlation

Answer: x = 5.118

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 = 42/6 = 7
ȳ = 63/6 = 10.5

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 93

Regression line

b = 93/70 = 1.329
a = 10.5 − 1.329·7 = 1.2
ŷ = 1.2 + 1.329·x

Solve for x

x = (8 − 1.2)/1.329 = 5.118

Answer

x = 5.118
15.Chi-square tests

Answer: χ² = 0.1538

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

smokers: 40·49/140 = 14, 40·91/140 = 26
non-smokers: 100·49/140 = 35, 100·91/140 = 65

Chi-square statistic

χ² = (15 − 14)²/14 + (25 − 26)²/26 + (34 − 35)²/35 + (66 − 65)²/65
χ² = 0.1538

Answer

χ² = 0.1538

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

16.Chi-square tests

Answer: χ²crit = 15.09

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 = 6 − 1 = 5

Table

χ²crit(df = 5, α = 0.01) = 15.086

Answer

χ²crit = 15.086

Reject H₀ if the computed χ² is larger than this value.

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