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

Why are behaviour categories in an observation operationalised before data collection?

  1. ASo observers record the same behaviours consistently, improving reliability
  2. BSo the observation becomes an experiment
  3. CSo the hypothesis can be changed after the data are seen
  4. DSo participants know exactly what is being recorded
2.Research methods in psychologyHard

A memory study uses a repeated measures design where everyone learns list A then list B. Recall is better for list B. What is the most likely flaw?

  1. AThe use of an ordinal level of measurement
  2. BOrder effects such as practice, because conditions were not counterbalanced
  3. CToo large a sample size
  4. DIndividual differences between the groups
3.Descriptive statisticsMedium
2025303540455055

Find the sample standard deviation s of the data: 22, 36, 46, 56, 20, 36.

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

Exam scores are normally distributed with mean 50 and standard deviation 15. Find the z-score z of a score of 65.

μ = 50σ = 15x = 65
z =
5.ProbabilityMedium
n = 7k = 3N = ?

In how many ways N can a committee of 3 people be chosen from a group of 7? The order does not matter.

n = 7k = 3
N =
6.ProbabilityHard
xp(x)00.310.120.2530.35μ

X has the distribution shown, and Y = 0.5X + 20. Find the standard deviation σY of Y.

x: 0, 1, 2, 3p(x): 0.3, 0.1, 0.25, 0.35Y = 0.5X + 20
σY =
7.Normal distributionMedium
164186.75P = ?

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 164 and 186.75 cm? Give P as a decimal.

μ = 178σ = 7x1 = 164x2 = 186.75
P =
8.Normal distributionHard
μ = 32x = ?97.5%

The delivery times are normally distributed with mean 32 min and standard deviation 6 min. Find the value x below which 97.5% of them lie (the 97.5th percentile).

μ = 32σ = 6P = 97.5%
x = min
9.Confidence intervalsMedium
μ₁ − μ₂0L = ?Ux̄₁ − x̄₂E95%

The battery life of two phone models is compared. Model 1: 60 phones, mean 40 h, known standard deviation 7 h. Model 2: 40 phones, mean 29 h, known standard deviation 3 h. Find the lower limit of the 95% confidence interval for μ₁ − μ₂.

n₁ = 60x̄₁ = 40 hσ₁ = 7 hn₂ = 40x̄₂ = 29 hσ₂ = 3 h95%
lower limit = h
10.Confidence intervalsHard
σ = 16, n = ?x̄ − Ex̄ + Ex̄E = 599%

The population standard deviation is 16. 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 5?

σ = 16E = 599%
n =
11.Hypothesis tests & confidence intervalsMedium
−2.1452.145t = ?df = 14

15 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 −3.87 and the standard deviation of the differences is 7.5. Find the paired test statistic t for H₀: μd = 0.

n = 15d̄ = −3.87sd = 7.5
t =
12.Hypothesis tests & confidence intervalsHard
−1.961.96z = ?

In a trial, 88 of 250 patients on treatment A recover, and 31 of 150 patients on treatment B. Using the pooled proportion, find the test statistic z for H₀: p₁ = p₂.

x₁ = 88n₁ = 250x₂ = 31n₂ = 150
z =
13.Linear regression & correlationMedium
xy332439540647750859xy46830405060

The table gives 6 paired observations of x = advertising spend in k€ and y = sales in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 6Σx = 33Σy = 267Σx² = 199Σy² = 12335Σxy = 1556
r =
14.Linear regression & correlationHard
xy2494626738841098xy5106080100x = 3

The table gives 5 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. A new observation (3, 57) is made. Find its residual e = y − ŷ.

n = 5(3, 57)
e =
15.Chi-square testsMedium
no defectminormajorTotalday shift28422898night shift18244183Total466669181

The contingency table shows observed counts. For a chi-square test of independence, find the expected count E for the cell “day shift, minor”.

row total = 98column total = 66N = 181
E =
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 4 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 = 0.05.

α = 0.05
χ²crit =

3Solutions

1.Research methods in psychology

Answer: A · So observers record the same behaviours consistently, improving reliability

Why this is the answer

Clear, observable categories (for example "hits with an open hand") reduce subjective judgement and raise inter-observer reliability.

Answer

A · So observers record the same behaviours consistently, improving reliability
2.Research methods in psychology

Answer: B · Order effects such as practice, because conditions were not counterbalanced

Why this is the answer

When all participants experience conditions in the same order, improvement may reflect practice rather than the list. Counterbalancing (ABBA or half AB, half BA) controls this.

Answer

B · Order effects such as practice, because conditions were not counterbalanced
3.Descriptive statistics

Answer: s = 13.8

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 = 216/6 = 36

Squared deviations

Σ(x − x̄)² = (−14)² + 0² + 10² + 20² + (−16)² + 0² = 952

Standard deviation

s = √(952/5) = 13.8

Answer

s = 13.8
4.Descriptive statistics

Answer: z = 1

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 = (65 − 50)/15 = 1

Answer

z = 1
5.Probability

Answer: N = 35

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.

Combinations

C(7, 3) = (7·6·5)/(3·2·1) = 35

Answer

N = 35
6.Probability

Answer: σY = 0.618

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.

Spread of X

μ = 1.65, Var(X) = 1.528, σX = 1.236

Linear change

σY = |0.5|·1.236 = 0.618

Answer

σY = 0.618

The mean does move: E(Y) = 0.5·1.65 + 20 = 20.82.

7.Normal distribution

Answer: P = 0.8716

Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.

z-scores

z₁ = (164 − 178)/7 = −2
z₂ = (186.75 − 178)/7 = 1.25

Area between

P = Φ(1.25) − Φ(−2) = 0.8944 − 0.02275 = 0.8716

Answer

P = 0.8716
8.Normal distribution

Answer: x = 43.76 min

Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.

z for the percentile

Φ(z) = 0.975 → z = 1.96

Back to x

x = 32 + 1.96·6 = 43.76 min

Answer

x = 43.76 min
9.Confidence intervals

Answer: lower limit = 9 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 = 40 − 29 = 11

Standard error

SE = √(7²/60 + 3²/40) = 1.021

Margin of error

E = 1.96·1.021 = 2

Interval

11 ± 2 → [9, 13]

Answer

lower limit = 9 h

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

10.Confidence intervals

Answer: n = 68

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 = (2.576·16/5)² = 67.95

Round up

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

n = 68

Answer

n = 68
11.Hypothesis tests & confidence intervals

Answer: t = −1.998

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 = 7.5/√15 = 1.936

Test statistic

t = −3.87/1.936 = −1.998

Answer

t = −1.998

With df = 14, the two-sided 5% critical value is 2.145; |t| is smaller, so H₀ is not rejected.

12.Hypothesis tests & confidence intervals

Answer: z = 3.078

Hypothesis test: the test statistic = (estimate − value under H₀)/(standard error under H₀). The p-value is the probability, assuming H₀ is true, of a statistic at least as extreme as the one observed: the area in one tail for a one-sided H₁, twice the tail area for a two-sided H₁. Reject H₀ when p ≤ α.

Sample proportions

p̂₁ = 88/250 = 0.352
p̂₂ = 31/150 = 0.2067
p̂ = 119/400 = 0.2975

Standard error under H₀

SE = √(0.2975·0.7025·(1/250 + 1/150)) = 0.04722

Test statistic

z = (0.352 − 0.2067)/0.04722 = 3.078

Answer

z = 3.078

|z| > 1.96, so H₀ is rejected at the 5% level (two-sided).

13.Linear regression & correlation

Answer: r = 0.9822

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 = 1556 − 33·267/6 = 87.5
Sxx = 199 − 33²/6 = 17.5
Syy = 12335 − 267²/6 = 453.5

Correlation coefficient

r = 87.5/√(17.5·453.5) = 0.9822

Answer

r = 0.9822
14.Linear regression & correlation

Answer: e = 1.8

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 = 30/5 = 6
ȳ = 366/5 = 73.2

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 240

Regression line

b = 240/40 = 6
a = 73.2 − 6·6 = 37.2
ŷ = 37.2 + 6·x

Predicted value

ŷ = 37.2 + 6·3 = 55.2

Residual

e = 57 − 55.2 = 1.8

Answer

e = 1.8

The point lies above the line.

15.Chi-square tests

Answer: E = 35.73

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 count

E = 98·66/181 = 35.73

Answer

E = 35.73
16.Chi-square tests

Answer: χ²crit = 7.815

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 = 4 − 1 = 3

Table

χ²crit(df = 3, α = 0.05) = 7.815

Answer

χ²crit = 7.815

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

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