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

In multiple regression predicting exam scores from study hours and sleep, R² = 0.36. What does this mean?

  1. AStudy hours cause 36% of exam success
  2. BThe predictors together explain 36% of the variance in exam scores
  3. CThe correlation between the predictors is 0.36
  4. D36% of students passed the exam
3.Descriptive statisticsMedium
2025303540455055

Find the sample standard deviation s of the data: 25, 57, 27, 39, 17, 46, 22.

n = 7
s =
4.Descriptive statisticsHard
z −2−1012

A quantity is normally distributed with mean 500 and standard deviation 60. What proportion P of values is below 380?

μ = 500σ = 60x = 380
P =
5.ProbabilityMedium
xp(x)00.310.320.130.0540.25μ = 1.65

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

x: 0, 1, 2, 3, 4p(x): 0.3, 0.3, 0.1, 0.05, 0.25
σ =
6.ProbabilityHard
xp(x)10.220.1530.0540.3550.25μ

X has the distribution shown, and Y = 2X + 4. Find the standard deviation σY of Y.

x: 1, 2, 3, 4, 5p(x): 0.2, 0.15, 0.05, 0.35, 0.25Y = 2X + 4
σY =
7.Normal distributionMedium
174.85183.6P = ?

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 174.85 and 183.6 cm? Give P as a decimal.

μ = 178σ = 7x1 = 174.85x2 = 183.6
P =
8.Normal distributionHard
0.02 mm0D − d?

Hole diameters D are normal with mean 25.02 mm and standard deviation 0.01 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.015 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.02 mm, σ = 0.01 mmshaft: μ = 25 mm, σ = 0.015 mm
P =
9.Confidence intervalsMedium
n = 7, s = 2LU = ?x̄ = 38.5E99%

A sample of 7 items has mean 38.5 and standard deviation s = 2. The population is roughly normal. Find the upper limit of the 99% confidence interval for the mean.

n = 7x̄ = 38.5s = 299%t(6) = 3.707
upper limit =
10.Confidence intervalsHard
σ = 31, n = ?x̄ − Ex̄ + Ex̄E = 495%

The population standard deviation is 31. 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 4?

σ = 31E = 495%
n =
11.Hypothesis tests & confidence intervalsMedium
−2.0932.093t = ?df = 19

Two teaching methods are compared. Group 1: 6 students, mean score 86, standard deviation 12. Group 2: 15 students, mean score 80.6, standard deviation 2.5. Assume equal population variances and use the pooled two-sample t test. Find the test statistic t for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (group 1 minus group 2).

n₁ = 6x̄₁ = 86s₁ = 12n₂ = 15x̄₂ = 80.6s₂ = 2.5
t =
12.Hypothesis tests & confidence intervalsHard
σ = 12, n = ?x̄ − Ex̄ + Ex̄E = 2.595%

How large must a sample be so that the 95% confidence interval for a mean has a margin of error of at most 2.5? The standard deviation is 12. Find the smallest sample size n.

σ = 12E = 2.595%
n =
13.Linear regression & correlationMedium
xy17902011423124261462916432178xy202530100150

The table gives 6 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 = 6Σx = 147Σy = 816Σx² = 3759Σy² = 116368Σxy = 20910
r =
14.Linear regression & correlationHard
xy131229327422525623xy2462530x = 1.5

The table gives 6 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. A new observation (1.5, 33) is made. Find its residual e = y − ŷ.

n = 6(1.5, 33)
e =
15.Chi-square testsMedium
busbikecarTotalstudents31172068staff17174377Total483463145

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

row total = 77column total = 63N = 145
E =
16.Chi-square testsHard

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). How many degrees of freedom df does the test have?

k = 4
df =

3Solutions

1.Research methods in psychology

Answer: B · 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

B · Defining variables in a precise, measurable way
2.Research methods in psychology

Answer: B · The predictors together explain 36% of the variance in exam scores

Why this is the answer

R² shows how much of the variability in the outcome is accounted for by all predictors together. It does not establish causation.

Answer

B · The predictors together explain 36% of the variance in exam scores
3.Descriptive statistics

Answer: s = 14.48

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 = 233/7 = 33.29

Squared deviations

Σ(x − x̄)² = (−8.286)² + 23.71² + (−6.286)² + 5.714² + (−16.29)² + 12.71² + (−11.29)² = 1257

Standard deviation

s = √(1257/6) = 14.48

Answer

s = 14.48
4.Descriptive statistics

Answer: P = 0.02275

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 = −2

Standard normal table

Φ(−2) = 0.02275
P = 0.02275

Answer

P = 0.02275
5.Probability

Answer: σ = 1.558

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 + 1·0.3 + 2·0.1 + 3·0.05 + 4·0.25 = 1.65

Mean of the squares

E(X²) = 0·0.3 + 1·0.3 + 4·0.1 + 9·0.05 + 16·0.25 = 5.15

Variance

Var(X) = 5.15 − 1.65² = 2.428
σ = √2.428 = 1.558

Answer

σ = 1.558
6.Probability

Answer: σY = 2.973

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 = 3.3, Var(X) = 2.21, σX = 1.487

Linear change

σY = |2|·1.487 = 2.973

Answer

σY = 2.973

The mean does move: E(Y) = 2·3.3 + 4 = 10.6.

7.Normal distribution

Answer: P = 0.4618

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 = (174.85 − 178)/7 = −0.45
z₂ = (183.6 − 178)/7 = 0.8

Area between

P = Φ(0.8) − Φ(−0.45) = 0.7881 − 0.3264 = 0.4618

Answer

P = 0.4618
8.Normal distribution

Answer: P = 0.8665

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 = 25.02 − 25 = 0.02 mm, σ = 0.01803 mm

Standardise

z = (0 − 0.02)/0.01803 = −1.11

Probability

P = 1 − Φ(−1.11) = 0.8665

Answer

P = 0.8665
9.Confidence intervals

Answer: upper limit = 41.3

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 = 6, t = 3.707

Margin of error

E = 3.707·2/√7 = 2.802

Interval

38.5 ± 2.802 → [35.7, 41.3]

Answer

upper limit = 41.3
10.Confidence intervals

Answer: n = 231

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 = (1.96·31/4)² = 230.7

Round up

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

n = 231

Answer

n = 231
11.Hypothesis tests & confidence intervals

Answer: t = 1.715

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.

Pooled variance

sp² = (5·12² + 14·2.5²)/19 = 42.5
sp = 6.519

Standard error

SE = 6.519·√(1/6 + 1/15) = 3.149

Test statistic

t = (86 − 80.6)/3.149 = 1.715

Answer

t = 1.715

With df = 19, the two-sided 5% critical value is 2.093; |t| is smaller, so H₀ is not rejected.

12.Hypothesis tests & confidence intervals

Answer: n = 89

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).

Solve for n

n = (1.96·12/2.5)² = 88.51

Round up

n = 89

Answer

n = 89
13.Linear regression & correlation

Answer: r = 0.9962

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 = 20910 − 147·816/6 = 918
Sxx = 3759 − 147²/6 = 157.5
Syy = 116368 − 816²/6 = 5392

Correlation coefficient

r = 918/√(157.5·5392) = 0.9962

Answer

r = 0.9962
14.Linear regression & correlation

Answer: e = 3.576

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 = 21/6 = 3.5
ȳ = 157/6 = 26.17

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = −28.5

Regression line

b = −28.5/17.5 = −1.629
a = 26.17 − (−1.629)·3.5 = 31.87
ŷ = 31.87 − 1.629·x

Predicted value

ŷ = 31.87 + (−1.629)·1.5 = 29.42

Residual

e = 33 − 29.42 = 3.576

Answer

e = 3.576

The point lies above the line.

15.Chi-square tests

Answer: E = 33.46

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 = 77·63/145 = 33.46

Answer

E = 33.46
16.Chi-square tests

Answer: df = 3

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

Answer

df = 3

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