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

Demand characteristics are:

  1. AStatistical requirements for a test
  2. BThe formal demands an ethics committee makes before a study can begin
  3. CCharacteristics of the sample
  4. DCues that make participants guess the aim and change their behaviour
2.Research methods in psychologyHard

In qualitative research, reflexivity means that the researcher:

  1. AReflects on how their own background and assumptions may shape the analysis
  2. BConverts themes into numerical data
  3. CRepeats the study to check reliability
  4. DUses a random sample to remove bias
3.Descriptive statisticsMedium
10152025303540455055

Find the sample standard deviation s of the data: 10, 40, 54, 33, 59.

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

Exam scores are normally distributed with mean 70 and standard deviation 20. Find the z-score z of a score of 100.

μ = 70σ = 20x = 100
z =
5.ProbabilityMedium
n = 9k = 2N = ?

In how many ways N can a committee of 2 people be chosen from a group of 9? The order does not matter.

n = 9k = 2
N =
6.ProbabilityHard
xp(x)10.120.3530.240.35μ

X has the distribution shown, and Y = 5X − 3. Find the standard deviation σY of Y.

x: 1, 2, 3, 4p(x): 0.1, 0.35, 0.2, 0.35Y = 5X − 3
σY =
7.Normal distributionMedium
−3−2−10123zshaded area to the right of z = 0.9505

Drag z so that the shaded area under the standard normal curve, to the right of z, is 0.9505. Find z to two decimals.

A = 0.9505
z =
8.Normal distributionHard
μ = 65x = ?95%

The exam scores are normally distributed with mean 65 points and standard deviation 10 points. Find the value x below which 95% of them lie (the 95th percentile).

μ = 65σ = 10P = 95%
x = points
9.Confidence intervalsMedium
n = 7, s = 2.5LUx̄ = 102E = ?90%

A sample of 7 items has mean 102 and standard deviation s = 2.5. The population is roughly normal. Find the margin of error E of a 90% confidence interval for the mean.

n = 7x̄ = 102s = 2.590%t(6) = 1.943
E =
10.Confidence intervalsHard
μ₁ − μ₂, t(15) = 1.7530L = ?Ux̄₁ − x̄₂E90%

Two fertilisers are tested. Plot group 1: 11 plots, mean yield 73.5 kg, standard deviation 8.5 kg. Plot group 2: 6 plots, mean 68.5 kg, standard deviation 3.5 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the lower limit of the 90% confidence interval for μ₁ − μ₂.

n₁ = 11x̄₁ = 73.5s₁ = 8.5n₂ = 6x̄₂ = 68.5s₂ = 3.590%t(15) = 1.753
lower limit = kg
11.Hypothesis tests & confidence intervalsMedium
t −2.06402.064

A sample of 25 students has a mean score of 198.6 and a sample standard deviation of 8. Find the test statistic t for H₀: μ = 200.

μ₀ = 200x̄ = 198.6s = 8n = 25
t =
12.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 4α = 0.05βσ = 26, power 80%, n = ?

A one-sided z test at α = 0.05 must detect a shift of the mean of 4 units with a power of 80%. The standard deviation is 26. Find the minimum sample size n.

α = 0.05power = 80%δ = 4σ = 26zα = 1.645zβ = 0.842
n =
13.Linear regression & correlationMedium
xy3285387439521159xy51030405060

The table gives 5 paired observations of x = advertising spend in k€ and y = sales in k€. Find Pearson's correlation coefficient r.

n = 5
r =
14.Linear regression & correlationHard
xy2624706888931011412117xy5106080100120ŷ = 114

The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 114?

n = 6ŷ = 114
x =
15.Chi-square testsMedium
improvednot improvedTotalnew drug332760placebo122840Total4555100

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 = 1002 × 2
χ² =
16.Chi-square testsHard

A chi-square goodness-of-fit test checks whether counts in 7 categories follow a Poisson distribution whose mean is estimated from the same data. How many degrees of freedom df does the test have?

k = 7
df =

3Solutions

1.Research methods in psychology

Answer: D · Cues that make participants guess the aim and change their behaviour

Why this is the answer

When participants work out the hypothesis they may act to confirm or sabotage it, threatening validity. Deception and blind procedures reduce this.

Answer

D · Cues that make participants guess the aim and change their behaviour
2.Research methods in psychology

Answer: A · Reflects on how their own background and assumptions may shape the analysis

Why this is the answer

Reflexivity acknowledges the researcher's influence on data collection and interpretation, and is a key quality criterion in qualitative work.

Answer

A · Reflects on how their own background and assumptions may shape the analysis
3.Descriptive statistics

Answer: s = 19.38

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 = 196/5 = 39.2

Squared deviations

Σ(x − x̄)² = (−29.2)² + 0.8² + 14.8² + (−6.2)² + 19.8² = 1503

Standard deviation

s = √(1503/4) = 19.38

Answer

s = 19.38
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 = (100 − 70)/20 = 1.5

Answer

z = 1.5
5.Probability

Answer: N = 36

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(9, 2) = (9·8)/(2·1) = 36

Answer

N = 36
6.Probability

Answer: σY = 5.148

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 = 2.8, Var(X) = 1.06, σX = 1.03

Linear change

σY = |5|·1.03 = 5.148

Answer

σY = 5.148

The mean does move: E(Y) = 5·2.8 − 3 = 11.

7.Normal distribution

Answer: z = −1.65

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 − 0.9505 = 0.0495

Read the z-table backwards

Find 0.0495 inside the table and read off z.

z = −1.65

Answer

z = −1.65
8.Normal distribution

Answer: x = 81.45 points

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) = 0.95 → z = 1.645

Back to x

x = 65 + 1.645·10 = 81.45 points

Answer

x = 81.45 points
9.Confidence intervals

Answer: E = 1.836

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 = 1.943

Margin of error

E = 1.943·2.5/√7 = 1.836

Answer

E = 1.836
10.Confidence intervals

Answer: lower limit = −1.431 kg

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.

Pooled standard deviation

sp² = (10·8.5² + 5·3.5²)/15 = 52.25
sp = 7.228

Margin of error

E = 1.753·7.228·√(1/11 + 1/6) = 6.431

Interval

5 ± 6.431 → [−1.431, 11.43]

Answer

lower limit = −1.431 kg
11.Hypothesis tests & confidence intervals

Answer: t = −0.875

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 = 8/√25 = 8/5 = 1.6

Test statistic

t = (198.6 − 200)/1.6 = −0.875

Answer

t = −0.875
t0.025; df = 24 = 2.064

|t| < 2.064, so H₀ is not rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: n = 262

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 = ((1.645 + 0.842)·26/4)² = 261.3

Round up

n = 262

Answer

n = 262
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.

Means

x̄ = 35/5 = 7
ȳ = 220/5 = 44

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 152
Syy = Σ(y − ȳ)² = 582

Correlation coefficient

r = 152/√(40·582) = 0.9962

Answer

r = 0.9962
14.Linear regression & correlation

Answer: x = 10.96

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 = 544/6 = 90.67

Sums of squares and products

Sxx = Σ(x − x̄)² = 70
Sxy = Σ(x − x̄)(y − ȳ) = 412

Regression line

b = 412/70 = 5.886
a = 90.67 − 5.886·7 = 49.47
ŷ = 49.47 + 5.886·x

Solve for x

x = (114 − 49.47)/5.886 = 10.96

Answer

x = 10.96
15.Chi-square tests

Answer: χ² = 6.061

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

new drug: 60·45/100 = 27, 60·55/100 = 33
placebo: 40·45/100 = 18, 40·55/100 = 22

Chi-square statistic

χ² = (33 − 27)²/27 + (27 − 33)²/33 + (12 − 18)²/18 + (28 − 22)²/22
χ² = 6.061

Answer

χ² = 6.061

df = 1. At α = 0.05 the critical value is 3.841, so H₀ (independence) is rejected.

16.Chi-square tests

Answer: df = 5

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 = 7 − 1 − 1 = 5

Answer

df = 5

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