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.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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

Experimenter bias is reduced most directly by:

  1. AUsing a researcher who does not know the hypothesis or conditions to collect data
  2. BUsing a one-tailed hypothesis
  3. CUsing a larger sample
  4. DUsing only volunteers
2.Research methods in psychologyHard

Interpretative phenomenological analysis (IPA) mainly aims to:

  1. ACount the frequency of words in texts
  2. BExplore how individuals make sense of their own lived experiences
  3. CTest causal hypotheses with large samples
  4. DMeasure reaction times accurately
3.Descriptive statisticsMedium
354045505560

Find the sample standard deviation s of the data: 35, 43, 57, 55, 32, 60.

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

A quantity is normally distributed with mean 100 and standard deviation 10. What proportion P of values is below 110?

μ = 100σ = 10x = 110
P =
5.ProbabilityMedium
xp(x)10.3520.1530.1540.2550.1μ = 2.6

For the distribution shown, find the variance Var(X) of X.

x: 1, 2, 3, 4, 5p(x): 0.35, 0.15, 0.15, 0.25, 0.1
Var(X) =
6.ProbabilityHard
n = 6, p = 0.6: P(X = 5) = ?k0123456

A test has 6 multiple-choice questions. A student who guesses gets each one right with probability 0.6, independently. What is the probability P of exactly 5 correct answers?

n = 6p = 0.6k = 5
P =
7.Normal distributionMedium
750 g742.1 gX + Y?

X is normal with mean 250 g and standard deviation 4 g; Y is independent and normal with mean 500 g and standard deviation 9 g. Find the probability P that X + Y is more than 742.1 g. Round z to two decimals.

μX = 250, σX = 4μY = 500, σY = 9X + Y > 742.1 g
P =
8.Normal distributionHard
μ = 500x = ?99%

The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 99% of them lie (the 99th percentile).

μ = 500σ = 40P = 99%
x = h
9.Confidence intervalsMedium
n = 9, s = 13.5LUx̄ = 51.5E = ?90%

A sample of 9 items has mean 51.5 and standard deviation s = 13.5. The population is roughly normal. Find the margin of error E of a 90% confidence interval for the mean.

n = 9x̄ = 51.5s = 13.590%t(8) = 1.86
E =
10.Confidence intervalsHard
μ₁ − μ₂, t(14) = 2.1450LUx̄₁ − x̄₂E = ?95%

Two fertilisers are tested. Plot group 1: 6 plots, mean yield 83.5 kg, standard deviation 6 kg. Plot group 2: 10 plots, mean 74.5 kg, standard deviation 7 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the margin of error E of the 95% confidence interval for μ₁ − μ₂.

n₁ = 6x̄₁ = 83.5s₁ = 6n₂ = 10x̄₂ = 74.5s₂ = 795%t(14) = 2.145
E = kg
11.Hypothesis tests & confidence intervalsMedium
−1.961.96z = ?

Two production lines make the same part. Line 1: a sample of 80 parts has mean length 67.5 mm, and the line's known standard deviation is 4 mm. Line 2: a sample of 30 parts has mean 64.8 mm, known standard deviation 6 mm. Find the test statistic z for H₀: μ₁ = μ₂, taking the difference as x̄₁ − x̄₂ (line 1 minus line 2).

n₁ = 80x̄₁ = 67.5 mmσ₁ = 4 mmn₂ = 30x̄₂ = 64.8 mmσ₂ = 6 mm
z =
12.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 3α = 0.01βσ = 15, power 90%, n = ?

A one-sided z test at α = 0.01 must detect a shift of the mean of 3 units with a power of 90%. The standard deviation is 15. Find the minimum sample size n.

α = 0.01power = 90%δ = 3σ = 15zα = 2.326zβ = 1.282
n =
13.Linear regression & correlationMedium
xy17541956217323812580xy2025607080

The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Find the coefficient of determination r² of the least-squares line.

n = 5
r² =
14.Linear regression & correlationHard
xy243349455566661xy2465060ŷ = 57

The table gives 5 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 = 57?

n = 5ŷ = 57
x =
15.Chi-square testsMedium
improvednot improvedTotalnew drug174360placebo312960Total4872120

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 = 1202 × 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 10 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 · Using a researcher who does not know the hypothesis or conditions to collect data

Why this is the answer

Experimenters who know the hypothesis may unintentionally influence participants or recordings. Single- or double-blind procedures and standardised instructions reduce this.

Answer

A · Using a researcher who does not know the hypothesis or conditions to collect data
2.Research methods in psychology

Answer: B · Explore how individuals make sense of their own lived experiences

Why this is the answer

IPA studies small, homogeneous samples in depth, examining how participants interpret significant experiences, with the researcher interpreting their sense-making.

Answer

B · Explore how individuals make sense of their own lived experiences
3.Descriptive statistics

Answer: s = 11.98

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 = 282/6 = 47

Squared deviations

Σ(x − x̄)² = (−12)² + (−4)² + 10² + 8² + (−15)² + 13² = 718

Standard deviation

s = √(718/5) = 11.98

Answer

s = 11.98
4.Descriptive statistics

Answer: P = 0.8413

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 = 1

Standard normal table

Φ(1) = 0.8413
P = 0.8413

Answer

P = 0.8413
5.Probability

Answer: Var(X) = 2.04

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 = 1·0.35 + 2·0.15 + 3·0.15 + 4·0.25 + 5·0.1 = 2.6

Mean of the squares

E(X²) = 1·0.35 + 4·0.15 + 9·0.15 + 16·0.25 + 25·0.1 = 8.8

Variance

Var(X) = 8.8 − 2.6² = 2.04

Answer

Var(X) = 2.04
6.Probability

Answer: P = 0.1866

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 = 5) = 6·0.65·0.41 = 6·0.07776·0.4 = 0.1866

Answer

P = 0.1866
7.Normal distribution

Answer: P = 0.7881

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 sum

μ = 750 g, σ = √(4² + 9²) = 9.849 g

Standardise

z = (742.1 − 750)/9.849 = −0.8

Upper tail

P = 1 − Φ(−0.8) = 1 − 0.2119 = 0.7881

Answer

P = 0.7881
8.Normal distribution

Answer: x = 593.1 h

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) = 0.99 → z = 2.326

Back to x

x = 500 + 2.326·40 = 593.05 h

Answer

x = 593.05 h
9.Confidence intervals

Answer: E = 8.37

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 = 8, t = 1.86

Margin of error

E = 1.86·13.5/√9 = 8.37

Answer

E = 8.37
10.Confidence intervals

Answer: E = 7.377 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 = (5·6² + 9·7²)/14 = 44.36
sp = 6.66

Margin of error

E = 2.145·6.66·√(1/6 + 1/10) = 7.377

Answer

E = 7.377 kg
11.Hypothesis tests & confidence intervals

Answer: z = 2.282

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.

Standard error of the difference

SE = √(4²/80 + 6²/30) = √1.4 = 1.183

Test statistic

z = (67.5 − 64.8)/1.183 = 2.282

Answer

z = 2.282

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

12.Hypothesis tests & confidence intervals

Answer: n = 326

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 + 1.282)·15/3)² = 325.4

Round up

n = 326

Answer

n = 326
13.Linear regression & correlation

Answer: r² = 0.8786

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 = 105/5 = 21
ȳ = 344/5 = 68.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 154
Syy = Σ(y − ȳ)² = 674.8

Coefficient of determination

r² = 154²/(40·674.8) = 0.8786

Answer

r² = 0.8786

About 87.86% of the variation in y is explained by the line.

14.Linear regression & correlation

Answer: x = 4.415

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 = 20/5 = 4
ȳ = 274/5 = 54.8

Sums of squares and products

Sxx = Σ(x − x̄)² = 10
Sxy = Σ(x − x̄)(y − ȳ) = 53

Regression line

b = 53/10 = 5.3
a = 54.8 − 5.3·4 = 33.6
ŷ = 33.6 + 5.3·x

Solve for x

x = (57 − 33.6)/5.3 = 4.415

Answer

x = 4.415
15.Chi-square tests

Answer: χ² = 6.806

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: 60·48/120 = 24, 60·72/120 = 36
placebo: 60·48/120 = 24, 60·72/120 = 36

Chi-square statistic

χ² = (17 − 24)²/24 + (43 − 36)²/36 + (31 − 24)²/24 + (29 − 36)²/36
χ² = 6.806

Answer

χ² = 6.806

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

16.Chi-square tests

Answer: χ²crit = 21.67

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 = 10 − 1 = 9

Table

χ²crit(df = 9, α = 0.01) = 21.666

Answer

χ²crit = 21.666

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

ExamProblems · examproblems.com/courses/research-methods-psychology