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

A study is preregistered. This means that:

  1. AThe hypotheses, methods and analysis plan are recorded publicly before data collection
  2. BThe study has already been published
  3. CThe results were checked and approved by a second, independent statistician before publication
  4. DParticipants register their consent online
2.Research methods in psychologyHard

In thematic analysis, the analyst mainly:

  1. ATests a numerical hypothesis
  2. BCounts word frequencies only
  3. CCalculates the mean score of each theme to compare them statistically
  4. DIdentifies recurring patterns of meaning across qualitative data
3.Descriptive statisticsMedium
152025303540

Find the sample standard deviation s of the data: 15, 28, 31, 37, 40, 20.

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

A quantity is normally distributed with mean 170 and standard deviation 10. What proportion P of values is above 185?

μ = 170σ = 10x = 185
P =
5.ProbabilityMedium
xp(x)00.3510.220.2530.0540.15μ = 1.45

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

x: 0, 1, 2, 3, 4p(x): 0.35, 0.2, 0.25, 0.05, 0.15
Var(X) =
6.ProbabilityHard
n = 11, p = 0.5: P(X ≤ 1) = ?k01234567891011

11 components are tested, and each one fails with probability 0.5, independently. What is the probability P that at most one fails?

n = 11p = 0.5
P =
7.Normal distributionMedium
−3−2−10123zshaded area between −z and z = 0.2510

Drag z so that the shaded area between −z and z under the standard normal curve is 0.2510. Find z to two decimals.

A = 0.2510
z =
8.Normal distributionHard
4 s0X − Y?

Runner A's lap time X and runner B's lap time Y are independent and normal: X has mean 52 s and standard deviation 4 s, Y has mean 48 s and standard deviation 3 s. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 52 s, σX = 4 sμY = 48 s, σY = 3 s
P =
9.Confidence intervalsMedium
x̄E₁ = 6.799%, n = 100x̄E₂ = ?95%, n = 100

A 99% confidence interval for a mean, based on 100 observations and a known σ, has a margin of error of 6.7. What is the margin of error E₂ of a 95% interval from the same data?

99%n₁ = 100E₁ = 6.795%
E₂ =
10.Confidence intervalsHard
σ = 6, n = ?x̄ − Ex̄ + Ex̄E = 2.599%

The population standard deviation is 6. 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 2.5?

σ = 6E = 2.599%
n =
11.Hypothesis tests & confidence intervalsMedium
−2.2282.228t = ?df = 10

11 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 −8.88 and the standard deviation of the differences is 9.2. Find the paired test statistic t for H₀: μd = 0.

n = 11d̄ = −8.88sd = 9.2
t =
12.Hypothesis tests & confidence intervalsHard
H₀H₁μ₀x̄cμ₀ + 2α = 0.05βσ = 16, power 90%, n = ?

A one-sided z test at α = 0.05 must detect a shift of the mean of 2 units with a power of 90%. The standard deviation is 16. Find the minimum sample size n.

α = 0.05power = 90%δ = 2σ = 16zα = 1.645zβ = 1.282
n =
13.Linear regression & correlationMedium
xy3495597609701181xy51050607080

The table gives 5 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 = 5Σx = 35Σy = 319Σx² = 285Σy² = 20943Σxy = 2383
r =
14.Linear regression & correlationHard
xy155261371474574678xy246607080x = 5.5

The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 5.5.

n = 6x = 5.5
ŷ =
15.Chi-square testsMedium
lowmediumhighTotalurban32211568rural2681549Total582930117

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

row total = 49column total = 58N = 117
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 test of independence is carried out on a contingency table with 3 rows and 5 columns (not counting the totals). 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 · The hypotheses, methods and analysis plan are recorded publicly before data collection

Why this is the answer

Preregistration separates confirmatory from exploratory analyses and reduces practices such as p-hacking and HARKing.

Answer

A · The hypotheses, methods and analysis plan are recorded publicly before data collection
2.Research methods in psychology

Answer: D · Identifies recurring patterns of meaning across qualitative data

Why this is the answer

Thematic analysis involves familiarisation, coding, generating, reviewing and naming themes and writing up, focusing on meaning rather than statistics.

Answer

D · Identifies recurring patterns of meaning across qualitative data
3.Descriptive statistics

Answer: s = 9.649

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 = 171/6 = 28.5

Squared deviations

Σ(x − x̄)² = (−13.5)² + (−0.5)² + 2.5² + 8.5² + 11.5² + (−8.5)² = 465.5

Standard deviation

s = √(465.5/5) = 9.649

Answer

s = 9.649
4.Descriptive statistics

Answer: P = 0.06681

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 = 1.5

Standard normal table

Φ(1.5) = 0.9332
P = 1 − 0.9332 = 0.06681

Answer

P = 0.06681
5.Probability

Answer: Var(X) = 1.947

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 + 1·0.2 + 2·0.25 + 3·0.05 + 4·0.15 = 1.45

Mean of the squares

E(X²) = 0·0.35 + 1·0.2 + 4·0.25 + 9·0.05 + 16·0.15 = 4.05

Variance

Var(X) = 4.05 − 1.45² = 1.947

Answer

Var(X) = 1.947
6.Probability

Answer: P = 0.005859

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.

Zero and one failure

P(X = 0) = 0.511 = 0.0004883
P(X = 1) = 11·0.5·0.510 = 0.005371

Add them

P = 0.0004883 + 0.005371 = 0.005859

Answer

P = 0.005859
7.Normal distribution

Answer: z = 0.32

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 + 0.2510)/2 = 0.6255

Read the z-table backwards

Find 0.6255 inside the table and read off z.

z = 0.32

Answer

z = 0.32
8.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 difference

μ = 52 − 48 = 4 s, σ = 5 s

Standardise

z = (0 − 4)/5 = −0.8

Probability

P = 1 − Φ(−0.8) = 0.7881

Answer

P = 0.7881
9.Confidence intervals

Answer: E₂ = 5.098

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.

Ratio of margins

E₂/E₁ = (1.96/2.576)·√(100/100) = 0.7609

New margin

E₂ = 6.7·0.7609 = 5.098

Answer

E₂ = 5.098
10.Confidence intervals

Answer: n = 39

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 = (2.576·6/2.5)² = 38.22

Round up

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

n = 39

Answer

n = 39
11.Hypothesis tests & confidence intervals

Answer: t = −3.201

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 = 9.2/√11 = 2.774

Test statistic

t = −8.88/2.774 = −3.201

Answer

t = −3.201

With df = 10, the two-sided 5% critical value is 2.228; |t| is larger, so H₀ is rejected.

12.Hypothesis tests & confidence intervals

Answer: n = 549

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 + 1.282)·16/2)² = 548.3

Round up

n = 549

Answer

n = 549
13.Linear regression & correlation

Answer: r = 0.9758

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 = 2383 − 35·319/5 = 150
Sxx = 285 − 35²/5 = 40
Syy = 20943 − 319²/5 = 590.8

Correlation coefficient

r = 150/√(40·590.8) = 0.9758

Answer

r = 0.9758
14.Linear regression & correlation

Answer: ŷ = 77.8

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 = 21/6 = 3.5
ȳ = 413/6 = 68.83

Sums of squares and products

Sxx = Σ(x − x̄)² = 17.5
Sxy = Σ(x − x̄)(y − ȳ) = 78.5

Regression line

b = 78.5/17.5 = 4.486
a = 68.83 − 4.486·3.5 = 53.13
ŷ = 53.13 + 4.486·x

Prediction

ŷ = 53.13 + 4.486·5.5 = 77.8

Answer

ŷ = 77.8
15.Chi-square tests

Answer: E = 24.29

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 = 49·58/117 = 24.29

Answer

E = 24.29
16.Chi-square tests

Answer: χ²crit = 15.51

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 = (3 − 1)·(5 − 1) = 8

Table

χ²crit(df = 8, α = 0.05) = 15.507

Answer

χ²crit = 15.507

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

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