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)·p^k(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

Convert the value to a z-score: z = (x − μ)/σ.
Look up Φ(z), the area to the left of z.
For the area to the right use 1 − Φ(z); for an interval subtract the two left areas.
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

Decide between z (population σ known) and t (only the sample s, with n − 1 degrees of freedom).
Find the critical value for the confidence level.
Margin of error: E = critical value × standard deviation/√n.
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

State H₀ and H₁ and decide whether the test is one- or two-sided.
Calculate the test statistic: (x̄ − μ₀)/(σ/√n), or with s and t.
Compare with the critical value, or find the p-value.
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: S_xx = Σ(x − x̄)² = Σx² − (Σx)²/n, S_xy = Σxy − ΣxΣy/n
Slope: b = S_xy/S_xx
Intercept: a = ȳ − b·x̄
Correlation: r = S_xy/√(S_xxS_yy)
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 > χ²_crit

2Practice problems

1.Research methods in psychologyMedium

Which kind of data is collected with an open question in an interview?

ANominal quantitative data only
BQualitative data
CInterval data
DRatio data

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?

AToo large a sample size
BOrder effects such as practice, because conditions were not counterbalanced
CThe use of an ordinal level of measurement
DIndividual differences between the groups

3.Descriptive statisticsMedium

Find the sample standard deviation s of the data: 48, 59, 37, 17, 53, 27, 56.

n = 7

s =

4.Descriptive statisticsHard

Exam scores are normally distributed with mean 60 and standard deviation 8. Find the z-score z of a score of 68.

μ = 60σ = 8x = 68

z =

5.ProbabilityMedium

A bag holds 6 red and 6 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

6 / 6

P =

6.ProbabilityHard

A game costs €3 to play. You win €12 with probability 0.25 and €30 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €3€12: p = 0.25€30: p = 0.05

E(G) = €

7.Normal distributionMedium

The IQ scores are normally distributed with mean 100 points and standard deviation 15 points. What fraction lies between 78.25 and 107.5 points? Give P as a decimal.

μ = 100σ = 15x₁ = 78.25x₂ = 107.5

P =

8.Normal distributionHard

The bag weights are normally distributed with mean 250 g and standard deviation 5 g. Find the value x below which 97.5% of them lie (the 97.5th percentile).

μ = 250σ = 5P = 97.5%

x = g

9.Confidence intervalsMedium

A sample of 8 items has mean 50.5 and standard deviation s = 1.5. The population is roughly normal. Find the upper limit of the 95% confidence interval for the mean.

n = 8x̄ = 50.5s = 1.595%t(7) = 2.365

upper limit =

10.Confidence intervalsHard

A pilot study suggests that about 0.2 of products are defective. How many products n must be inspected so that a 95% confidence interval for the proportion has a margin of error of at most 0.02?

p ≈ 0.2E = 0.0295%

n =

11.Hypothesis tests & confidence intervalsMedium

10 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 −1.67 and the standard deviation of the differences is 3.3. Find the paired test statistic t for H₀: μ_d = 0.

n = 10d̄ = −1.67s_d = 3.3

t =

12.Hypothesis tests & confidence intervalsHard

A sample of 64 measurements has mean 498.9. The population standard deviation is 4. Find the upper limit U of the 99% confidence interval for the mean.

x̄ = 498.9σ = 4n = 6499%

U =

13.Linear regression & correlationMedium

The table gives 6 paired observations of x = car age in years and y = price in k€. Find the coefficient of determination r² of the least-squares line.

n = 6

r² =

14.Linear regression & correlationHard

The table gives 6 paired observations of x = hours studied and y = exam score. Use the least-squares line ŷ = a + b·x. A new observation (4, 53) is made. Find its residual e = y − ŷ.

n = 6(4, 53)

e =

15.Chi-square testsMedium

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

A chi-square test of independence is carried out on a contingency table with 2 rows and 6 columns (not counting the totals). How many degrees of freedom df does the test have?

2 × 6

df =

3Solutions

1.Research methods in psychology

Answer: B · Qualitative data

Why this is the answer

Open questions produce rich qualitative data that is harder to analyse statistically than closed-question responses.

Answer

B · Qualitative data

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 = 15.89

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 = 297/7 = 42.43

Squared deviations

Σ(x − x̄)² = 5.571² + 16.57² + (−5.429)² + (−25.43)² + 10.57² + (−15.43)² + 13.57² = 1516

Standard deviation

s = √(1516/6) = 15.89

Answer

s = 15.89

4.Descriptive statistics

Answer: z = 1

Standardise

z = (68 − 60)/8 = 1

Answer

z = 1

5.Probability

Answer: P = 0.2273

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.

First and second draw

P = 6/12 · 5/11 = 30/132 = 0.2273

Answer

P = 0.2273

6.Probability

Answer: E(G) = 1.5 €

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 σ_Y = |a|·σ_X; adding a constant does not change the spread.

Net outcomes

G: 9 (p = 0.25), 27 (p = 0.05), −3 (p = 0.7)

Expected value

E(G) = 9·0.25 + 27·0.05 − 3·0.7 = 1.5 €

Answer

E(G) = 1.5 €

A positive value means you win money on average.

7.Normal distribution

Answer: P = 0.6179

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 = (78.25 − 100)/15 = −1.45

z₂ = (107.5 − 100)/15 = 0.5

Area between

P = Φ(0.5) − Φ(−1.45) = 0.6915 − 0.07353 = 0.6179

Answer

P = 0.6179

8.Normal distribution

Answer: x = 259.8 g

z for the percentile

Φ(z) = 0.975 → z = 1.96

Back to x

x = 250 + 1.96·5 = 259.8 g

Answer

x = 259.8 g

9.Confidence intervals

Answer: upper limit = 51.75

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 = 7, t = 2.365

Margin of error

E = 2.365·1.5/√8 = 1.254

Interval

50.5 ± 1.254 → [49.25, 51.75]

Answer

upper limit = 51.75

10.Confidence intervals

Answer: n = 1537

Confidence interval for a proportion: p̂ ± z·√(p̂(1 − p̂)/n), where p̂ = x/n. To reach a margin E you need n ≥ z²·p(1 − p)/E², rounded up; without a prior estimate use p = 0.5, which gives the largest n.

Solve for n

n = 1.96²·0.2·0.8/0.02² = 1537

Round up

n = 1537

Answer

n = 1537

11.Hypothesis tests & confidence intervals

Answer: t = −1.6

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 = 3.3/√10 = 1.044

Test statistic

t = −1.67/1.044 = −1.6

Answer

t = −1.6

With df = 9, the two-sided 5% critical value is 2.262; |t| is smaller, so H₀ is not rejected.

12.Hypothesis tests & confidence intervals

Answer: U = 500.2

Margin of error

E = 2.576·4/√64 = 1.288

Interval limit

U = 498.9 + 1.288 = 500.2

Answer

U = 500.2

13.Linear regression & correlation

Answer: r² = 0.9295

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 = 113/6 = 18.83

Sums of squares and products

S_xx = Σ(x − x̄)² = 17.5

S_xy = Σ(x − x̄)(y − ȳ) = −40.5

S_yy = Σ(y − ȳ)² = 100.8

Coefficient of determination

r² = −40.5²/(17.5·100.8) = 0.9295

Answer

r² = 0.9295

About 92.95% of the variation in y is explained by the line.

14.Linear regression & correlation

Answer: e = −5.005

Means

x̄ = 48/6 = 8

ȳ = 419/6 = 69.83

Sums of squares and products

S_xx = Σ(x − x̄)² = 70

S_xy = Σ(x − x̄)(y − ȳ) = 207

Regression line

b = 207/70 = 2.957

a = 69.83 − 2.957·8 = 46.18

ŷ = 46.18 + 2.957·x

Predicted value

ŷ = 46.18 + 2.957·4 = 58

Residual

e = 53 − 58 = −5.005

Answer

e = −5.005

The point lies below the line.

15.Chi-square tests

Answer: χ² = 1.607

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

smokers: 80·84/120 = 56, 80·36/120 = 24

non-smokers: 40·84/120 = 28, 40·36/120 = 12

Chi-square statistic

χ² = (53 − 56)²/56 + (27 − 24)²/24 + (31 − 28)²/28 + (9 − 12)²/12

χ² = 1.607

Answer

χ² = 1.607

df = 1. At α = 0.05 the critical value is 3.841, so H₀ (independence) is not rejected.

16.Chi-square tests

Answer: df = 5

Degrees of freedom

df = (2 − 1)·(6 − 1) = 5

Answer

df = 5

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