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 problems Back 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)·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

The same 25 participants complete a memory test before and after caffeine. The scores are interval and roughly normal. Which test is appropriate?

AChi-square test
BPaired (related) samples t-test
CMann–Whitney U test
DIndependent samples t-test

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?

AIndividual differences between the groups
BToo large a sample size
COrder effects such as practice, because conditions were not counterbalanced
DThe use of an ordinal level of measurement

3.Descriptive statisticsMedium

Find the sample standard deviation s of the data: 30, 15, 30, 20, 58, 15.

n = 6

s =

4.Descriptive statisticsHard

A quantity is normally distributed with mean 500 and standard deviation 40. What proportion P of values is above 516?

μ = 500σ = 40x = 516

P =

5.ProbabilityMedium

A bag holds 4 red and 7 blue balls. You draw two balls without putting the first one back. What is the probability P that both are red?

4 / 7

P =

6.ProbabilityHard

A game costs €10 to play. You win €20 with probability 0.25 and €500 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €10€20: p = 0.25€500: p = 0.05

E(G) = €

7.Normal distributionMedium

The masses of bags of cement are independent and normal with mean 25 kg and standard deviation 0.5 kg. Find the probability P that the total mass of 5 of them exceeds 125.6 kg. Round z to two decimals.

μ = 25 kgσ = 0.5 kgn = 5> 125.6 kg

P =

8.Normal distributionHard

Hole diameters D are normal with mean 25.04 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 = 25.04 mm, σ = 0.01 mmshaft: μ = 25 mm, σ = 0.015 mm

P =

9.Confidence intervalsMedium

A sample of 16 items has mean 69 and standard deviation s = 19.5. The population is roughly normal. Find the upper limit of the 95% confidence interval for the mean.

n = 16x̄ = 69s = 19.595%t(15) = 2.131

upper limit =

10.Confidence intervalsHard

The population standard deviation is 27. 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?

σ = 27E = 299%

n =

11.Hypothesis tests & confidence intervalsMedium

A sample of 25 students has a mean score of 107.6 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 120.

μ₀ = 120x̄ = 107.6s = 20n = 25

t =

12.Hypothesis tests & confidence intervalsHard

A one-sided z test of H₀: μ = 200 against H₁: μ < 200 is done at α = 0.01 with a sample of 36; the standard deviation is known to be 12. In reality the mean is 193. Find the power 1 − β of the test in percent.

μ₀ = 200μ = 193σ = 12n = 36α = 0.01

1 − β = %

13.Linear regression & correlationMedium

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

The table gives 5 paired observations of x = car age in years and y = price in k€. Use the least-squares line ŷ = a + b·x. Predict ŷ for x = 2.5.

n = 5x = 2.5

ŷ =

15.Chi-square testsMedium

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

row total = 46column total = 35N = 86

E =

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 · Paired (related) samples t-test

Why this is the answer

A paired t-test compares means from the same participants under two conditions, using the differences within each person.

Answer

B · Paired (related) samples t-test

2.Research methods in psychology

Answer: C · 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

C · Order effects such as practice, because conditions were not counterbalanced

3.Descriptive statistics

Answer: s = 16.19

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 = 168/6 = 28

Squared deviations

Σ(x − x̄)² = 2² + (−13)² + 2² + (−8)² + 30² + (−13)² = 1310

Standard deviation

s = √(1310/5) = 16.19

Answer

s = 16.19

4.Descriptive statistics

Answer: P = 0.3446

Standardise

z = 0.4

Standard normal table

Φ(0.4) = 0.6554

P = 1 − 0.6554 = 0.3446

Answer

P = 0.3446

5.Probability

Answer: P = 0.1091

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 = 4/11 · 3/10 = 12/110 = 0.1091

Answer

P = 0.1091

6.Probability

Answer: E(G) = 20 €

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: 10 (p = 0.25), 490 (p = 0.05), −10 (p = 0.7)

Expected value

E(G) = 10·0.25 + 490·0.05 − 10·0.7 = 20 €

Answer

E(G) = 20 €

A positive value means you win money on average.

7.Normal distribution

Answer: P = 0.2946

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) = μ_X ± μ_Y. Variances always add, also for a difference: Var(X ± Y) = σ_X² + σ_Y², so σ = √(σ_X² + σ_Y²). 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 total

μ_T = 125 kg, σ_T = 0.5·√5 = 1.118 kg

Standardise

z = (125.6 − 125)/1.118 = 0.54

Probability

P = 1 − Φ(0.54) = 0.2946

Answer

P = 0.2946

8.Normal distribution

Answer: P = 0.9868

Distribution of the difference

μ = 25.04 − 25 = 0.04 mm, σ = 0.01803 mm

Standardise

z = (0 − 0.04)/0.01803 = −2.22

Probability

P = 1 − Φ(−2.22) = 0.9868

Answer

P = 0.9868

9.Confidence intervals

Answer: upper limit = 79.39

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.

Critical value

df = n − 1 = 15, t = 2.131

Margin of error

E = 2.131·19.5/√16 = 10.39

Interval

69 ± 10.39 → [58.61, 79.39]

Answer

upper limit = 79.39

10.Confidence intervals

Answer: n = 1210

Solve for n

n = (2.576·27/2)² = 1209

Round up

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

n = 1210

Answer

n = 1210

11.Hypothesis tests & confidence intervals

Answer: t = −3.1

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 = 20/√25 = 20/5 = 4

Test statistic

t = (107.6 − 120)/4 = −3.1

Answer

t = −3.1

t_{0.025; df = 24} = 2.064

|t| > 2.064, so H₀ is rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: 1 − β = 87.98 %

Power of a test: power = 1 − β is the probability of rejecting H₀ when it is false. Find the cutoff x̄c from μ₀ and z_α, then compute the probability of landing beyond x̄c using the true mean: z = (x̄c − μ)/(σ/√n).

Critical sample mean

SE = 12/√36 = 2

x̄_c = 200 − 2.326·2 = 195.3

z under the true mean

z = (195.3 − 193)/2 = 1.174

Power

power = P(Z < 1.174) = 87.98%

1 − β = 87.98%

Answer

1 − β = 87.98%

13.Linear regression & correlation

Answer: r² = 0.9831

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 = 110/5 = 22

ȳ = 603/5 = 120.6

Sums of squares and products

S_xx = Σ(x − x̄)² = 90

S_xy = Σ(x − x̄)(y − ȳ) = 507

S_yy = Σ(y − ȳ)² = 2905

Coefficient of determination

r² = 507²/(90·2905) = 0.9831

Answer

r² = 0.9831

About 98.31% of the variation in y is explained by the line.

14.Linear regression & correlation

Answer: ŷ = 20.25

Means

x̄ = 15/5 = 3

ȳ = 95/5 = 19

Sums of squares and products

S_xx = Σ(x − x̄)² = 10

S_xy = Σ(x − x̄)(y − ȳ) = −25

Regression line

b = −25/10 = −2.5

a = 19 − (−2.5)·3 = 26.5

ŷ = 26.5 − 2.5·x

Prediction

ŷ = 26.5 + (−2.5)·2.5 = 20.25

Answer

ŷ = 20.25

15.Chi-square tests

Answer: E = 18.72

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 = 46·35/86 = 18.72

Answer

E = 18.72

16.Chi-square tests

Answer: df = 5

Degrees of freedom

df = (2 − 1)·(6 − 1) = 5

Answer

df = 5

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