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 statistic is appropriate for the relationship between two sets of ranks, such as judges' rankings of paintings?

ASpearman's rho
BPearson's r
CIndependent t-test
DChi-square

2.Research methods in psychologyHard

A researcher wants to test whether a new teaching method causes better grades, but cannot randomise students. Which design gives the strongest causal evidence in this situation?

AMeasuring only the class using the new method after teaching
BComparing matched classes measured before and after, with a comparison group
CCorrelating grades with how much students like the teacher
DAsking students whether they think the method worked

3.Descriptive statisticsMedium

Find the sample standard deviation s of the data: 56, 17, 33, 14, 26, 29.

n = 6

s =

4.Descriptive statisticsHard

A quantity is normally distributed with mean 70 and standard deviation 10. What proportion P of values is below 65?

μ = 70σ = 10x = 65

P =

5.ProbabilityMedium

In how many ways N can a committee of 4 people be chosen from a group of 14? The order does not matter.

n = 14k = 4

N =

6.ProbabilityHard

A test has 9 multiple-choice questions. A student who guesses gets each one right with probability 0.25, independently. What is the probability P of exactly 4 correct answers?

n = 9p = 0.25k = 4

P =

7.Normal distributionMedium

X is normal with mean 120 mm and standard deviation 0.3 mm; Y is independent and normal with mean 80 mm and standard deviation 0.4 mm. Find the probability P that X + Y is more than 201 mm. Round z to two decimals.

μ_X = 120, σ_X = 0.3μ_Y = 80, σ_Y = 0.4X + Y > 201 mm

P =

8.Normal distributionHard

Hole diameters D are normal with mean 25.04 mm and standard deviation 0.02 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.02 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 = 0.02 mmshaft: μ = 25 mm, σ = 0.02 mm

P =

9.Confidence intervalsMedium

A sample of 13 items has mean 46 and standard deviation s = 8.5. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.

n = 13x̄ = 46s = 8.595%t(12) = 2.179

lower limit =

10.Confidence intervalsHard

The population standard deviation is 20. How large a sample n do you need so that a 95% confidence interval for the mean has a margin of error of at most 2?

σ = 20E = 295%

n =

11.Hypothesis tests & confidence intervalsMedium

A sample of 25 students has a mean score of 205 and a sample standard deviation of 12. Find the test statistic t for H₀: μ = 200.

μ₀ = 200x̄ = 205s = 12n = 25

t =

12.Hypothesis tests & confidence intervalsHard

How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 2? The standard deviation is 5. Find the smallest sample size n.

σ = 5E = 290%

n =

13.Linear regression & correlationMedium

The table gives 5 paired observations of x = load in kN and y = extension in mm. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 30Σy = 89Σx² = 220Σy² = 1801Σxy = 626

r =

14.Linear regression & correlationHard

The table gives 6 paired observations of x = temperature in °C and y = ice creams sold. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 131?

n = 6ŷ = 131

x =

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 “students, car”.

row total = 81column total = 68N = 164

E =

16.Chi-square testsHard

A chi-square test of independence is carried out on a contingency table with 2 rows and 11 columns (not counting the totals). How many degrees of freedom df does the test have?

2 × 11

df =

3Solutions

1.Research methods in psychology

Answer: A · Spearman's rho

Why this is the answer

Spearman's rank correlation measures the strength of a monotonic relationship between two ranked or ordinal variables.

Answer

A · Spearman's rho

2.Research methods in psychology

Answer: B · Comparing matched classes measured before and after, with a comparison group

Why this is the answer

A non-equivalent groups pretest-posttest design controls for initial differences and for change over time, though it is weaker than a randomised experiment.

Answer

B · Comparing matched classes measured before and after, with a comparison group

3.Descriptive statistics

Answer: s = 14.99

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 = 175/6 = 29.17

Squared deviations

Σ(x − x̄)² = 26.83² + (−12.17)² + 3.833² + (−15.17)² + (−3.167)² + (−0.1667)² = 1123

Standard deviation

s = √(1123/5) = 14.99

Answer

s = 14.99

4.Descriptive statistics

Answer: P = 0.3085

Standardise

z = −0.5

Standard normal table

Φ(−0.5) = 0.3085

P = 0.3085

Answer

P = 0.3085

5.Probability

Answer: N = 1001

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(14, 4) = (14·13·12·11)/(4·3·2·1) = 1001

Answer

N = 1001

6.Probability

Answer: P = 0.1168

Binomial formula

P(X = 4) = 126·0.25⁴·0.75⁵ = 126·0.003906·0.2373 = 0.1168

Answer

P = 0.1168

7.Normal distribution

Answer: P = 0.02275

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 sum

μ = 200 mm, σ = √(0.3² + 0.4²) = 0.5 mm

Standardise

z = (201 − 200)/0.5 = 2

Upper tail

P = 1 − Φ(2) = 1 − 0.9772 = 0.02275

Answer

P = 0.02275

8.Normal distribution

Answer: P = 0.9207

Distribution of the difference

μ = 25.04 − 25 = 0.04 mm, σ = 0.02828 mm

Standardise

z = (0 − 0.04)/0.02828 = −1.41

Probability

P = 1 − Φ(−1.41) = 0.9207

Answer

P = 0.9207

9.Confidence intervals

Answer: lower limit = 40.86

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 = 12, t = 2.179

Margin of error

E = 2.179·8.5/√13 = 5.137

Interval

46 ± 5.137 → [40.86, 51.14]

Answer

lower limit = 40.86

10.Confidence intervals

Answer: n = 385

Solve for n

n = (1.96·20/2)² = 384.2

Round up

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

n = 385

Answer

n = 385

11.Hypothesis tests & confidence intervals

Answer: t = 2.083

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 = 12/√25 = 12/5 = 2.4

Test statistic

t = (205 − 200)/2.4 = 2.083

Answer

t = 2.083

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: n = 17

Solve for n

n = (1.645·5/2)² = 16.91

Round up

n = 17

Answer

n = 17

13.Linear regression & correlation

Answer: r = 0.9879

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

S_xy = 626 − 30·89/5 = 92

S_xx = 220 − 30²/5 = 40

S_yy = 1801 − 89²/5 = 216.8

Correlation coefficient

r = 92/√(40·216.8) = 0.9879

Answer

r = 0.9879

14.Linear regression & correlation

Answer: x = 25.57

Means

x̄ = 135/6 = 22.5

ȳ = 681/6 = 113.5

Sums of squares and products

S_xx = Σ(x − x̄)² = 157.5

S_xy = Σ(x − x̄)(y − ȳ) = 898.5

Regression line

b = 898.5/157.5 = 5.705

a = 113.5 − 5.705·22.5 = −14.86

ŷ = −14.86 + 5.705·x

Solve for x

x = (131 − (−14.86))/5.705 = 25.57

Answer

x = 25.57

15.Chi-square tests

Answer: E = 33.59

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 = 81·68/164 = 33.59

Answer

E = 33.59

16.Chi-square tests

Answer: df = 10

Degrees of freedom

df = (2 − 1)·(11 − 1) = 10

Answer

df = 10

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