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

Experts check whether the items of a maths exam cover all topics in the syllabus. This assesses:

  1. ATest-retest reliability
  2. BContent validity
  3. CPredictive validity
  4. DEcological validity
2.Research methods in psychologyHard

A researcher measures the effect of noise on concentration. Participants in the noise condition happened to be tested in the morning and the quiet condition in the afternoon. What is the problem?

  1. ATime of day is a confounding variable
  2. BThe data are at nominal level
  3. CTime of day is the dependent variable
  4. DThe study has too many conditions
3.Descriptive statisticsMedium
2025303540455055

Find the sample standard deviation s of the data: 59, 34, 16, 54, 35, 42.

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

Exam scores are normally distributed with mean 70 and standard deviation 5. Find the z-score z of a score of 72.

μ = 70σ = 5x = 72
z =
5.ProbabilityMedium
xp(x)00.3510.320.35μ = 1

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

x: 0, 1, 2p(x): 0.35, 0.3, 0.35
Var(X) =
6.ProbabilityHard
€p(x)−100.65400.34900.05

A game costs €10 to play. You win €50 with probability 0.3 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€50: p = 0.3€500: p = 0.05
E(G) = €
7.Normal distributionMedium
32 min35.8 minX + Y?

X is normal with mean 12 min and standard deviation 1.5 min; Y is independent and normal with mean 20 min and standard deviation 2 min. Find the probability P that X + Y is more than 35.8 min. Round z to two decimals.

μX = 12, σX = 1.5μY = 20, σY = 2X + Y > 35.8 min
P =
8.Normal distributionHard
μ = 250x = ?5%

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

μ = 250σ = 5P = 5%
x = g
9.Confidence intervalsMedium
x̄E₁ = 10.690%, n = 64x̄E₂ = ?99%, n = 64

A 90% confidence interval for a mean, based on 64 observations and a known σ, has a margin of error of 10.6. What is the margin of error E₂ of a 99% interval from the same data?

90%n₁ = 64E₁ = 10.699%
E₂ =
10.Confidence intervalsHard
μ₁ − μ₂, t(29) = 2.7560LUx̄₁ − x̄₂E = ?99%

Two fertilisers are tested. Plot group 1: 14 plots, mean yield 57.5 kg, standard deviation 1.5 kg. Plot group 2: 17 plots, mean 65 kg, standard deviation 3 kg. Assume equal variances (pooled t, df = n₁ + n₂ − 2). Find the margin of error E of the 99% confidence interval for μ₁ − μ₂.

n₁ = 14x̄₁ = 57.5s₁ = 1.5n₂ = 17x̄₂ = 65s₂ = 399%t(29) = 2.756
E = kg
11.Hypothesis tests & confidence intervalsMedium
t −1.99801.998

A sample of 64 students has a mean score of 194.5 and a sample standard deviation of 20. Find the test statistic t for H₀: μ = 200.

μ₀ = 200x̄ = 194.5s = 20n = 64
t =
12.Hypothesis tests & confidence intervalsHard
σ = 8, n = ?x̄ − Ex̄ + Ex̄E = 390%

How large must a sample be so that the 90% confidence interval for a mean has a margin of error of at most 3? The standard deviation is 8. Find the smallest sample size n.

σ = 8E = 390%
n =
13.Linear regression & correlationMedium
xy126221322416512xy24152025

The table gives 5 paired observations of x = car age in years and y = price in k€. Their sums are given. Find Pearson's correlation coefficient r from these sums.

n = 5Σx = 15Σy = 97Σx² = 55Σy² = 2001Σxy = 258
r =
14.Linear regression & correlationHard
xy294156218281037xy51010203040ŷ = 25

The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. For which value x does the line predict ŷ = 25?

n = 5ŷ = 25
x =
15.Chi-square testsMedium
no defectminormajorTotalday shift19293280night shift213168120Total4060100200

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 = 2002 × 3
χ² =
16.Chi-square testsHard
passfailTotalage 18–30204060age 31–50154560age 51+3169100Total66154220dfα = 0.05α = 0.0113.8416.63525.9919.21037.81511.34549.48813.277511.07015.086612.59216.812714.06718.475815.50720.090916.91921.6661018.30723.209

The table shows observed counts. Test at α = 0.01 whether the two variables are independent (no continuity correction). Compute the test statistic χ², then compare it with the critical value from the table.

N = 2203 × 2α = 0.01
χ² =

3Solutions

1.Research methods in psychology

Answer: B · Content validity

Why this is the answer

Content validity is the extent to which a measure covers all relevant aspects of the construct, usually judged by experts.

Answer

B · Content validity
2.Research methods in psychology

Answer: A · Time of day is a confounding variable

Why this is the answer

Because noise and testing time vary together, any difference could be due to either. Random allocation of testing times or holding time constant would remove the confound.

Answer

A · Time of day is a confounding variable
3.Descriptive statistics

Answer: s = 15.48

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 = 240/6 = 40

Squared deviations

Σ(x − x̄)² = 19² + (−6)² + (−24)² + 14² + (−5)² + 2² = 1198

Standard deviation

s = √(1198/5) = 15.48

Answer

s = 15.48
4.Descriptive statistics

Answer: z = 0.4

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 = (72 − 70)/5 = 0.4

Answer

z = 0.4
5.Probability

Answer: Var(X) = 0.7

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 + 1·0.3 + 2·0.35 = 1

Mean of the squares

E(X²) = 0·0.35 + 1·0.3 + 4·0.35 = 1.7

Variance

Var(X) = 1.7 − 1² = 0.7

Answer

Var(X) = 0.7
6.Probability

Answer: E(G) = 30 €

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.

Net outcomes

G: 40 (p = 0.3), 490 (p = 0.05), −10 (p = 0.65)

Expected value

E(G) = 40·0.3 + 490·0.05 − 10·0.65 = 30 €

Answer

E(G) = 30 €

A positive value means you win money on average.

7.Normal distribution

Answer: P = 0.06426

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 = 32 min, σ = √(1.5² + 2²) = 2.5 min

Standardise

z = (35.8 − 32)/2.5 = 1.52

Upper tail

P = 1 − Φ(1.52) = 1 − 0.9357 = 0.06426

Answer

P = 0.06426
8.Normal distribution

Answer: x = 241.8 g

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) = 0.05 → z = −1.645

Back to x

x = 250 + −1.645·5 = 241.78 g

Answer

x = 241.78 g
9.Confidence intervals

Answer: E₂ = 16.6

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 = (2.576/1.645)·√(64/64) = 1.566

New margin

E₂ = 10.6·1.566 = 16.6

Answer

E₂ = 16.6
10.Confidence intervals

Answer: E = 2.431 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 = (13·1.5² + 16·3²)/29 = 5.974
sp = 2.444

Margin of error

E = 2.756·2.444·√(1/14 + 1/17) = 2.431

Answer

E = 2.431 kg
11.Hypothesis tests & confidence intervals

Answer: t = −2.2

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/√64 = 20/8 = 2.5

Test statistic

t = (194.5 − 200)/2.5 = −2.2

Answer

t = −2.2
t0.025; df = 63 = 1.998

|t| > 1.998, so H₀ is rejected at the 5% level (two-sided).

12.Hypothesis tests & confidence intervals

Answer: n = 20

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).

Solve for n

n = (1.645·8/3)² = 19.24

Round up

n = 20

Answer

n = 20
13.Linear regression & correlation

Answer: r = −0.9558

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 = 258 − 15·97/5 = −33
Sxx = 55 − 15²/5 = 10
Syy = 2001 − 97²/5 = 119.2

Correlation coefficient

r = −33/√(10·119.2) = −0.9558

Answer

r = −0.9558
14.Linear regression & correlation

Answer: x = 6.87

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 = 30/5 = 6
ȳ = 110/5 = 22

Sums of squares and products

Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 138

Regression line

b = 138/40 = 3.45
a = 22 − 3.45·6 = 1.3
ŷ = 1.3 + 3.45·x

Solve for x

x = (25 − 1.3)/3.45 = 6.87

Answer

x = 6.87
15.Chi-square tests

Answer: χ² = 5.34

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

day shift: 80·40/200 = 16, 80·60/200 = 24, 80·100/200 = 40
night shift: 120·40/200 = 24, 120·60/200 = 36, 120·100/200 = 60

Chi-square statistic

χ² = (19 − 16)²/16 + (29 − 24)²/24 + (32 − 40)²/40 + (21 − 24)²/24 + (31 − 36)²/36 + (68 − 60)²/60
χ² = 5.34

Answer

χ² = 5.34

df = 2. At α = 0.05 the critical value is 5.991, so H₀ (independence) is not rejected.

16.Chi-square tests

Answer: χ² = 1.079

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

age 18–30: 18, 42
age 31–50: 18, 42
age 51+: 30, 70

Chi-square statistic

χ² = (20 − 18)²/18 + (40 − 42)²/42 + (15 − 18)²/18 + (45 − 42)²/42 + (31 − 30)²/30 + (69 − 70)²/70
χ² = 1.079

Decision

df = 2, χ²crit = 9.210
1.079 < 9.210

Answer

χ² = 1.079

χ² is below the critical value, so H₀ is not rejected: there is no significant evidence of a relationship.

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